CHAPTER 2
Derivatives
Review of Prerequisite Skills,
pp. 62–63
1. a. a5 3 a3 5 a513
5 a8
b. A22a2 B 3 5 (22)3 Aa2 B 3
5 28Aa233 B
5 28a6
7
4p 3 6p9
24p719
c.
5
12p15
12p15
5 2p16215
5 2p
4 25
26 22
d. Aa b B Aa b B 5 Aa426 B Ab2522 B
5 a22b27
1
5 2 7
ab
e. A3e6 B A2e3 B 4 5 (3)Ae6 B A24 B Ae3 B 4
5 (3)A24 BAe6 B Ae334 B
5 (3)(16)Ae6112 B
5 48e18
24
3
A3a B C2a (2b)3 D
(3)(2)A21B 3 Aa2413 B Ab3 B
f.
5
12a5b2
12a5b2
2125
B Ab322 B
26Aa
5
12
26
21Aa B AbB
5
2
b
52 6
2a
1
2
1
2
2. a. Ax2 B Ax3 B 5 x2 1 3
7
5 x6
2
2
2
b. A8x6 B 3 5 83x63 3
5 4x4
1
1
3
( a2)( a3)
"a "
a
c.
5
1
a2
"a
1
5 a3
3. A perpendicular line will have a slope that is the
negative reciprocal of the slope of the given line:
21
a. slope 5 2
3
52
3
2
Calculus and Vectors Solutions Manual
21
2 12
52
21
c. slope 5 5
b. slope 5
3
3
5
21
d. slope 5
21
51
52
24 2 (22)
23 2 9
22
5
212
1
5
6
The equation of the desired line is therefore
y 1 4 5 16 (x 1 3) or x 2 6y 2 21 5 0.
b. The equation 3x 2 2y 5 5 can be rewritten as
2y 5 3x 2 5 or y 5 32x 2 52, which has slope 32.
The equation of the desired line is therefore
y 1 5 5 32 (x 1 2) or 3x 2 2y 2 4 5 0.
c. The line perpendicular to y 5 34 x 2 6 will have
21
slope m 5 3 5 2 43. The equation of the desired line
4. a. This line has slope m 5
4
is therefore y 1 3 5 2 43 (x 2 4) or 4x 1 3y 2 7 5 0.
5. a. (x 2 3y)(2x 1 y) 5 2x2 1 xy 2 6xy 2 3y2
5 2x2 2 5xy 2 3y2
2
b. (x 2 2)(x 2 3x 1 4)
5 x3 2 3x2 1 4x 2 2x2 1 6x 2 8
5 x3 2 5x2 1 10x 2 8
c. (6x 2 3)(2x 1 7) 5 12x2 1 42x 2 6x 2 21
5 12x2 1 36x 2 21
d. 2(x 1 y) 2 5(3x 2 8y) 5 2x 1 2y 2 15x 1 40y
5 213x 1 42y
2
2
e. (2x 2 3y) 1 (5x 1 y)
5 4x2 2 12xy 1 9y2 1 25x2 1 10xy 1 y2
5 29x2 2 2xy 1 10y2
f. 3x(2x 2 y)2 2 x(5x 2 y)(5x 1 y)
5 3x(4x2 2 4xy 1 y2 ) 2 x(25x2 2 y2 )
5 12x3 2 12x2y 1 3xy2 2 25x3 1 xy2
5 213x3 2 12x2y 1 4xy2
2-1
3x(x 1 2)
5x3
15x4 (x 1 2)
3
5
x2
2x(x 1 2)
2x3 (x 1 2)
15
5 x423
2
15
5 x
2
x 2 0, 22
y
(y 2 5)2
3
b.
(y 1 2)(y 2 5)
4y3
y(y 2 5)(y 2 5)
5 3
4y (y 1 2)(y 2 5)
y25
5 2
4y (y 1 2)
y 2 22, 0, 5
4
9
4
2(h 1 k)
c.
4
5
3
h 1 k 2(h 1 k)
h1k
9
8(h 1 k)
5
9(h 1 k)
8
5
9
h 2 2k
(x 1 y)(x 2 y) (x 1 y)3
d.
4
5(x 2 y)
10
(x 1 y)(x 2 y)
10
5
3
5(x 2 y)
(x 1 y)3
10(x 1 y)(x 2 y)
5
5(x 2 y)(x 1 y)3
2
5
(x 1 y)2
x 2 2y, 1y
x27
5x
(x 2 7)(x 2 1)
(5x)(2x)
1
5
1
e.
2x
x21
2x(x 2 1)
2x(x 2 1)
x2 2 7x 2 x 1 7 1 10x2
5
2x(x 2 1)
11x2 2 8x 1 7
5
2x(x 2 1)
x 2 0, 1
x12
x11
2
f.
x22
x13
(x 1 2)(x 2 2)
(x 1 1)(x 1 3)
2
5
(x 2 2)(x 1 3)
(x 1 3)(x 2 2)
x2 1 x 1 3x 1 3 2 x2 1 4
5
(x 1 3)(x 2 2)
4x 1 7
5
(x 1 3)(x 2 2)
x 2 23, 2
7. a. 4k2 2 9 5 (2k 1 3)(2k 2 3)
6. a.
2-2
b. x2 1 4x 2 32 5 x2 1 8x 2 4x 2 32
5 x(x 1 8) 2 4(x 1 8)
5 (x 2 4)(x 1 8)
c. 3a2 2 4a 2 7 5 3a2 2 7a 1 3a 2 7
5 a(3a 2 7) 1 1(3a 2 7)
5 (a 1 1)(3a 2 7)
d. x4 2 1 5 (x2 1 1)(x2 2 1)
5 (x2 1 1)(x 1 1)(x 2 1)
3
3
e. x 2 y 5 (x 2 y)(x2 1 xy 1 y2 )
f. r4 2 5r 2 1 4 5 r4 2 4r 2 2 r 2 1 4
5 r 2 (r 2 2 4) 2 1(r 2 2 4)
5 (r2 2 1)(r2 2 4)
5 (r 1 1)(r 2 1)(r 1 2)(r 2 2)
8. a. Letting f(a) 5 a3 2 b3, f(b) 5 b3 2 b3
50
So b is a root of f (a), and so by the factor theorem,
a 2 b is a factor of a3 2 b3. Polynomial long
division provides the other factor:
a2 1 ab 1 b2
a 2 bq a3 1 0a2 1 0a 2 b3
a3 2 a2b
a2b 1 0a 2 b3
a2b 2 ab2
ab2 2 b3
ab2 2 b3
0
So a3 2 b3 5 (a 2 b)(a2 1 ab 1 b2 ).
b. Using long division or recognizing a pattern from
the work in part a.:
a5 2 b5 5 (a 2 b)(a4 1 a3b 1 a2b2 1 ab3 1 b4 ).
c. Using long division or recognizing a pattern from
the work in part a.: a7 2 b7
5 (a 2 b)(a6 1 a5b 1 a4b2 1 a3b3
1 a2b4 1 ab5 1 b6 ).
d. Using the pattern from the previous parts:
an 2 bn 5 (a 2 b)(an21 1 an22b 1 an23b2 1 c
1 a2bn23 1 abn22 1 bn21 ).
9. a. f(2) 5 22(24 ) 1 3(22 ) 1 7 2 2(2)
5 232 1 12 1 7 2 4
5 217
b. f(21) 5 22(21)4 1 3(21)2 1 7 2 2(21)
5 22 1 3 1 7 1 2
5 10
1 4
1 2
1
1
c. f a b 5 22a b 1 3a b 1 7 2 2a b
2
2
2
2
3
1
52 1 1721
8
4
53
5
8
Chapter 2: Derivatives
1
d. f(20.25) 5 f a2 b
4
1 4
1 2
1
5 2a2 b 1 3a2 b 1 7 2 2a2 b
4
4
4
1
3
1
52
1
171
128
16
2
983
5
128
8 7.68
3
3 !2
5
10. a.
!2
( !2)( !2)
3 !2
5
2
4 2 !2
(4 2 !2)( !3)
5
b.
!3
( !3)(!3)
4!3 2 !6
5
3
2 1 3 !2
(2 1 3 !2)(3 1 4 !2)
5
c.
3 2 4 !2
(3 2 4 !2)(3 1 4 !2)
6 1 9!2 1 8!2 1 12(2)
5
32 2 A4 !2B 2
30 1 17 !2
5
9 2 16(2)
30 1 17!2
52
23
3!2 2 4!3
(3!2 2 4 !3)(3 !2 2 4 !3)
5
d.
3!2 1 4!3
(3 !2 1 4 !3)(3"2 2 4 !3)
(3!2)2 2 24!6 1 (4 !3)2
5
(3!2)2 2 (4!3)2
9(2) 2 24!6 1 16(3)
5
9(2) 2 16(3)
66 2 24 !6
52
30
11 2 4 !6
52
5
11. a. f(x) 5 3x2 2 2x
When a 5 2,
f(a 1 h) 2 f(a)
f(2 1 h) 2 f(2)
5
h
h
2
3(2 1 h) 2 2(2 1 h) 2 C3(2)2 2 2(2)D
5
h
3(4 1 4h 1 h2 ) 2 4 2 2h 2 8
5
h
12 1 12h 1 3h2 2 2h 2 12
5
h
Calculus and Vectors Solutions Manual
3h2 1 10h
h
5 3h 1 10
This expression can be used to determine the slope of
the secant line between (2, 8) and (2 1 h, f(2 1 h)).
b. For h 5 0.01: 3(0.01) 1 10 5 10.03
c. The value in part b. represents the slope of the
secant line through (2, 8) and (2.01, 8.1003).
5
2.1 The Derivative Function, pp. 73–75
1. A function is not differentiable at a point where its
graph has a cusp, a discontinuity, or a vertical tangent:
a. The graph has a cusp at x 5 22, so f is
differentiable on 5xPR Z x 2 226.
b. The graph is discontinuous at x 5 2, so f is
differentiable on 5xPR Z x 2 26.
c. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
d. The graph has a cusp at x 5 1, so f is
differentiable on 5xPR Z x 2 16.
e. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
f. The function does not exist for x , 2, but has
no cusps, discontinuities, or vertical tangents
elsewhere, so f is differentiable on 5xPR Z x . 26.
2. The derivative of a function represents the slope of
the tangent line at a given value of the independent
variable or the instantaneous rate of change of the
function at a given value of the independent variable.
3.
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
f(x) 5 5x 2 2
f(a 1 h) 5 5(a 1 h) 2 2
5 5a 1 5h 2 2
f(a 1 h) 2 f(a) 5 5a 1 5h 2 2 2 (5a 2 2)
5 5h
b.
f(x) 5 x2 1 3x 2 1
f(a 1 h) 5 (a 1 h)2 1 3(a 1 h) 2 1
5 a2 1 2ah 1 h2 1 3a
1 3h 2 1
f(a 1 h) 2 f(a) 5 a2 1 2ah 1 h2 1 3a 1 3h
2 1 2 (a2 1 3a 2 1)
5 2ah 1 h2 1 3h
4. a.
2-3
f(x) 5 x3 2 4x 1 1
f(a 1 h) 5 (a 1 h)3 2 4(a 1 h) 1 1
5 a3 1 3a2h 1 3ah2 1 h3
2 4a 2 4h 1 1
f(a 1 h) 2 f(a) 5 a3 1 3a2h 1 3ah2 1 h3 2 4a
2 4h 1 1 2 (a3 2 4a 1 1)
5 3a2h 1 3ah2 1 h3 2 4h
f(x) 5 x2 1 x 2 6
d.
f(a 1 h) 5 (a 1 h)2 1 (a 1 h) 2 6
5 a2 1 2ah 1 h2 1 a 1 h 2 6
f(a 1 h) 2 f(a) 5 a2 1 2ah 1 h2 1 a 1 h 2 6
2 (a2 1 a 2 6)
5 2ah 1 h2 1 h
e.
f(x) 5 27x 1 4
f(a 1 h) 5 27(a 1 h) 1 4
5 27a 2 7h 1 4
f(a 1 h) 2 f(a) 5 27a 2 7h 1 4 2 (27a 1 4)
5 27h
f(x) 5 4 2 2x 2 x2
f.
f(a 1 h) 5 4 2 2(a 1 h) 2 (a 1 h)2
5 4 2 2a 2 2h 2 a2 2 2ah 2 h2
f(a 1 h) 2 f(a) 5 4 2 2a 2 2h 2 a2 2 2ah
2 h2 2 4 1 2a 1 a2
5 22h 2 h2 2 2ah
f(1 1 h) 2 f(1)
5. a. f r (1) 5 lim
hS0
h
(1 1 h)2 2 12
5 lim
hS0
h
1 1 2h 1 h2 2 1
5 lim
hS0
h
2h 1 h2
5 lim
hS0
h
5 lim (2 1 h)
c.
hS0
52
f(3 1 h) 2 f(3)
b. f r(3) 5 lim
hS0
h
(3 1 h)2 1 3(3 1 h) 1 1
5 lim c
hS0
h
2
(3 1 3(3) 1 1)
d
2
h
9 1 6h 1 h2 1 9 1 3h 1 1 2 19
5 lim
hS0
h
9h 1 h2
5 lim
hS0
h
5 lim (9 1 h)
hS0
c.
f(0 1 h) 2 f(0)
hS0
h
!h 1 1 2 !0 1 1
5 lim
hS0
h
!h 1 1 2 1
5 lim
hS0
h
( ! h 1 1 2 1)( ! h 1 1 1 1)
5 lim
hS0
h( ! h 1 1 1 1)
f r (0) 5 lim
( "h 1 1) 2 2 1
hS0 h( !h 1 1 1 1)
5 lim
h1121
hS0 h( !h 1 1 1 1)
1
5 lim
hS0 ( !h 1 1 1 1)
1
5 lim
hS0 ( !1 1 1)
1
5
2
f(21 1 h) 2 f(21)
d. f r(21) 5 lim
hS0
h
5 lim
5 lim
hS0
5 lim
hS0
5
5
2 21
21 1 h
h
5
21 1 h 1 5
h
5
5(21 1 h)
1 21 1 h
21
1
h
5 lim
h
5 2 5 1 5h
5 lim
hS0 h(21 1 h)
5h
5 lim
hS0 h(21 1 h)
5
5 lim
hS0 (21 1 h)
5
5
21
5 25
f(x 1 h) 2 f(x)
6. a. f r (x) 5 lim
hS0
h
25(x 1 h) 2 8 2 (25x 2 8)
5 lim
hS0
h
25x 2 5h 2 8 1 5x 1 8
5 lim
hS0
h
hS0
59
2-4
Chapter 2: Derivatives
25h
hS0 h
5 lim 25
5 lim
hS0
5 25
f(x 1 h) 2 f(x)
b. f r(x) 5 lim
hS0
h
2(x 1 h)2 1 4(x 1 h)
5 lim c
hS0
h
A2x2 1 4xB
d
h
2x2 1 4xh 1 2h2 1 4x
5 lim c
hS0
h
4h 2 2x2 2 4x
1
d
h
4xh 1 2h2 1 4h
5 lim
hS0
h
5 lim (4x 1 2h 1 4)
2
hS0
5 4x 1 4
f(x 1 h) 2 f(x)
c. f r(x) 5 lim
hS0
h
6(x 1 h)3 2 7(x 1 h)
5 lim c
hS0
h
(6x3 2 7x)
2
d
h
6x3 1 18x2h 1 18xh2 1 6h3
5 lim c
hS0
h
27x 2 7h 2 6x3 1 7x
1
d
h
18x2h 1 18xh2 1 6h3 2 7h
5 lim
hS0
h
2
5 lim (18x 1 18xh 1 6h2 2 7)
hS0
5 18x2 2 7
f(x 1 h) 2 f(x)
d. f r(x) 5 lim
hS0
h
5 lim
hS0
5 lim
hS0
A !3x 1 3h 1 2 B 2 2 A !3x 1 2 B 2
hA !3x 1 3h 1 2 1 !3x 1 2 B
3x 1 3h 1 2 2 3x 2 2
hS0 hA !3x 1 3h 1 2 1 !3x 1 2B
3
5 lim
hS0 !3x 1 3h 1 2 1 !3x 1 2
3
5
2 !3x 1 2
7. a. Let y 5 f(x), then
f(x 1 h) 2 f(x)
dy
5 f r(x) 5 lim
dx
hS0
h
6 2 7(x 1 h) 2 (6 2 7x)
5 lim
hS0
h
6 2 7x 2 7h 2 6 1 7x
5 lim
hS0
h
27h
5 lim
hS0
h
5 lim 27
5 lim
hS0
5 27
b. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
5 lim
!3x 1 3h 1 2 2 !3x 1 2
hS0
h
A !3x 1 3h 1 2 2 !3x 1 2 B
5 lim c
hS0
h
3
A !3x 1 3h 1 2 1 !3x 1 2B
d
A !3x 1 3h 1 2 1 !3x 1 2B
Calculus and Vectors Solutions Manual
h
hS0
5 lim
hS0
(x 1 h 1 1)(x 2 1)
£ (x 1 h 2 1)(x 2 1)
h
(x 1 1)(x 1 h 2 1)
(x 2 1)(x 1 h 2 1) §
2
h
x2
5 lim £
2
5 lim
1 hx 1 x 2 x 2 h 2 1
(x 1 h 2 1)(x 2 1)
h
hS0
!3(x 1 h) 1 2 2 !3x 1 2
h
5 lim
x1h11
x11
2x21
x1h21
x2 1 hx 2 x 1 x 1 h 2 1
(x 1 h 2 1)(x 2 1)
§
h
22h
(x 1 h 2 1)(x 2 1)
h
22
5 lim
hS0 (x 1 h 2 1)(x 2 1)
2
52
(x 2 1)2
hS0
2-5
c. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
3(x 1 h)2 2 3x2
5 lim
hS0
h
3x2 1 6xh 1 3h2 2 3x2
5 lim
hS0
h
6xh 1 3h2
5 lim
hS0
h
5 lim 6x 1 3h
9. a.
5 6x
8. Let y 5 f(x), then the slope of the tangent at
each point x can be found by calculating f r(x)
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
2(x 1 h)2 2 4(x 1 h) 2 2x2 1 4x
5 lim
hS0
h
2
2x 1 4xh 1 2h2 2 4x 2 4h
5 lim c
hS0
h
2
22x 1 4x
1
d
h
4xh 1 h2 2 4h
5 lim
hS0
h
5 lim 4x 1 h 2 4
b. Let y 5 f(x), then the slope of the tangent at
each point x can be found by calculating f(x)
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
(x 1 h)3 2 x3
5 lim
hS0
h
x3 1 3x2h 1 3xh2 1 h3 2 x3
5 lim
hS0
h
3x2h 1 3xh2 1 h3
5 lim
hS0
h
5 lim 3x2 1 3xh 1 h2
hS0
hS0
5 4x 2 4
So the slope of the tangent at x 5 0 is
f r(0) 5 4(0) 2 4
5 24
At x 5 1, the slope of the tangent is
f r(1) 5 4(1) 2 4
50
At x 5 2, the slope of the tangent is
f r(2) 5 4(2) 2 4
54
y
4
3
2
1
x
0
–2 –1
1 2 3 4
–1
–2
2-6
12
10
8
6
4
2
–6 –4 –2 0
–2
–4
y
x
2 4 6
hS0
5 3x2
So the slope of the tangent at x 5 22 is
f r(22) 5 3(22)2
5 12
At x 5 21, the slope of the tangent is
f r(21) 5 3(21)2
53
At x 5 0, the slope of the tangent is
f r(0) 5 3(0)2
50
At x 5 1, the slope of the tangent is
f r(1) 5 3(1)2
53
At x 5 2, the slope of the tangent is
f r(2) 5 3(2)2
5 12
c.
y
12
10
8
6
4
2
x
0
–6 –4 –2
2 4 6
–2
Chapter 2: Derivatives
d. The graph of f(x) is a cubic. The graph of f r(x)
seems to be a parabola.
10. The velocity the particle at time t is given by sr(t)
s(t 1 h) 2 s(t)
sr(t) 5 lim
hS0
h
2 (t 1 h)2 1 8(t 1 h) 2 (2t2 1 8t)
5 lim
hS0
h
2t2 2 2th 2 h2 1 8t 1 8h 1 t2 2 8t
5 lim
hS0
h
22th 2 h2 1 8h
5 lim
hS0
h
5 lim 2 2t 2 h 1 8
hS0
5 22t 1 8
So the velocity at t 5 0 is
sr(0) 5 22(0) 1 8
5 8 m>s
At t 5 4, the velocity is
sr(4) 5 22(4) 1 8
5 0 m>s
At t 5 6, the velocity is
sr(6) 5 22(6) 1 8
5 24 m>s
f(x 1 h) 2 f(x)
11. f r(x) 5 lim
hS0
h
!x 1 h 1 1 2 !x 1 1
hS0
h
5 lim
5 lim c
hS0
A !x 1 h 1 1 2 !x 1 1 B
h
3
A !x 1 h 1 1 1 !x 1 1 B
d
A !x 1 h 1 1 1 !x 1 1 B
A !x 1 h 1 1 B 2 2 A !x 1 1 B 2
hS0 hA !x 1 h 1 1 1 !x 1 1 B
5 lim
x1h112x21
hS0 hA !x 1 h 1 1 1 !x 1 1B
5 lim
h
hS0 hA !x 1 h 1 1 1 !x 1 1B
5 lim
5 lim
hS0
5
1
A !x 1 h 1 1 1 !x 1 1 B
1
2 !x 1 1
The equation x 2 6y 1 4 5 0 can be rewritten as
y 5 16 x 1 23, so this line has slope 16. The value of x
where the tangent to f(x) has slope 16 will satisfy
f r(x) 5 16.
Calculus and Vectors Solutions Manual
1
1
5
2 !x 1 1
6
6 5 2 !x 1 1
32 5 A !x 1 1B 2
95x11
85x
f(8) 5 !8 1 1
5 !9
53
So the tangent passes through the point (8, 3), and its
equation is y 2 3 5 16 (x 2 8) or x 2 6y 1 10 5 0.
12. a. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
c2c
5 lim
hS0
h
0
5 lim
hS0 h
50
b. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
(x 1 h) 2 x
5 lim
hS0
h
h
5 lim
hS0 h
5 lim 1
hS0
51
c. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
m(x 1 h) 1 b 2 mx 2 b
5 lim
hS0
h
mx 1 mh 1 b 2 mx 2 b
5 lim
hS0
h
mh
5 lim
hS0 h
5 lim m
hS0
5m
d. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
a(x 1 h)2 1 b(x 1 h) 1 c
5 lim c
hS0
h
(ax2 1 bx 1 c)
2
d
h
2-7
5 lim c
ax2 1 2axh 1 ah2 1 bx 1 bh
hS0
h
2ax2 2 bx 2 c
1
d
h
2axh 1 ah2 1 bh
5 lim
hS0
h
5 lim (2ax 1 ah 1 b)
hS0
5 2ax 1 b
13. The slope of the function at a point x is given by
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
(x 1 h)3 2 x3
5 lim
hS0
h
3
x 1 3x2h 1 3xh2 1 h3 2 x3
5 lim
hS0
h
2
2
3x h 1 3xh 1 h3
5 lim
hS0
h
2
5 lim 3x 1 3xh 1 h2
hS0
2
5 3x
Since 3x2 is nonnegative for all x, the original
function never has a negative slope.
14. h(t) 5 18t 2 4.9t2
h(t 1 k) 2 h(t)
a. hr(t) 5 lim
kS0
k
18(t 1 k) 2 4.9(t 1 k)2
5 lim
kS0
k
A18t 2 4.9t2 B
2
k
18t 1 18k 2 4.9t 2 2 9.8tk 2 4.9k2
5 lim
kS0
k
18t 1 4.9t2
2
k
18k 2 9.8tk 2 4.9k2
5 lim
kS0
k
5 lim (18 2 9.8t 2 4.9k)
kS0
5 18 2 9.8t 2 4.9(0)
5 18 2 9.8t
Then hr(2) 5 18 2 9.8(2) 5 21.6 m>s.
b. hr(2) measures the rate of change in the height
of the ball with respect to time when t 5 2.
15. a. This graph has positive slope for x , 0, zero
slope at x 5 0, and negative slope for x . 0, which
corresponds to graph e.
b. This graph has positive slope for x , 0, zero
slope at x 5 0, and positive slope for x . 0, which
corresponds to graph f.
2-8
c. This graph has negative slope for x , 22,
positive slope for 22 , x , 0, negative slope for
0 , x , 2, positive slope for x . 2, and zero slope
at x 5 22, x 5 0, and x 5 2, which corresponds to
graph d.
16. This function is defined piecewise as f(x) 5 2x2
for x , 0, and f(x) 5 x2 for x $ 0. The derivative
will exist if the left-side and right-side derivatives are
the same at x 5 0:
f(0 1 h) 2 f(0)
2 (0 1 h)2 2 A20 2 B
lim2
5 lim2
hS0
h
hS0
h
2h2
5 lim2
hS0
h
5 lim2 (2h)
hS0
50
lim1
hS0
f(0 1 h) 2 f(0)
(0 1 h)2 2 A02 B
5 lim1
h
hS0
h
h2
5 lim1
hS0 h
5 lim1 (h)
hS0
50
Since the limits are equal for both sides, the derivative
exists and f r(0) 5 0.
17. Since f r(a) 5 6 and f(a) 5 0,
f(a 1 h) 2 f(a)
6 5 lim
hS0
h
f(a 1 h) 2 0
6 5 lim
hS0
h
f(a 1 h)
3 5 lim
hS0
2h
18.
y
6
4
2
x
–1
–2
1 2 3 4 5
f(x) is continuous.
f(3) 5 2
But f r(3) 5 `.
(Vertical tangent)
19. y 5 x2 2 4x 2 5 has a tangent parallel to
2x 2 y 5 1.
Let f(x) 5 x2 2 4x 2 5. First, calculate
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
Chapter 2: Derivatives
5 lim c
(x 1 h)2 2 4(x 1 h) 2 5
hS0
h
2
Ax 2 4x 2 5B
2
d
h
x2 1 2xh 1 h2 2 4x 2 4h 2 5
5 lim c
hS0
h
2x2 1 4x 1 5
1
d
h
2xh 1 h2 2 4h
5 lim
hS0
h
5 lim (2x 1 h 2 4)
hS0
5 2x 1 0 2 4
5 2x 2 4
Thus, 2x 2 4 is the slope of the tangent to the curve
at x. We want the tangent parallel to 2x 2 y 5 1.
Rearranging, y 5 2x 2 1.
If the tangent is parallel to this line,
2x 2 4 5 2
x53
When x 5 3, y 5 (3)2 2 4(3) 2 5 5 28.
The point is (3, 28).
20. f(x) 5 x2
The slope of the tangent at any point Ax, x2 B is
f(x 1 h) 2 f(x)
f r 5 lim
hS0
h
(x 1 h)2 2 x2
5 lim
hS0
h
(x 1 h 1 x)(x 1 h 2 x)
5 lim
hS0
h
h(2x 1 h)
5 lim
hS0
h
5 lim (2x 1 h)
hS0
5 2x 1 0
5 2x
Let (a, a2 ) be a point of tangency. The equation of
the tangent is
y 2 a2 5 (2a)(x 2 a)
y 5 (2a)x 2 a2
Suppose the tangent passes through (1, 23).
Substitute x 5 1 and y 5 23 into the equation of
the tangent:
23 5 (2a)(1) 2 a2
2
a 2 2a 2 3 5 0
(a 2 3)(a 1 1) 5 0
a 5 21, 3
So the two tangents are y 5 22x 2 1 or
2x 1 y 1 1 5 0 and y 5 6x 2 9 or 6x 2 y 2 9 5 0.
Calculus and Vectors Solutions Manual
2.2 The Derivatives of Polynomial
Functions, pp. 82–84
1. Answers may vary. For example:
d
constant function rule:
(5) 5 0
dx
d 3
Ax B 5 3x2
power rule:
dx
d
A4x3 B 5 12x2
constant multiple rule:
dx
d 2
Ax 1 xB 5 2x 1 1
sum rule:
dx
d 3
Ax 2 x2 1 3xB 5 3x2 2 2x 1 3
difference rule:
dx
d
d
2. a. f r(x) 5 (4x) 2 (7)
dx
dx
d
d
5 4 (x) 2
(7)
dx
dx
5 4Ax0 B 2 0
54
d
d
b. f r(x) 5 Ax3 B 2 Ax2 B
dx
dx
5 3x2 2 2x
d
d
d
c. f r(x) 5 A2x2 B 1 (5x) 1 (8)
dx
dx
dx
d
d
d
5 2 Ax2 B 1 5 (x) 1
(8)
dx
dx
dx
5 2 (2x) 1 5 1 0
5 22x 1 5
d 3
x)
d. f r(x) 5 ("
dx
d 1
5 ( x3 )
dx
1 1
5 ( x(3 21))
3
1 2
5 (x23)
3
1
5 3 2
3"x
d x 4
e. f r(x) 5 aa b b
dx 2
1 4d 4
5a b
Ax B
2 dx
1
5 A4x3 B
16
x3
5
4
2-9
d 23
(x )
dx
5 (23)( x2321)
5 23x24
d
3. a. hr(x) 5 ((2x 1 3)(x 1 4))
dx
d
5
A2x2 1 8x 1 3x 1 12B
dx
d
d
d
5
A2x2 B 1
(11x) 1
(12)
dx
dx
dx
d
d
d
5 2 Ax2 B 1 11 (x) 1
(12)
dx
dx
dx
5 2(2x) 1 11(1) 1 0
5 4x 1 11
d
b. f r(x) 5 A2x3 1 5x2 2 4x 2 3.75B
dx
d
d
d
5
(2x3 ) 1
A5x2 B 2
(4x)
dx
dx
dx
d
2
(3.75)
dx
d
d
d
5 2 Ax3 B 1 5 Ax2 B 2 4 (x)
dx
dx
dx
d
2
(3.75)
dx
5 2A3x2 B 1 5(2x) 2 4(1) 2 0
5 6x2 1 10x 2 4
ds
d
c.
5 At 2 At 2 2 2tB B
dt
dt
d
5 At 4 2 2t 3 B
dt
d
d
5 At 4 B 2 A2t 3 B
dt
dt
d 4
d
5 At B 2 2 At 3 B
dt
dt
5 4t 3 2 2A3t 2 B
5 4t 3 2 6t 2
dy
d 1 5 1 3 1 2
d.
5
a x 1 x 2 x 1 1b
dx
dx 5
3
2
d 1 5
d 1 3
d 1 2
d
5
a xb1
a xb2
a xb1
(1)
dx 5
dx 3
dx 2
dx
1 d
1 d
1 d
5 a b Ax5 B 1 a b Ax3 B 2 a b Ax2 B
5 dx
3 dx
2 dx
d
1
(1)
dx
1
1
1
5 A5x4 B 1 A3x2 B 2 (2x) 1 0
5
3
2
5 x4 1 x2 2 x
f. f r(x) 5
2-10
d
A5Ax2 B 4 B
dx
d
5 5 Ax234 B
dx
d
5 5 Ax8 B
dx
5 5A8x7 B
5 40x7
d t5 2 3t2
f. sr(t) 5 a
b
dt
2t
1 d
5 a b At 4 2 3tB
2 dt
1 d
d
5 a b a At4 B 2 (3t)b
2 dt
dt
1 d
d
5 a b a At4 B 2 3 (t)b
2 dt
dt
1
5 a b A4t 3 2 3(1)B
2
3
5 2t 3 2
2
dy
d
5
4. a.
5
A3x3 B
dx
dx
d 5
5 3 ( x3)
dx
5
5
5 a b3( xA3 21))
3
2
5 5x3
dy
6
d
1
b.
5
a4x22 2 b
x
dx
dx
d
d
1
5 4 ( x22) 2 6 Ax21 B
dx
dx
21
1
5 4a
b ( x22 21) 2 6(21)Ax2121 B
2
3
5 22x22 1 6x22
dy
d 6
2
5
a 3 1 2 2 3b
c.
dx
dx x
x
d
d
d
5 6 Ax23 B 1 2 Ax22 B 2
(3)
dx
dx
dx
5 6(23)Ax2321 B 1 2(22)Ax2221 B 2 0
e. gr(x) 5
5 218x24 2 4x23
218
4
5 4 2 3
x
x
Chapter 2: Derivatives
d.
dy
d
5
9x22 1 3"x
dx
dx
d
d 1
5 9 Ax22 B 1 3 ( x2)
dx
dx
(
)
(
1
1
5 9(22)Ax2221 B 1 3a b ( x2 21)
2
3
1
5 218x23 1 x22
2
dy
d
e.
5
"x 1 6"x3 1 "2
dx
dx
d 1
d 3
d
5 ( x2) 1 6 ( x2) 1
"2
dx
dx
dx
1 1
3
3
5 ( x2 21) 1 6a b ( x2 21) 1 0
2
2
1
1
1
5 ( x22) 1 9x2
2
(
5 2t 2 6(1) 1 0
5 2t 2 6
d 3
6. a. f r(x) 5
x 2 "x
dx
d 3
d 1
5
Ax B 2 ( x2)
dx
dx
1 1
5 3x2 2 ( x2 21)
2
1 1
5 3x2 2 x22
2
1
1
so f r(a) 5 f r(4) 5 3(4)2 2 (4)22
2
1 1
5 3(16) 2
2 "4
1 1
5 48 2 a b a b
2 2
5 47.75
d
2
7 2 6"x 1 5x3
b. f r(x) 5
dx
d
d 1
d 2
5
(7) 2 6 ( x2) 1 5 ( x3)
dx
dx
dx
1
2
1
2
5 0 2 6a b ( x2 21) 1 5a b ( x3 21)
2
3
10
1
1
5 23x22 1 a b ( x23)
3
10
1
1
so f r(a) 5 f r(64) 5 23( 6422) 1 a b ( 6423)
3
1
10 1
5 23a b 1 a b
8
3 4
11
5
24
dy
d
5
A3x4 B
7. a.
dx
dx
d
5 3 Ax4 B
dx
5 3A4x3 B
5 12x3
The slope at (1, 3) is found by substituting x 5 1 into
)
( )
dy
d 1 1 "x
5
a
b
x
dx
dx
1
d 1
d x2
5
a b1
a b
dx x
dx x
d 21
d
1
5
(x ) 1 ( x22)
dx
dx
21 212 21
5 (21)x2121 1
(x )
2
1 3
5 2x22 2 x22
2
ds
d
5. a.
5 A22t 2 1 7tB
dt
dt
d
d
5 (22)a At2 B b 1 7a (t)b
dt
dt
5 (22)(2t) 1 7(1)
5 24t 1 7
ds
d
1
b.
5 a18 1 5t 2 t 3 b
dt
dt
3
d
d
1 d
5 (18) 1 5 (t) 2 a b At 3 B
dt
dt
3 dt
1
5 0 1 5(1) 2 a b A3t 2 B
3
5 5 2 t2
d
ds
c.
5 A(t 2 3)2B
dt
dt
d
5 At 2 2 6t 1 9B
dt
d
d
d
5 At 2 B 2 (6) (t) 1 (9)
dt
dt
dt
f.
Calculus and Vectors Solutions Manual
(
)
)
dy
the equation for dx. So the slope 5 12(1)3
5 12
dy
d 1
b
5
a
b.
dx
dx x25
d 5
5
Ax B
dx
5 5x4
2-11
The slope at (21, 21) is found by substituting x 5 21
The slope at x 5 4 is found by substituting x 5 4
dy
into the equation for dx. So the
into the equation for dx. So the slope is (4) 2 5 12.
16
c. y 5 2
x
dy
d 16
5
a b
dx
dx x2
d
5 16 Ax22 B
dx
5 16(22)x2221
5 232x23
The slope at x 5 22 is found by substituting
dy
slope 5 5(21)4
55
dy
d 2
c.
5
a b
dx
dx x
d
5 2 Ax21 B
dx
5 2(21)x2121
5 22x22
The slope at (22 , 21) is found by substituting
dy
x 5 22 into the equation for dx. So the
slope 5 22(22)22
1
52
2
dy
d
d.
5 ("16x3)
dx
dx
d 3
5 "16 ( x2)
dx
3 3
5 4a bx2 21
2
1
2
5 6x
The slope at (4, 32) is found by substituting x 5 4
into the equation for
1
dy
. So the
dx
slope 5 6(4)2
5 12
8. a. y 5 2x3 1 3x
dy
d
5
A2x3 1 3xB
dx
dx
d
d
5 2 Ax3 B 1 3 (x)
dx
dx
5 2A3x2 B 1 3(1)
5 6x2 1 3
The slope at x 5 1 is found by substituting x 5 1
dy
into the equation for dx. So the slope is
6(1)2 1 3 5 9.
b. y 5 2"x 1 5
dy
d
5 (2"x 1 5)
dx
dx
d 1
d
5 2 ( x2 ) 1
(5)
dx
dx
1
1
5 2a b ( x2 21) 1 0
2
21
5 x2
2-12
21
x 5 22 into the equation for
23
232(22)
5
(232)
(22)3
dy
.
dx
So the slope is
5 4.
d. y 5 x23 (x21 1 1)
5 x24 1 x23
dy
d 24
5
(x 1 x23 )
dx
dx
5 24x25 2 3x24
3
4
52 52 4
x
x
The slope at x 5 1 is found by substituting
dy
x 5 1 into the equation for dx. So the slope is
2 145 2 134 5 27.
dy
d
1
5
a2x 2 b
x
dx
dx
d
d 21
5 2 (x) 2
Ax B
dx
dx
5 2(1) 2 (21)x2121
5 2 1 x22
The slope at x 5 0.5 is found by substituting
9. a.
dy
x 5 0.5 into the equation for dx.
So the slope is 2 1 (0.5)22 5 6.
The equation of the tangent line is therefore
y 1 1 5 6(x 2 0.5) or 6x 2 y 2 4 5 0.
dy
d 3
4
b.
5
a 2 3b
dx
dx x2
x
d
d
5 3 Ax22 B 2 4 Ax23 B
dx
dx
5 3(22)x2221 2 4(23)x2321
5 12x24 2 6x23
The slope at x 5 21 is found by substituting
dy
x 5 21 into the equation for dx. So the slope is
12(21)24 2 6(21)23 5 18.
Chapter 2: Derivatives
The equation of the tangent line is therefore
y 2 7 5 18(x 1 1) or 18x 2 y 1 25 5 0.
d
dy
c.
5 ("3x3)
dx
dx
d 3
5 "3 ( x2)
dx
3 32 21
5 "3a bx
2
1
3"3x2
5
2
The slope at x 5 3 is found by substituting x 5 3
dy
into the equation for dx.
1
3"3(3)2
9
So the slope is
5 .
2
2
The equation of the tangent line is therefore
y 2 9 5 92 (x 2 3) or 9x 2 2y 2 9 5 0.
dy
d 1 2 1
d.
5
a ax 1 bb
x
dx
dx x
d
1
5
ax 1 2 b
dx
x
d
d 22
5
(x) 1
Ax B
dx
dx
5 1 1 (22)x2221
5 1 2 2x23
The slope at x 5 1 is found by substituting
dy
into the equation for dx.
So the slope is 1 2 2(1)23 5 21.
The equation of the tangent line is therefore
y 2 2 5 2 (x 2 1) or x 1 y 2 3 5 0.
dy
d
e.
5 (("x 2 2)(3"x 1 8))
dx
dx
d
5 (3("x)2 1 8"x 2 6"x 2 16)
dx
d
5 (3x 1 2"x 2 16)
dx
d
d 1
d
5
(3x) 1 2 ( x2) 2
(16)
dx
dx
dx
1 1
5 3(1) 1 2a bx2 21 2 0
2
1
5 3 1 x22
The slope at x 5 4 is found by substituting x 5 4
The equation of the tangent line is therefore
y 5 3.5(x 2 4) or 7x 2 2y 2 28 5 0.
dy
d "x 2 2
f.
5
a 3
b
dx
dx
"x
1
d x2 2 2
5
a
b
1
dx
x3
d 1 1
1
5 ( x2 2 3 2 2x23)
dx
d 1
d
1
5 ( x6) 2 2 Ax23 B
dx
dx
1 1
1
1
5 ( x6 21) 2 2a2 bx23 21 2 0
6
3
2 243
1 256
5 (x ) 1 x
6
3
The slope at x 5 1 is found by substituting x 5 1
dy
into the equation for dx.
5
4
So the slope is 16 (1)26 1 23 (1)23 5 56.
The equation of the tangent line is therefore
y 1 1 5 56 (x 2 1) or 5x 2 6y 2 11 5 0.
10. A normal to the graph of a function at a point is
a line that is perpendicular to the tangent at the
given point.
3
4
y 5 2 2 3 at P(21, 7)
x
x
Slope of the tangent is 18, therefore, the slope of
the normal is 2 181 .
1
Equation is y 2 7 5 2 (x 1 1).
18
x 1 18y 2 125 5 0
3
1
11. y 5 3 5 3x23
"x
Parallel to x 1 16y 1 3 2 0
Slope of the line is 2 161 .
dy
4
5 2x23
dx
1
4
x23 5
16
1
1
4 5
3
16
x
4
x3 5 16
3
x 5 (16)4 5 8
dy
into the equation for dx.
1
So the slope is 3 1 (4)22 5 3.5.
Calculus and Vectors Solutions Manual
2-13
1
5 x21 : y 5 x3
x
dy
1 dy
52 2:
5 3x2
dx
x dx
1
Now, 2 2 5 3x2
x
1
x4 5 2 .
3
No real solution. They never have the same slope.
dy
5 2x
13. y 5 x2,
dx
The slope of the tangent at A(2, 4) is 4 and at
B Q 2 18 , 641 R is 2 14.
Since the product of the slopes is 21, the tangents
at A(2, 4) and B Q 2 18 , 641 R will be perpendicular.
12. y 5
4 y
3
2
1
–3 –2 –1 0
–1
x
1 2 3
14. y 5 2x2 1 3x 1 4
dy
5 22x 1 3
dx
dy
55
For
dx
5 5 22x 1 3
x 5 21.
The point is (21, 0).
6
5
4
3
2
1
–2 –1 0
–1
–2
y
x
1
2 3 4
15. y 5 x3 1 2
dy
5 3x2, slope is 12
dx
x2 5 4
x 5 2 or x 5 22
Points are (2, 10) and (22, 26).
2-14
16. y 5 15x5 2 10x, slope is 6
dy
5 x4 2 10 5 6
dx
x4 5 16
x2 5 4 or x2 5 24
x 5 62 non-real
Tangents with slope 6 are at the points Q 2, 2 685 R
and Q2 2, 685 R .
17. y 5 2x2 1 3
a. Equation of tangent from A(2, 3):
If x 5 a, y 5 2x2 1 3.
Let the point of tangency be PAa, 2a2 1 3B.
dy
dy
Now, dx 5 4x and when x 5 a, dx 5 4a.
The slope of the tangent is the slope of AP.
2a2
5 4a.
a22
2a2 5 4a2 2 8a
2
2a 2 8a 5 0
2a(a 2 4) 5 0
a 5 0 or a 5 4.
Point (2, 3):
Slope is 0.
Slope is 16.
Equation of tangent is
Equation of tangent is
y 2 3 5 0.
y 2 3 5 16(x 2 2) or
16x 2 y 2 29 5 0.
b. From the point B(2, 27):
2a2 1 10
Slope of BP:
5 4a
a22
2a2 1 10 5 4a2 2 8a
2
2a 2 8a 2 10 5 0
a2 2 4a 2 5 5 0
(a 2 5)(a 1 1) 5 0
a55
a 5 21
Slope is 4a 5 20.
Slope is 4a 5 24.
Equation is
Equation is
y 1 7 5 20(x 2 2)
y 1 7 5 24(x 2 2)
or 20x 2 y 2 47 5 0.
or 4x 1 y 2 1 5 0.
a
18. ax 2 4y 1 21 5 0 is tangent to y 5 x2 at x 5 22.
Therefore, the point of tangency is a22, 4 b,
This point lies on the line, therefore,
a
a(22) 2 4a b 1 21 5 0
4
23a 1 21 5 0
a 5 7.
a
Chapter 2: Derivatives
19. a. When h 5 200,
d 5 3.53"200
8 49.9
Passengers can see about 49.9 km.
1
b. d 5 3.53!h 5 3.53h2
1 1
dr 5 3.53a h 2 2 b
2
3.53
5
2!h
When h 5 200,
3.53
dr 5
2 !200
8 0.12
The rate of change is about 0.12 km>m.
20. d(t) 5 4.9t2
a. d(2) 5 4.9(2)2 5 19.6 m
d(5) 5 4.9(5)2 5 122.5 m
The average rate of change of distance with respect
to time from 2 s to 5 s is
Dd
122.5 2 19.6
5
Dt
522
5 34.3 m>s
b. dr(t) 5 9.8t
Thus, dr(4) 5 9.8(4) 5 39.2 m>s.
c. When the object hits the ground, d 5 150.
Set d(t) 5 150:
4.9t2 5 150
1500
t2 5
49
10
t 5 6 "15
7
10
Since t $ 0, t 5 "15
7
Then,
10
10
dra "15b 5 9.8a "15b
7
7
8 54.2 m>s
21. v(t) 5 sr(t) 5 2t 2 t2
0.5 5 2t 2 t2
t2 2 2t 1 0.5 5 0
2t2 2 4t 1 1 5 0
4 6 "8
t5
4
t 8 1.71, 0.29
The train has a velocity of 0.5 km>min at about
0.29 min and 1.71 min.
Calculus and Vectors Solutions Manual
22. v(t) 5 Rr(t) 5 210t
v(2) 5 220
The velocity of the bolt at t 5 2 is 220 m>s.
23.
y
3 (0, 3)
2
1
–3 –2 –1 0
–1
–2
–3
1
x
2 3
Let the coordinates of the points of tangency be
AAa,23a2 B.
dy
5 26x, slope of the tangent at A is 26a
dx
23a2 2 3
5 26a
a
23a2 2 3 5 26a2
3a2 5 3
a 5 1 or a 5 21
Coordinates of the points at which the tangents
touch the curve are (1, 23) and (21, 23).
24. y 5 x3 2 6x2 1 8x, tangent at A(3, 23)
dy
5 3x2 2 12x 1 8
dx
When x 5 3,
dy
5 27 2 36 1 8 5 21
dx
The slope of the tangent at A(3, 23) is 21.
Equation will be
y 1 3 5 21(x 2 3)
y 5 2x.
2x 5 x3 2 6x2 1 8x
x3 2 6x2 1 9x 5 0
xAx2 2 6x 1 9B 5 0
x(x 2 3)2 5 0
x 5 0 or x 5 3
Coordinates are B(0, 0).
Slope of PA:
3
2
1
–1 0
–11
–2
–3
y
x
1
2 3 4
2-15
25. a. i. f(x) 5 2x 2 5x2
f r(x) 5 2 2 10x
Set f r(x) 5 0:
2 2 10x 5 0
10x 5 2
1
x5
5
Then,
1
1
1 2
f a b 5 2a b 2 5a b
5
5
5
2
1
5 2
5
5
1
5
5
Thus the point is Q 15, 15 R .
ii. f(x) 5 4x2 1 2x 2 3
f r(x) 5 8x 1 2
Set f r(x) 5 0:
8x 1 2 5 0
8x 5 22
1
x52
4
Then,
1
1 2
1
f a2 b 5 4a2 b 1 2a2 b 2 3
4
4
4
1
2
12
5 2 2
4
4
4
13
52
4
Thus the point is Q 2 14, 2 134 R .
iii. f(x) 5 x3 2 8x2 1 5x 1 3
f r(x) 5 3x2 2 16x 1 5
Set f r(x) 5 0:
3x2 2 16x 1 5 5 0
2
3x 2 15x 2 x 1 5 5 0
3x(x 2 5) 2 (x 2 5) 5 0
(3x 2 1)(x 2 5) 5 0
1
x5 ,5
3
1 3
1 2
1
1
f a b 5 a b 2 8a b 1 5a b 1 3
3
3
3
3
1
24
45
81
5
2
1
1
27
27
27
27
103
5
27
f(5) 5 (5)3 2 8(5)2 1 5(5) 1 3
5 25 2 200 1 25 1 3
5 247
Thus the two points are Q 13, 103
27 R and (5, 247).
b. At these points, the slopes of the tangents are
zero, meaning that the rate of change of the value
of the function with respect to the domain is zero.
These points are also local maximum or minimum
points.
26. "x 1 "y 5 1
P(a, b) is on the curve, therefore a $ 0, b $ 0.
!y 5 1 2 !x
y 5 1 2 2 !x 1 x
1
dy
1
5 2 ? 2x22 1 1
dx
2
21 1 !a
1
At x 5 a, slope is 2
115
.
!a
!a
But !a 1 !b 5 1
2 !b 5 !a 2 1.
"b
b
Therefore, slope is 2
52
.
Åa
"a
27. f(x) 5 xn, f r(x) 5 nxn21
Slope of the tangent at x 5 1 is f r(1) 5 n,
The equation of the tangent at (1, 1) is:
y 2 1 5 n(x 2 1)
nx 2 y 2 n 1 1 5 0
Let y 5 0, nx 5 n 2 1
n21
1
x5
512 .
n
n
1
1
The x-intercept is 1 2 ; as n S `, S 0, and
n
n
the x-intercept approaches 1. As n S `, the slope
of the tangent at (1, 1) increases without bound, and
the tangent approaches a vertical line having equation
x 2 1 5 0.
28. a.
y
9
8
7
6
5
4
3
2
1
–2 –1 0
2-16
x
1
2 3 4
Chapter 2: Derivatives
f(x) 5 e
x2, if x , 3
x 1 6, if x $ 3
f r(3) does not exist.
b.
7
6
5
4
3
2
1
2x, if x , 3
1, if x $ 3
y
x
–3 –2 –1 0
–1
f(x) 5 e
f '(x) 5 e
1 2 3
3x2 2 6, if x , 2"2 or x . "2
6 2 3x2, if 2"2 , x , "2
f r(x) 5 e
6x, if x , 2"2 or x . "2
26x, if 2"2 # x # "2
f r "2 and f r 2"2 do not exist.
( )
(
c.
3
2
1
–3 –2 –1 0
–1
)
y
c. h(x) 5 (3x 1 2)(2x 2 7)
hr(x) 5 (3x 1 2)(2) 1 (3)(2x 2 7)
5 12x 2 17
d. h(x) 5 A5x7 1 1B Ax2 2 2xB
hr(x) 5 A5x7 1 1B (2x 2 2) 1 A35x6 BAx2 2 2xB
5 45x8 2 80x7 1 2x 2 2
e. s(t) 5 At2 1 1B A3 2 2t2 B
sr(t) 5 At2 1 1B (24t) 1 (2t)A3 2 2t2 B
5 28t3 1 2t
x23
f. f(x) 5
x13
f(x) 5 (x 2 3)(x 1 3)21
f r(x) 5 (x 2 3)(21)(x 1 3)22 1 (1)(x 1 3)21
5 (x 1 3)22 (2x 1 3 1 x 1 3)
6
5
(x 1 3)2
2. a. y 5 (5x 1 1)3 (x 2 4)
dy
5 (5x 1 1)3 (1) 1 3(5x 1 1)2 (5)(x 2 4)
dx
5 (5x 1 1)3 1 15(5x 1 1)2 (x 2 4)
b. y 5 A3x2 1 4BA3 1 x3 B 5
dy
5 A3x2 1 4B (5)A3 1 x3 B 4 A3x2 B
dx
1 (6x)A3 1 x3 B 5
5 15x2 (3x2 1 4)(3 1 x3 )4 1 6x(3 1 x3 )5
x
1 2 3
x 2 1, if x $ 1
1 2 x, if 0 # x , 1
f(x) 5 μ
x 1 1, if 21 , x , 0
2x 2 1, if x # 21
since
since
since
since
1, if x . 1
21, if 0 , x , 1
f'(x) 5 μ
1, if 21 , x , 0
21, if x , 21
f r(0), f r(21), and f r(1) do not exist.
2.3 The Product Rule, pp. 90–91
1. a. h(x) 5 x(x 2 4)
hr(x) 5 x(1) 1 (1)(x 2 4)
5 2x 2 4
b. h(x) 5 x2 (2x 2 1)
hr(x) 5 x2 (2) 1 (2x)(2x 2 1)
5 6x2 2 2x
Calculus and Vectors Solutions Manual
Zx 2 1 Z 5 x 2 1
Zx 2 1 Z 5 1 2 x
Z2x 2 1 Z 5 x 1 1
Z2x21 Z 5 2x 2 1
y 5 A1 2 x2 B 4 (2x 1 6)3
dy
5 4A1 2 x2 B 3 (22x)(2x 1 6)3
dx
1 A1 2 x2 B 4 3(2x 1 6)2 (2)
5 28xA1 2 x2 B 3 (2x 1 6)3
1 6A1 2 x2 B 4 (2x 1 6)2
d. y 5 Ax2 2 9B 4 (2x 2 1)3
dy
5 Ax2 2 9B 4 (3)(2x 2 1)2 (2)
dx
1 4Ax2 2 9B 3 (2x)(2x 2 1)3
5 6(x2 2 9)4 (2x 2 1)2
1 8x(x2 2 9)3 (2x 2 1)3
c.
2-17
3. It is not appropriate or necessary to use the product
rule when one of the factors is a constant or when it
would be easier to first determine the product of the
factors and then use other rules to determine the
derivative. For example, it would not be best to
use the product rule for f(x) 5 3Ax2 1 1B or
g(x) 5 (x 1 1) (x 2 1).
4. F(x) 5 3b(x)43c(x)4
F r(x) 5 3b(x)43cr(x)4 1 3br(x)43c(x)4
5. a. y 5 (2 1 7x)(x 2 3)
dy
5 (2 1 7x)(1) 1 7(x 2 3)
dx
At x 5 2,
dy
5 (2 1 14) 1 7(21)
dx
5 16 2 7
59
b.
y 5 (1 2 2x)(1 1 2x)
dy
5 (1 2 2x)(2) 1 (22)(1 1 2x)
dx
1
At x 5 ,
2
dy
5 (0)(2) 2 2(2)
dx
5 24
c.
y 5 A3 2 2x 2 x2 B Ax2 1 x 2 2B
dy
5 A3 2 2x 2 x2 B A2x 1 1B
dx
1 (22 2 2x)Ax2 1 x 2 2B
At x 5 22,
dy
5 (3 1 4 2 4)(24 1 1)
dx
1 (22 1 4)(4 2 2 2 2)
5 (3)(23) 1 (2)(0)
5 29
d. y 5 x3 (3x 1 7)2
dy
5 3x2 (3x 1 7)2 1 x36(3x 1 7)
dx
At x 5 22,
dy
5 12(1)2 1 (28)(6)(1)
dx
5 12 2 48
5 236
e.
y 5 (2x 1 1)5 (3x 1 2)4, x 5 21
dy
5 5(2x 1 1)4 (2)(3x 1 2)4
dx
1 (2x 1 1)54(3x 1 2)3 (3)
At x 5 21,
2-18
dy
5 5(21)4 (2)(21)4
dx
1 (21)5 (4)(21)3 (3)
5 10 1 12
5 22
f.
y 5 x(5x 2 2)(5x 1 2)
5 xA25x2 2 4B
dy
5 x(50x) 1 (25x2 2 4)(1)
dx
At x 5 3,
dy
5 3(150) 1 (25 ? 9 2 4)
dx
5 450 1 221
5 671
6. Tangent to y 5 Ax3 2 5x 1 2B A3x2 2 2xB
at (1, 22)
dy
5 A3x2 2 5B A3x2 2 2xB
dx
1 Ax3 2 5x 1 2B (6x 2 2)
when x 5 1,
dy
5 (22)(1) 1 (22)(4)
dx
5 22 1 28
5 210
Slope of the tangent at (1, 22) is 210.
The equation is y 1 2 5 210(x 2 1);
10x 1 y 2 8 5 0.
7. a. y 5 2(x 2 29)(x 1 1)
dy
5 2(x 2 29)(1) 1 2(1)(x 1 1)
dx
2x 2 58 1 2x 1 2 5 0
4x 2 56 5 0
4x 5 56
x 5 14
Point of horizontal tangency is (14, 2450).
b. y 5 Ax2 1 2x 1 1B Ax2 1 2x 1 1B
5 Ax2 1 2x 1 1B 2
dy
5 2Ax2 1 2x 1 1B (2x 1 2)
dx
Ax2 1 2x 1 1B (2x 1 2) 5 0
2(x 1 1)(x 1 1)(x 1 1) 5 0
x 5 21
Point of horizontal tangency is (21, 0).
8. a. y 5 (x 1 1)3 (x 1 4)(x 2 3)2
dy
5 3(x 1 1)2 (x 1 4)(x 2 3)2
dx
1 (x 1 1)3 (1)(x 2 3)2
1 (x 1 1)3 (x 1 4)32(x 2 3)4
Chapter 2: Derivatives
b. y 5 x2 A3x2 1 4B 2 A3 2 x3 B 4
dy
5 2xA3x2 1 4B 2 A3 2 x3 B 4
dx
1 x2 32A3x2 1 4B (6x)4 A3 2 x3 B 4
1 x2 A3x2 1 4B 2 34A3 2 x3 B 3 A23x2 B4
t 2
9. V(t) 5 75a1 2 b , 0 # t # 24
24
75 L 3 60% 5 45 L
t 2
45
5 a1 2 b
Set
75
24
t
3
512
6
24
Å5
3
t5a6
2 1b (224)
Å5
t 8 42.590 (inadmissable) or t 8 5.4097
t 2
V(t) 5 75a1 2 b
24
t
t
V(t) 5 75a1 2 b a1 2 b
24
24
t
1
Vr(t) 5 75 c a1 2 b a2 b
24
24
1
t
1 a2 b a1 2 b d
24
24
t
1
5 (75)(2)a1 2 b a2 b
24
24
Vr(5.4097) 5 24.84 L>h
10. Determine the point of tangency, and then find the
negative reciprocal of the slope of the tangent. Use
this information to find the equation of the normal.
h(x) 5 2x(x 1 1)3 Ax2 1 2x 1 1B 2
hr(x) 5 2(x 1 1)3 Ax2 1 2x 1 1B 2
1 (2x)(3)(x 1 1)2 Ax2 1 2x 1 1B 2
1 2x(x 1 1)3 2Ax2 1 2x 1 1B (2x 1 2)
hr(22) 5 2(21)3 (1)2
1 2(22)(3)(21)2 (1)2
1 2(22)(21)3 (2)(1)(22)
5 22 2 12 2 16
5 230
11.
a. f(x) 5 g1 (x)g2 (x)g3 (x) c gn21 (x)gn (x)
f r(x) 5 g1r(x)g2 (x)g3 (x) c gn21 (x)gn (x)
1 g1 (x)g2r(x)g3 (x) c gn21 (x)gn (x)
1 g1 (x)g2 (x)g3r(x) c gn21 (x)gn (x)
1 c 1 g1 (x)g2 (x)g3 (x) cgn21 (x)gnr(x)
Calculus and Vectors Solutions Manual
f(x) 5 (1 1 x)(1 1 2x)(1 1 3x) c
(1 1 nx)
f r(x) 5 1(1 1 2x)(1 1 3x) c(1 1 nx)
1 (1 1 x)(2)(1 1 3x) c(1 1 nx)
1 (1 1 x)(1 1 2x)(3) c(1 1 nx)
1 c 1 (1 1 x)(1 1 2x)(1 1 3x)
c (n)
f r(0) 5 1(1)(1)(1) c (1)
1 1(2)(1)(1) c (1)
1 1(1)(3)(1) c (1)
1 c 1 (1)(1)(1) c (n)
5112131c1n
n(n 1 1)
f r(0) 5
2
12. f(x) 5 ax2 1 bx 1 c
f r(x) 5 2ax 1 b
(1)
Horizontal tangent at (21, 28)
f r(x) 5 0 at x 5 21
22a 1 b 5 0
Since (2, 19) lies on the curve,
4a 1 2b 1 c 5 19
(2)
Since (21, 28) lies on the curve,
a 2 b 1 c 5 28
(3)
4a 1 2b 1 c 5 19
23a 2 3b 5 227
a1b59
22a 1 b 5 0
3a 5 9
a 5 3, b 5 6
3 2 6 1 c 5 28
c 5 25
The equation is y 5 3x2 1 6x 2 5.
13.
y
3
2
1
x
0
–3 –2 –1
1 2 3
–1
b.
a. x 5 1 or x 5 21
b. f r(x) 5 2x, x , 21 or x . 1
f r(x) 5 22x, 21 , x , 1
6
4
2
–3 –2 –1 0
–2
–4
–6
y
x
1 2 3
2-19
c. f r(22) 5 2(22) 5 24
f r(0) 5 22(0) 5 0
f r(3) 5 2(3) 5 6
16
14. y 5 2 2 1
x
dy
32
52 3
dx
x
Slope of the line is 4.
32
2 3 54
x
4x3 5 232
x3 5 28
x 5 22
16
y5
21
4
53
Point is at (22, 3).
Find intersection of line and curve:
4x 2 y 1 11 5 0
y 5 4x 1 11
Substitute,
16
4x 1 11 5 2 2 1
x
4x3 1 11x2 5 16 2 x2 or 4x3 1 12x2 2 16 5 0.
Let x 5 22
RS 5 4(22)3 1 12(22)2 2 16
50
Since x 5 22 satisfies the equation, therefore it is
a solution.
When x 5 22, y 5 4(22) 1 11 5 3.
Intersection point is (22, 3). Therefore, the line is
tangent to the curve.
Mid-Chapter Review, pp. 92–93
1. a.
6
4
2
y
x
–2 0
–2
–4
–6
2 4 6
˛
x
Slope of Tangent
f 9(x)
0
25
1
23
2
21
3
1
4
3
5
5
y
c.
6
4
2
–2 0
–2
–4
–6
x
2 4 6
d. f(x) is quadratic; f r(x) is linear.
(6(x 1 h) 1 15) 2 (6x 1 15)
2. a. f r(x) 5 lim
hS0
h
6h
5 lim
hS0 h
5 lim 6
˛
hS0
56
A2(x 1 h)2 2 4B 2 A2x2 2 4B
b. f r(x) 5 lim
hS0
h
(x 1 h) 2 2 x2
5 lim2
hS0
h
((x 1 h) 2 x)((x 1 h) 1 x)
5 lim2
hS0
h
h(2x 1 h)
5 lim2
hS0
h
5 lim2(2x 1 h)
hS0
5 4x
((x 1 h)2 2 5(x 1 h)) 2 (x2 2 5x)
b. f r(x) 5 lim
hS0
h
x2 1 2hx 1 h2 2 5x 2 5h 2 x2 1 5x
5 lim
hS0
h
h2 1 2hx 2 5h
5 lim
hS0
h
˛
˛
˛
2-20
h(h 1 2x 2 5)
hS0
h
5 2x 2 5
Use the derivative function to calculate the slopes of
the tangents.
5 lim
c. f r(x) 5 lim
5
5
2x15
(x 1 h) 1 5
h
5(x 1 5) 2 5((x 1 h) 1 5)
5 lim
hS0
((x 1 h) 1 5)(x 1 5)h
25h
5 lim
h S 0 ((x 1 h) 1 5) (x 1 5)h
hS0
Chapter 2: Derivatives
25
h S 0 ((x 1 h) 1 5) (x 1 5)
25
5
(x 1 5) 2
!(x 1 h) 2 2 2 !x 2 2
d. f r(x) 5 lim
hS0
h
5 lim
5 lim c
hS0
!(x 1 h) 2 2 2 !x 2 2
h
3
!(x 1 h) 2 2 1 !x 2 2
d
!(x 1 h) 2 2 1 !x 2 2
((x 1 h) 2 2) 2 (x 2 2)
1 h) 2 2 1 !x 2 2B
hS0 hA !(x
5 lim
h
hS0 hA !(x 1 h) 2 2 1 !x 2 2B
5 lim
1
hS0 !(x 1 h) 2 2 1 !x 2 2
1
5
2 !x 2 2
3. a. yr 5 2x 2 4
When x 5 1,
yr 5 2(1) 2 4
5 22.
When x 5 1,
y 5 (1)2 2 4(1) 1 3
5 0.
Equation of the tangent line:
y 2 0 5 22(x 2 1), or y 5 22x 1 2
b.
y
6
4
2
x
0
–4 –2
2 4 6
–2
–4
–6
5 lim
dy
5 24x3
dx
dy
1
b.
5 5x22
dx
5
5
!x
c. gr(x) 5 26x24
6
52 4
x
4. a.
Calculus and Vectors Solutions Manual
dy
5 5 2 6x23
dx
6
552 3
x
dy
5 2(11t 1 1)(11)
e.
dt
5 242t 1 22
1
f. y 5 1 2
x
5 1 2 x21
dy
5 x22
dx
1
5 2
x
5. f r(x) 5 8x3
8x3 5 1
1
x3 5
8
1
x5
2
1
1 4
f a b 5 2a b
2
2
1
5
8
Equation of the tangent line:
3
1
1
y 2 5 1ax 2 b, or y 5 x 2
8
2
8
6. a. f r(x) 5 8x 2 7
b. f r(x) 5 26x2 1 8x 1 5
c. f(x) 5 5x22 2 3x23
f r(x) 5 210x23 1 9x24
10
9
52 3 1 4
x
x
1
1
d. f(x) 5 x2 1 x3
1 1
1 2
f r(x) 5 x22 1 x23
2
3
1
1
5 12 1 23
2x
3x
1
e. f(x) 5 7x22 2 3x2
3 1
f r(x) 5 214x23 2 x22
2
14
3
5 2 3 2 12
x
2x
f. f r(x) 5 4x22 1 5
4
5 215
x
d.
2-21
7. a. yr 5 26x 1 6
When x 5 1,
yr 5 26(1) 1 6
5 0.
When x 5 1,
y 5 23A12 B 1 6(1) 1 4
5 7.
Equation of the tangent line:
y 2 7 5 0(x 2 1), or
y57
1
b. y 5 3 2 2x2
1
yr 5 2x22
21
5
!x
When x 5 9,
21
yr 5
!9
1
52 .
3
When x 5 9,
y 5 3 2 2 !9
5 23.
Equation of the tangent line:
1
1
y 2 (23) 5 2 (x 2 9), or y 5 2 x
3
3
c. f r(x) 5 28x3 1 12x2 2 4x 2 8
f r(3) 5 28(3)3 1 12(3)2 2 4(3) 2 8
5 2216 1 108 2 12 2 8
5 2218
f(3) 5 22(3)4 1 4(3)3 2 2(3)2 2 8(3) 1 9
5 2162 1 108 2 18 2 24 1 9
5 287
Equation of the tangent line:
y 2 (287) 5 2128(x 2 3), or
y 5 2128x 1 297
d
8. a. f r(x) 5 A4x2 2 9xB A3x2 1 5B
dx
d
1 A4x2 2 9xB A3x2 1 5B
dx
5 (8x 2 9)A3x2 1 5B 1 A4x2 2 9xB (6x)
5 24x3 2 27x2 1 40x 2 45
1 24x3 2 54x2
5 48x3 2 81x2 1 40x 2 45
d
b. f r(t) 5 A23t2 2 7t 1 8B (4t 2 1)
dt
d
1 A23t2 2 7t 1 8B (4t 2 1)
dt
5 (26t 2 7)(4t 2 1)
1 A23t2 2 7t 1 8B (4)
2-22
5 224t2 2 28t 1 6t 1 7 2 12t2 2 28t 1 32
5 236t2 2 50t 1 39
dy
d
c.
5
A3x2 1 4x 2 6B (2x2 2 9)
dx
dx
d
1 A3x2 1 4x 2 6B A2x2 2 9B
dx
5 (6x 1 4)A2x2 2 9B 1 A3x2 1 4x 2 6B (4x)
5 12x3 2 54x 1 8x2 2 36 1 12x3
1 16x2 2 24x
5 24x3 1 24x2 2 78x 2 36
dy
d
d.
5
A3 2 2x3 B 2 A3 2 2x3 B
dx
dx
d
1 A3 2 2x3 B 2 A3 2 2x3 B
dx
d
5 c A3 2 2x3 B A3 2 2x3 B
dx
d
1 A3 2 2x3 B A3 2 2x3 B d A3 2 2x3 B
dx
1 A3 2 2x3 B 2 A26x2 B
5 S 2A26x2 B A3 2 2x3 B T A3 2 2x3 B
1 A3 2 2x3 B 2 A26x2 B
5 3A3 2 2x3 B 2 A26x2 B
5 A3 2 2x3 B 2 A218x2 B
5 A9 2 12x3 1 4x6 BA218x2 B
5 2162x2 1 216x5 2 72x8
d
9. yr 5 A5x2 1 9x 2 2B A2x2 1 2x 1 3B
dx
d
1 A5x2 1 9x 2 2B A2x2 1 2x 1 3B
dx
5 (10x 1 9)A2x2 1 2x 1 3B
1 A5x2 1 9x 2 2B (2 2 2x)
yr(1) 5 (10(1) 1 9)(2 (1)2 1 2(1) 1 3)
1 (5(1)2 1 9(1) 2 2)(2 2 2(1))
5 (19)(4)
5 76
Equation of the tangent line:
y 2 48 5 76(x 2 1), or 76x 2 y 2 28 5 0
dy
d
10.
5 2 (x 2 1)(5 2 x)
dx
dx
d
1 2(x 2 1) (5 2 x)
dx
5 2(5 2 x) 2 2(x 2 1)
5 12 2 4x
dy
The tangent line is horizontal when dx 5 0.
12 2 4x 5 0
12 5 4x
x53
Chapter 2: Derivatives
When x 5 3,
y 5 2((3) 2 1)(5 2 (3))
5 8.
Point where tangent line is horizontal: (3, 8)
dy
(5(x 1 h)2 2 8(x 1 h) 1 4)
11.
5 lim c
dx hS0
h
A5x2 2 8x 1 4B
2
d
h
5(x 1 h)2 2 5x2 2 8h
5 lim
hS0
h
5((x 1 h) 2 x)((x 1 h) 1 x) 2 8h
5 lim
hS0
h
5h(2x 1 h) 2 8h
5 lim
hS0
h
5 lim (5(2x 1 h) 2 8)
t1h
5 lim
3
5 lim
30 2
b
90
1 2
5 500a b
3
500
5
L
9
b. V(0) 5 500(1 2 0)2 5 500 L
500
V(60) 5
L
9
The average rate of change of volume with respect
to time from 0 min to 60 min is
500
DV
2 500
5 9
Dt
60 2 0
28
(500)
5 9
60
200
52
L>min
27
c. Calculate Vr(t):
V(t 1 h) 2 V(t)
Vr(t) 5 lim
hS0
h
5 lim
hS0
500a1 2 90 b 2 500a21 1 90 b
t
h
Calculus and Vectors Solutions Manual
h
500a2 b a2 2
h
90
2t 1 h
b
90
h
500
2t 1 h
5 lim 2
a2 2
b
hS0
90
90
250
2t
5
a2 2 b
9
90
2900 1 10t
5
81
hS0
2900 1 10(30)
81
200
52
L>min
27
4
13. V(r) 5 pr3
3
4
4
V(15) 5 p(15)3
a. V(10) 5 p(10)3
3
3
4
4
5 p(3375)
5 p(1000)
3
3
4000
p
5
5 4500p
3
Then, the average rate of change of volume with
respect to radius is
DV
4500p 2 4000
3 p
5
Dr
15 2 10
500p Q 9 2 83 R
5
5
19
5 100pa b
3
1900
p cm3>cm
5
3
b. First calculate Vr(r):
V(r 1 h) 2 V(r)
Vr(r) 5 lim
hS0
h
4
p3(r
1
h)3 2 r34
5 lim 3
hS0
h
4
3
p
r
1
3r2h 1 3rh2 1 h3 2 r3 R
Q
3
5 lim
hS0
h
4
2
2
3
3 p Q 3r h 1 3rh 1 h R
5 lim
hS0
h
Vr(30) 5
5 500a
2
t1h
t
a1 2 90 1 1 2 90 b
Then,
t 2
12. V(t) 5 500a1 2 b . 0 # t # 90
90
a. After 1 h, t 5 60, and the volume is
2
V(60) 5 500 Q 1 2 60
90 R
t1h
t
h
hS0
hS0
5 10x 2 8
500a1 2 90 2 1 1 90 b
2
2-23
4
5 lim pA3r2 1 3rh 1 h2 B
hS0 3
4
5 pA3r2 1 3r(0) 1 (0)2 B
3
5 4pr2
Then, Vr(8) 5 4p(8)2
5 4p(64)
5 256p cm3>cm
14. This statement is always true. A cubic polynomial
function will have the form f(x) 5 ax3 1 bx2 1
cx 1 d, a 2 0. So the derivative of this cubic is
f r(x) 5 3ax2 1 2bx 1 c, and since 3a 2 0, this
derivative is a quadratic polynomial function. For
example, if f(x) 5 x3 1 x2 1 1,
we get
f r(x) 5 3x2 1 2x,
and if
f(x) 5 2x3 1 3x2 1 6x 1 2,
we get
f r(x) 5 6x2 1 6x 1 6
x2a13b
15. y 5 a2b , a, bPI
x
Simplifying,
y 5 x2a13b2 (a2b) 5 xa14b
Then,
yr 5 (a 1 4b)a14b21
16. a. f(x) 5 26x3 1 4x 2 5x2 1 10
f r(x) 5 218x2 1 4 2 10x
Then, f r(x) 5 218(3)2 1 4 2 10(3)
5 2188
b. f r(3) is the slope of the tangent line to f(x) at
x 5 3 and the rate of change in the value of f(x)
with respect to x at x 5 3.
17. a. P(t) 5 100 1 120t 1 10t2 1 2t3
P(t) 5 100 1 120t 1 10t2 1 2t3
P(0) 5 100 1 120(0) 1 10(0)2 1 2(0)3
5 100 bacteria
b. At 5 h, the population is
P(5) 5 100 1 120(5) 1 10(5)2 1 2(5)3
5 1200 bacteria
c. Pr(t) 5 120 1 20t 1 6t2
At 5 h, the colony is growing at
Pr(5) 5 120 1 20(5) 1 6(5)2
5 370 bacteria> h
100
18. C(t) 5
,t.2
t
Simplifying, C(t) 5 100t21.
100
Then, Cr(t) 5 2100t22 5 2 2 .
t
2-24
Cr(5)
Cr(50)
Cr(100)
100
100
100
52
52
52
2
2
(5)
(50)
(100)2
1
100
100
52
52
52
25
2500
100
5 24
5 20.04
5 20.01
These are the rates of change of the percentage with
respect to time at 5, 50, and 100 min. The percentage
of carbon dioxide that is released per unit time from
the pop is decreasing. The pop is getting flat.
2.4 The Quotient Rule, pp. 97–98
1. For x, a, b real numbers,
xaxb 5 xa1b
For example,
x9x26 5 x3
Also,
Axa B b 5 xab
For example,
Ax2 B 3 5 x6
Also,
xa
5 xa2b, x 2 0
xb
For example,
x5
5 x2
x3
2.
Function
x2 1 3x
,
x
x20
f(x) 5
Rewrite
Differentiate
and Simplify,
If Necessary
f(x) 5 x 1 3
f r(x) 5 1
5
g(x) 5
3x3
,x20
x
1
,
10x5
x20
h(x) 5
y5
8x3 1 6x
,
2x
h(x) 5
1 25
x
10
hr(x) 5
21 26
x
2
y 5 4x2 1 3
dy
5 8x
dx
s5t13
ds
51
dt
x20
t2 2 9
s5
,t23
t23
1
gr(x) 5 2x23
2
g(x) 5 3x3
Chapter 2: Derivatives
3. In the previous problem, all of these rational
examples could be differentiated via the power rule
after a minor algebraic simplification.
A second approach would be to rewrite a rational
example
f(x)
h(x) 5
g(x)
using the exponent rules as
h(x) 5 f(x)(g(x))21,
and then apply the product rule for differentiation
(together with the power of a function rule to find
hr(x).
A third (and perhaps easiest) approach would be to
just apply the quotient rule to find hr(x).
(x 1 1)(1) 2 x(1)
4. a. hr(x) 5
Ax 1 1B 2
1
5
Ax 1 1B 2
(t 1 5)(2) 2 (2t 2 3)(1)
b. hr(t) 5
At 1 5B 2
13
5
At 1 5B 2
A2x2 2 1B A3x2 B 2 x3 A4xB
c. hr(x) 5
A2x2 2 1B 2
4
2x 2 3x2
5
A2x2 2 1B 2
Ax2 1 3B (0) 2 1(2x)
d. hr(x) 5
Ax2 1 3B 2
22x
5 2
Ax 1 3B 2
x(3x 1 5)
3x2 1 5x
e. y 5
5
2
A1 2 x B
1 2 x2
dy
(6x 1 5)A1 2 x2 B 2 A3x2 1 5xB (22x)
5
dx
A1 2 x2 B 2
6x 1 5 2 6x3 2 5x2 1 6x3 1 10x2
5
A1 2 x2 B 2
5x2 1 6x 1 5
5
A1 2 x2 B 2
dy
Ax2 1 3B A2x 2 1B 2 Ax2 2 x 1 1B (2x)
f.
5
dx
Ax2 1 3B 2
2x3 1 6x 2 x2 2 3 2 2x3 1 2x2 2 2x
5
Ax2 1 3B 2
x2 1 4x 2 3
5
Ax2 1 3B 2
Calculus and Vectors Solutions Manual
3x 1 2
, x 5 23
x15
dy
(x 1 5)(3) 2 (3x 1 2)(1)
5
dx
(x 1 5)2
At x 5 23:
dy
(2)(3) 2 (27)(1)
5
dx
A2B 2
13
5
4
x3
b. y 5 2
,x51
x 19
dy
Ax2 1 9B A3x2 B 2 Ax3 B (2x)
5
dx
Ax2 1 9B 2
At x 5 1:
dy
(10)(3) 2 (1)(2)
5
dx
A10B 2
28
5
100
7
5
25
x2 2 25
c. y 5 2
,x52
x 1 25
dy
2xAx2 1 25B 2 Ax2 2 25B (2x)
5
dx
Ax2 1 25B 2
At x 5 2:
dy
4(29) 2 (221)(4)
5
dx
A29B 2
116 1 84
5
292
200
5
841
(x 1 1)(x 1 2)
,x54
d. y 5
(x 2 1)(x 2 2)
x2 1 3x 1 2
5 2
x 2 3x 1 2
dy
(2x 1 3)Ax2 2 3x 1 2B
5
dx
Ax 2 1B 2 Ax 2 2B 2
Ax2 1 3x 1 2B (2x 2 3)
2
Ax 2 1B 2 Ax 2 2B 2
At x 5 4:
dy
(11)(6) 2 (30)(5)
5
dx
(9)(4)
84
52
36
7
52
3
5. a. y 5
2-25
x3
x2 2 6
dy
3x2 Ax2 2 6B 2 x3 A2xB
5
dx
Ax2 2 6B 2
At (3, 9):
3(9)(3) 2 (27)(6)
dy
5
dx
A3B 2
5 9 2 18
5 29
The slope of the tangent to the curve at (3, 9) is 29.
3x
7. y 5
x24
dy
3(x 2 4) 2 3x
12
5
52
2
dx
Ax 2 4B
Ax 2 4B 2
12
Slope of the tangent is 2 25.
6. y 5
Therefore,
12
2 5
(x 2 4)
dy
12
25
x 2 4 5 5 or x 2 4 5 25
x 5 9 or x 5 21
Points are Q 9, 275 R and Q 21, 35 R .
5x 1 2
8. f(x) 5
x12
(x 1 2)(5) 2 (5x 1 2)(1)
f r(x) 5
Ax 1 2B 2
8
f r(x) 5
Ax 1 2B 2
Since Ax 1 2B 2 is positive or zero for all xPR,
8
.0
(x 1 2)2
for x 2 22. Therefore, tangents to
5x 1 2
the graph of f(x) 5 x 1 2 do not have a negative
slope.
2x2
9. a. y 5
,x24
x24
dy
(x 2 4)(4x) 2 A2x2 B (1)
5
dx
Ax 2 4B 2
2
4x 2 16x 2 2x2
5
(x 2 4)2
2
2x 2 16x
5
(x 2 4)2
2x(x 2 8)
5
(x 2 4)2
dy
Curve has horizontal tangents when dx 5 0, or
when x 5 0 or 8. At x 5 0:
0
y5
24
50
At x 5 8:
2-26
2(8)2
4
5 32
So the curve has horizontal tangents at the points
(0, 0) and (8, 32).
x2 2 1
b. y 5 2
x 1x22
(x 2 1)(x 1 1)
5
(x 1 2)(x 2 1)
x11
5
,x21
x12
dy
(x 1 2) 2 (x 1 1)
5
dx
Ax 1 2B 2
1
5
Ax 1 2B 2
y5
Curve has horizontal tangents when dx 5 0.
No value of x will produce a slope of 0, so there
are no horizontal tangents.
4t
10. p(t) 5 1000a1 1 2
b
t 1 50
4At 2 1 50B 2 4t(2t)
pr(t) 5 1000a
b
At 2 1 50B 2
1000A200 2 4t 2 B
5
At 2 1 50B 2
1000(196)
5 75.36
pr(1) 5
A51B 2
1000(184)
5 63.10
pr(2) 5
A54B 2
Population is growing at a rate of 75.4 bacteria per
hour at t 5 1 and at 63.1 bacteria per hour at t 5 2.
x2 2 1
11. y 5
3x
1
1
5 x 2 x21
3
3
dy
1
1 22
5 1 x
dx
3
3
1
1
5 1 2
3
3x
At x 5 2:
A2B 2 2 1
y5
3(2)
1
5
2
and
dy
1
1
5 1
dx
3
3A2B 2
Chapter 2: Derivatives
1
1
1
3
12
5
5
12
So the equation of the tangent to the curve at x 5 2 is:
1
5
y 2 5 (x 2 2), or 5x 2 12y 2 4 5 0.
2
12
10(6 2 t)
, 0 # t # 6, t 5 0,
12. a. s(t) 5
t13
s(0) 5 20
The boat is initially 20 m from the dock.
(t 1 3)(21) 2 (6 2 t)(1)
b. v(t) 5 sr(t) 5 10 c
d
At 1 3B 2
290
v(t) 5
At 1 3B 2
At t 5 0, v(0) 5 210, the boat is moving towards
the dock at a speed of 10 m> s. When s(t) 5 0, the
boat will be at the dock.
10(6 2 t)
5 0, t 5 6.
t13
290
10
v(6) 5 2 5 2
9
9
The speed of the boat when it bumps into the dock
is 109 m> s.
13. a. i. t 5 0
1 1 2(0)
r(0) 5
110
5 1 cm
1 1 2t
ii.
5 1.5
11t
1 1 2t 5 1.5(1 1 t)
1 1 2t 5 1.5 1 1.5t
0.5t 5 0.5
t 5 1s
(1 1 t)(2) 2 (1 1 2t)(1)
iii. rr(t) 5
A1 1 tB 2
2 1 2t 2 1 2 2t
5
A1 1 tB 2
1
5
A1 1 tB 2
1
rr(1.5) 5
A1 1 1B 2
1
5
4
5 0.25 cm> s
b. No, the radius will never reach 2 cm, because
y 5 2 is a horizontal asymptote of the graph of the
function. Therefore, the radius approaches but never
equals 2 cm.
5
Calculus and Vectors Solutions Manual
ax 1 b
(x 2 1)(x 2 4)
(x 2 1)(x 2 4)(a)
f r(x) 5
Ax 2 1B 2 Ax 2 4B 2
d
(ax 1 b) 3(x 2 1)(x 2 4)4
dx
2
Ax 2 1B 2 Ax 2 4B 2
(x 2 1)(x 2 4)(a)
5
Ax 2 1B 2 Ax 2 4B 2
(ax 1 b)3(x 2 1) 1 (x 2 4)4
2
Ax 2 1B 2 Ax 2 4B 2
2
(x 2 5x 1 4)(a) 2 (ax 1 b)(2x 2 5)
5
(x 2 1)2 (x 2 4)2
2
2ax 2 2bx 1 4a 1 5b
5
(x 2 1)2 (x 2 4)2
Since the point (2, 21) is on the graph (as it’s on
the tangent line) we know that
21 5 f(2)
2a 1 b
5
(1)(22)
2 5 2a 1 b
b 5 2 2 2a
Also, since the tangent line is horizontal at (2, 21),
we know that
0 5 f r(2)
2aA2B 2 2 2b(2) 1 4a 1 5b
5
A1B 2 A22B 2
b50
0 5 2 2 2a
a51
So we get
x
f(x) 5
(x 2 1)(x 2 4)
Since the tangent line is horizontal at the point
(2, 21), the equation of this tangent line is
y 2 (21) 5 0(x 2 2), or y 5 21
Here are the graphs of both f(x) and this horizontal
tangent line:
x
f (x) =
(x – 1) (x –4)
y
8
6
4
2
x
0
–2 –1
1 2 3 4 5
–2
y =–1
–4
–6
14. f(x) 5
2-27
A2t 2 1 7B (5) 2 (5t)(4t)
A2t 2 1 7B 2
10t 2 1 35 2 20t 2
5
A2t 2 1 7B 2
210t 2 1 35
5
A2t 2 1 7B 2
Set cr(t) 5 0 and solve for t.
210t 2 1 35
50
(2t 2 1 7)2
210t 2 1 35 5 0
10t 2 5 35
t 2 5 3.5
15. cr(t) 5
t 5 6"3.5
t 8 61.87
To two decimal places, t 5 21.87 or t 5 1.87,
because sr(t) 5 0 for these values. Reject the
negative root in this case because time is positive
(t $ 0). Therefore, the concentration reaches its
maximum value at t 5 1.87 hours.
16. When the object changes direction, its velocity
changes sign.
At 2 1 8B (1) 2 t(2t)
sr(t) 5
At 2 1 8B 2
2
t 1 8 2 2t 2
5
(t 2 1 8)2
2t 2 1 8
5 2
(t 1 8)2
solve for t when sr(t) 5 0.
2t 2 1 8
50
At 2 1 8B 2
2t 2 1 8 5 0
t2 5 8
t 5 6"8
t 8 62.83
To two decimal places, t 5 2.83 or t 5 22.83,
because sr(t) 5 0 for these values. Reject the
negative root because time is positive (t $ 0).
The object changes direction when t 5 2.83 s.
d
ax 1 b
17. f(x) 5
,x22
c
cx 1 d
(cx 1 d)(a) 2 (ax 1 b)(c)
f r(x) 5
Acx 1 dB 2
ad 2 bc
f r(x) 5
Acx 1 dB 2
For the tangents to the graph of y 5 f(x) to have
positive slopes, f r(x) . 0. (cx 1 d)2 is positive for
all xPR. ad 2 bc . 0 will ensure each tangent has
a positive slope.
2-28
2.5 The Derivatives of Composite
Functions, pp. 105–106
1. f(x) 5 !x, g(x) 5 x2 2 1
a. f(g(1)) 5 f(1 2 1)
5 f(0)
50
b. g(f(1)) 5 g(1)
50
c. g(f(0)) 5 g(0)
5021
5 21
d. f(g(24)) 5 f(16 2 1)
5 f(15)
5 !15
e. f(g(x)) 5 f Ax2 2 1B
5 "x2 2 1
f. g(f(x)) 5 gA !xB
5 A !xB 2 2 1
5x21
2. a. f(x) 5 x2, g(x) 5 !x
(f + g)(x) 5 f(g(x))
5 f A !xB
5 A !xB 2
5x
Domain 5 5x $ 06
(g + f)(x) 5 g(f(x))
5 gAx2 B
5 "x2
5 Zx Z
Domain 5 5xPR6
The composite functions are not equal for negative
x-values (as (f + g) is not defined for these x), but
are equal for non-negative x-values.
1
b.
f(x) 5 , g(x) 5 x2 1 1
x
(f + g)(x) 5 f(g(x))
5 f Ax2 1 1B
1
5 2
x 11
Domain 5 5xPR6
(g + f)(x) 5 g(f(x))
1
5 ga b
x
1 2
5a b 11
x
Chapter 2: Derivatives
1
11
x2
Domain 5 5x 2 06
The composite functions are not equal here. For
instance, (f + g)(1) 5 12 and (g + f )(1) 5 2.
1
c.
f(x) 5 , g(x) 5 !x 1 2
x
( f + g)(x) 5 f(g(x))
5
5 f( !x 1 2)
1
5
!x 1 2
Domain 5 5x . 226
(g + f)(x) 5 g(f(x))
1
5 ga b
x
1
12
Åx
The domain is all x such that
5
1
12$0
x
and x 2 0, or equivalently
Domain 5 5x # 2 12 or x . 06
The composite functions are not equal here. For
instance, ( f + g)(2) 5 12 and (g + f )(2) 5 # 52.
3. If f(x) and g(x) are two differentiable functions
of x, and
h(x) 5 (f + g)(x)
5 f(g(x))
is the composition of these two functions, then
hr(x) 5 f r(g(x)) ? gr(x)
This is known as the “chain rule” for differentiation of
composite functions. For example, if f(x) 5 x10 and
g(x) 5 x2 1 3x 1 5, then h(x) 5 Ax2 1 3x 1 5B 10,
and so
hr(x) 5 f r(g(x)) ? gr(x)
5 10Ax2 1 3x 1 5B 9 (2x 1 3)
2
As another example, if f(x) 5 x3 and
2
g(x) 5 x2 1 1, then h(x) 5 Ax2 1 1B 3,
and so
2
1
hr(x) 5 Ax2 1 1B 23 (2x)
3
4. a. f(x) 5 (2x 1 3)4
f r(x) 5 4A2x 1 3B 3 (2)
5 8A2x 1 3B 3
b. g(x) 5 Ax2 2 4B 3
gr(x) 5 3Ax2 2 4B 2 (2x)
5 6xAx2 2 4B 2
Calculus and Vectors Solutions Manual
c. h(x) 5 A2x2 1 3x 2 5B 4
hr(x) 5 4A2x2 1 3x 2 5B 3 (4x 1 3)
d. f(x) 5 Ap2 2 x2 B 3
f r(x) 5 3Ap2 2 x2 B 2 (22x)
5 26xAp2 2 x2 B 2
e. y 5 "x2 2 3
1
5 Ax2 2 3B 2
1
1
yr 5 Ax2 2 3B 2 (2x)
2
x
5
2
"x 2 3
1
f. f(x) 5 2
Ax 2 16B 5
5 Ax2 2 16B 25
f r(x) 5 25Ax2 2 16B 26 (2x)
210x
5 2
Ax 2 16B 6
2
5. a. y 5 2 3
x
5 22x23
dy
5 (22)(23)x24
dx
6
5 4
x
1
b. y 5
x11
5 (x 1 1)21
dy
5 (21)(x 1 1)22 (1)
dx
21
5
Ax 1 1B 2
1
c. y 5 2
x 24
5 Ax2 2 4B 21
dy
5 (21)Ax2 2 4B 22 (2x)
dx
22x
5 2
Ax 2 4B 2
3
5 3A9 2 x2 B 21
d. y 5
9 2 x2
dy
6x
5
dx
A9 2 x2 B 2
2-29
1
5x 1 x
5 A5x2 1 xB 21
e. y 5
2
dy
5 (21)A5x2 1 xB 22 (10x 1 1)
dx
10x 1 1
52 2
A5x 1 xB 2
1
f. y 5 2
Ax 1 x 1 1B 4
5 Ax2 1 x 1 1B 24
dy
5 (24)Ax2 1 x 1 1B 25 (2x 1 1)
dx
8x 1 4
52 2
Ax 1 x 1 1B 5
h5g+f
6.
5 g(f(x))
h(21) 5 g(f(21))
5 g(1)
5 24
h(x) 5 g(f(x))
hr(x) 5 gr(f(x))f r(x)
hr(21) 5 gr(f(21))f r(21)
5 gr(1)(25)
5 (27)(25)
5 35
1
7. f(x) 5 (x 2 3)2, g(x) 5 , h(x) 5 f(g(x)),
x
1
f r(x) 5 2(x 2 3), gr(x) 5 2 2
x
hr(x) 5 f r(g(x))gr(x)
1
1
5 f ra b a2 2 b
x
x
1
1
5 2a 2 3b a2 2 b
x
x
2 1
5 2 2 a 2 3b
x x
8. a. f(x) 5 Ax 1 4B 3 Ax 2 3B 6
d
3Ax 1 4B 34 ? Ax 2 3B 6
f r(x) 5
dx
d
1 Ax 1 4B 3 3Ax 2 3B 64
dx
5 3Ax 1 4B 2 Ax 2 3B 6
1 Ax 1 4B 3 (6)Ax 2 3B 5
5 Ax 1 4B 2 Ax 2 3B 5
3 33(x 2 3) 1 6(x 1 4)4
5 Ax 1 4B 2 Ax 2 3B 5 (9x 1 15)
2-30
b. y 5 Ax2 1 3B 3 Ax3 1 3B 2
dy
d 2
5
3Ax 1 3B 34 ? Ax3 1 3B 2
dx
dx
d 3
1 Ax2 1 3B 3 ?
3Ax 1 3B 24
dx
5 3Ax2 1 3B 2 (2x)Ax3 1 3B 2
1 Ax2 1 3B 3 (2)Ax3 1 3BA3x2 B
5 6xAx2 1 3B 2 Ax3 1 3B 3Ax3 1 3B 1 xAx2 1 3B4
5 6xAx2 1 3B 2 Ax3 1 3B A2x3 1 3x 1 3B
3x2 1 2x
c. y 5 2
x 11
dy
A6x 1 2B Ax2 1 1B 2 A3x2 1 2xBA2xB
5
dx
Ax2 1 1B 2
3
2
6x 1 2x 1 6x 1 2 2 6x3 2 4x2
5
Ax2 1 1B 2
2
22x 1 6x 1 2
5
Ax2 1 1B 2
d. h(x) 5 x3 A3x 2 5B 2
d 3
d
hr(x) 5
3x 4 ? A3x 2 5B 2 1 x3 3A3x 2 5B 24
dx
dx
5 3x2 A3x 2 5B 2 1 x3 (2)(3x 2 5)(3)
5 3x2 (3x 2 5)3(3x 2 5) 1 2x4
5 3x2 (3x 2 5)(5x 2 5)
5 15x2 (3x 2 5)(x 2 1)
4
e. y 5 x A1 2 4x2 B 3
dy
d 4
d
5
3x 4A1 2 4x2 B 3 1 x4 ?
3A1 2 4x2 B 34
dx
dx
dx
5 4x3 A1 2 4x2 B 3 1 x4 (3)A1 2 4x2 B 2 A28xB
5 4x3 A1 2 4x2 B 2 3A1 2 4x2 B 2 6x24
5 4x3 A1 2 4x2 B 2 A1 2 10x2 B
x2 2 3 4
b
f. y 5 a 2
x 13
dy
x2 2 3 3 d x2 2 3
5 4a 2
b
c
d
dx
x 1 2 dx x2 1 3
x2 2 3 3 Ax2 1 3B (2x) 2 Ax2 2 3B (2x)
5 4a 2
b ?
x 13
Ax2 1 3B 2
x2 2 3 3
12x
5 4a 2
b ? 2
x 13
Ax 1 3B 2
48xAx2 2 3B 3
5
Ax2 1 3B 5
1
2
9. a. s(t) 5 t3 (4t 2 5)3
1
1
5 t3 3(4t 2 5)24 3
1
5 3t(4t 2 5)24 3
1
5 3t A16t2 2 40t 1 25B4 3
1
5 A16t3 2 40t2 1 25tB 3, t 5 8
Chapter 2: Derivatives
1
2
sr(t) 5 A16t3 2 40t2 1 25tB 23
3
3 A48t2 2 80t 1 25B
A48t2 2 80t 1 25B
5
2
3A16t3 2 40t2 1 25tB 3
Rate of change at t 5 8:
(48(8)2 2 80(8) 1 25)
sr(8) 5
2
3(16(8)3 2 40(8)2 1 25(8))3
2457
5
972
91
5
36
1
t2p 3
b. s(t) 5 a
b , t 5 2p
t 2 6p
2
1 t 2 p 23 d t 2 p
sr(t) 5 a
b ? c
d
3 t 2 6p
dt t 2 6p
2
1 t 2 6p 3 (t 2 6p) 2 (t 2 p)
5 a
b ?
3 t2p
(t 2 6p)2
2
1 t 2 6p 3
25p
5 a
b ?
3 t2p
(t 2 6p)2
Rate of change at t 5 2p:
1
25p
2
sr(2p) 5 (24)3 ?
3
16p2
3
5"2
52
24p
y 5 2x6
10. y 5 A1 1 x3 B 2
dy
dy
5 2A1 1 x3 B (3x2 )
5 12x5
dx
dx
For the same slope,
6x2 A1 1 x3 B 5 12x5
6x2 1 6x5 5 12x5
6x2 2 6x5 5 0
6x2 Ax3 2 1B 5 0
x 5 0 or x 5 1.
Curves have the same slope at x 5 0 and x 5 1.
11. y 5 A3x 2 x2 B 22
dy
5 22A3x 2 x2 B 23 (3 2 2x)
dx
At x 5 2,
dy
5 2236 2 44 23 (3 2 4)
dx
5 2(2)23
1
5
4
The slope of the tangent line at x 5 2 is 41.
Calculus and Vectors Solutions Manual
y 5 Ax3 2 7B 5 at x 5 2
dy
5 5Ax3 2 7B 4 A3x2 B
dx
When x 5 2,
dy
5 5(1)4 (12)
dx
5 60
Slope of the tangent is 60.
Equation of the tangent at (2, 1) is
y 2 1 5 60(x 2 2)
60x 2 y 2 119 5 0.
13. a. y 5 3u2 2 5u 1 2
u 5 x2 2 1, x 5 2
u53
du
dy
5 6u 2 5,
5 2x
du
dx
dy
dy
du
5
3
dx
du
dx
5 (6u 2 5)(2x)
5 (18 2 5)(4)
5 13(4)
5 52
1
b. y 5 2u3 1 3u2, u 5 x 1 x2, x 5 1
dy
dy du
5
?
dx
du dx
1
5 (6u2 1 6u)a1 1
b
2!x
At x 5 1:
1
u 5 1 1 12
52
dy
1
b
5 (6(2)2 1 6(2))a1 1
dx
2 !1
3
5 36 3
2
5 54
c. y 5 uAu2 1 3B 3, u 5 (x 1 3)2, x 5 22
du
dy
5 Au2 1 3B 3 1 6u2 Au2 1 3B 2,
5 2(x 1 3)
du
dx
dy
dy du
5
5 373 1 6(4)2432(1)4
dx
du dx
5 439 3 2
5 878
d. y 5 u3 2 5Au3 2 7uB 2,
12.
u 5 "x
1
5 x2, x 5 4
dy
dy du
5
?
dx
du dx
2-31
1 1
5 333u2 2 10Au3 2 7uB A3u2 2 7B4 ? a x2 b
2
1
5 33u2 2 10Au3 2 7uB A3u2 2 7B4 ?
2"x
At x 5 4:
u 5 "4
52
dy
1
5 33(2)2 2 10( (2)3 2 7(2))(3(2)2 2 7)4
dx
2(2)
5 78
14. h(x) 5 f(g(x)), therefore
hr(x) 5 f r(g(x)) 3 gr(x)
f(u) 5 u2 2 1, g(2) 5 3, gr(2) 5 21
Now, hr(2) 5 f r(g(2)) 3 gr(2)
5 f r(3) 3 gr(2).
Since f(u) 5 u2 2 1, f r(u) 5 2u, and f r(3) 5 6,
hr(2) 5 6(21)
5 26.
t 2
15. V(t) 5 50 000a1 2 b
30
1
t
Vr(t) 5 50 000 c2a1 2 b a2 b d
30
30
10
1
Vr(10) 5 50 000 c2a1 2 b a2 b d
30
30
2
1
5 50 000 c2a b a2 b d
3
30
8 2222
At t 5 10 minutes, the water is flowing out of the
tank at a rate of 2222 L> min.
16. The velocity function is the derivative of the
position function.
1
s(t) 5 At 3 1 t 2 B 2
1
1
v(t) 5 sr(t) 5 At 3 1 t 2 B 22 A3t 2 1 2tB
2
3t 2 1 2t
5
2"t 3 1 t 2
3(3)2 1 2(3)
v(3) 5
2"33 1 32
27 1 6
5
2"36
33
5
12
5 2.75
The particle is moving at 2.75 m/s.
17. a. h(x) 5 p(x)q(x)r(x)
hr(x) 5 pr(x)q(x)r(x) 1 p(x)qr(x)r(x)
1 p(x)q(x)rr(x)
2-32
b. h(x) 5 x(2x 1 7)4 (x 2 1)2
Using the result from part a.,
hr(x) 5 (1)(2x 1 7)4 (x 2 1)2
1 x34(2x 1 7)3 (2)4 (x 2 1)2
1 x(2x 1 7)4 32(x 2 1)4
hr(23) 5 1(16) 1 (23)34(1)(2)4 (16)
1 (23)(1)32(24)4
5 16 2 384 1 24
5 2344
18. y 5 Ax2 1 x 2 2B 3 1 3
dy
5 3Ax2 1 x 2 2B 2 (2x 1 1)
dx
At the point (1, 3), x 5 1 and the slope of the
tangent will be 3(1 1 1 2 2)2 (2 1 1) 5 0.
Equation of the tangent at (1, 3) is y 2 3 5 0.
Solving this equation with the function, we have
Ax2 1 x 2 2B 3 1 3 5 3
(x 1 2)3 (x 2 1)3 5 0
x 5 22 or x 5 1
Since 22 and 1 are both triple roots, the line with
equation y 2 3 5 0 will be a tangent at both x 5 1
and x 5 22. Therefore, y 2 3 5 0 is also a tangent
at (22, 3).
x2 (1 2 x)3
19. y 5
(1 1 x)3
12x 3
5 x2 c a
bd
11x
dy
12x 3
12x 2
5 2xa
b 1 3x2 a
b
dx
11x
11x
2 (1 1 x) 2 (1 2 x)(1)
3 c
d
(1 1 x)2
12x 3
12x 2
22
5 2xa
b 1 3x2 a
b c
d
11x
11x
(1 1 x)2
12x 2 12x
3x
5 2xa
b c
2
d
11x
11x
(1 1 x)2
1 2 x 2 1 2 x2 2 3x
5 2xa
b c
d
11x
(1 1 x)2
2xAx2 1 3x 2 1B (1 2 x)2
52
(1 1 x)4
Review Exercise, pp. 110–113
1. To find the derivative f r(x), the limit
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
must be computed, provided it exists. If this limit
does not exist, then the derivative of f (x) does not
Chapter 2: Derivatives
exist at this particular value of x. As an alternative
to this limit, we could also find f r(x) from the
definition by computing the equivalent limit
f(z) 2 f(x)
.
z2x
zSx
f r(x) 5 lim
These two limits are seen to be equivalent by
substituting z 5 x 1 h.
2. a. y 5 2x2 2 5x
dy
(2(x 1 h)2 2 5(x 1 h)) 2 A2x2 2 5xB
5 lim
dx hS0
h
2A(x 1 h)2 2 x2 B 2 5h
5 lim
hS0
h
2((x 1 h) 2 x)((x 1 h) 1 x) 2 5h
5 lim
hS0
h
2h(2x 1 h) 2 5h
5 lim
hS0
h
5 lim (2(2x 1 h) 2 5)
hS0
5 4x 2 5
b. y 5 !x 2 6
!(x 1 h) 2 6 2 !x 2 6
dy
5 lim
dx hS0
h
!(x 1 h) 2 6 2 !x 2 6
5 lim c
hS0
h
!(x 1 h) 2 6 1 !x 2 6
d
3
!(x 1 h) 2 6 1 !x 2 6
( (x 1 h) 2 6) 2 (x 2 6)
5 lim
hS0 h( !(x 1 h) 2 6 1 !x 2 6)
1
5 lim
hS0 !(x 1 h) 2 6 1 !x 2 6
1
5
2 !x 2 6
x
c. y 5
42x
x1h
x
2
dy
4 2 (x 1 h)
42x
5 lim
dx hS0
h
(x 1 h)(4 2 x) 2 x(4 2 (x 1 h))
(4 2 (x 1 h))(4 2 x)
5 lim
hS0
h
4h
5 lim
hS0 h(4 2 (x 1 h))(4 2 x)
4
5 lim
hS0 (4 2 (x 1 h))(4 2 x)
4
5
(4 2 x)2
Calculus and Vectors Solutions Manual
3. a. y 5 x2 2 5x 1 4
dy
5 2x 2 5
dx
3
b. f(x) 5 x4
3 1
f r(x) 5 x24
4
3
5 14
4x
7
c. y 5 4
3x
7
5 x24
3
dy
228 25
5
x
dx
3
28
52 5
3x
1
d. y 5 2
x 15
5 Ax2 1 5B 21
dy
5 (21)Ax2 1 5B 22 ? (2x)
dx
2x
52 2
Ax 1 5B 2
3
e. y 5
A3 2 x2 B 2
5 3A3 2 x2 B 22
dy
5 (26)A3 2 x2 B 23 ? (22x)
dx
12x
5
A3 2 x2 B 3
f. y 5 "7x2 1 4x 1 1
1
5 A7x2 1 4x 1 1B 2
dy
1
1
5 A7x2 1 4x 1 1B 22 A14x 1 4B
dx
2
7x 1 2
5
"7x2 1 4x 1 1
2x3 2 1
4. a. f(x) 5
x2
1
5 2x 2 2
x
5 2x 2 x22
f r(x) 5 2 1 2x23
2
521 3
x
2-33
b. g(x) 5 !xAx3 2 xB
1
5 x2 Ax3 2 xB
7
3
5 x2 2 x2
7 5
3 1
gr(x) 5 x2 2 x2
2
2
!x 2
5
A7x 2 3B
2
x
c. y 5
3x 2 5
dy
(3x 2 5)(1) 2 (x)(3)
5
dx
(3x 2 5)2
5
52
(3x 2 5)2
1
d. y 5 (x 2 1)2 (x 1 1)
1
1
1
yr 5 (x 2 1)2 1 (x 1 1)a b (x 2 1)22
2
x11
5 !x 2 1 1
2!x 2 1
2x 2 2 1 x 1 1
5
2!x 2 1
3x 2 1
5
2!x 2 1
2
e. f(x) 5 A !x 1 2B 23
1
2
5 ( x2 1 2) 23
1
22 12
5 1
f r(x) 5
x 1 2) 23 # x2 2
(
3
2
1
52
5
3 !xA !x 1 2B 3
x2 1 5x 1 4
f. y 5
x14
(x 1 4)(x 1 1)
5
x14
5 x 1 1, x 2 24
dy
51
dx
5. a. y 5 x4 (2x 2 5)6
yr 5 x4 36(2x 2 5)5 (2)4 1 4x3 (2x 2 5)6
5 4x3 (2x 2 5)5 33x 1 (2x 2 5)4
5 4x3 (2x 2 5)5 (5x 2 5)
5 20x3 (2x 2 5)5 (x 2 1)
b. y 5 x"x2 1 1
1
1
yr 5 x c Ax2 1 1B 22 (2x)d 1 (1)"x2 1 1
2
x2
5
1 "x2 1 1
"x2 1 1
2-34
(2x 2 5)4
(x 1 1)3
(x 1 1)34(2x 2 5)3 (2)
yr 5
(x 1 1)6
3(2x 2 5)4 (x 1 1)2
2
(x 1 1)6
(x 1 1)2 (2x 2 5)3 38x 1 8 2 6x 1 154
5
(x 1 1)6
3
(2x 2 5) (2x 1 23)
yr 5
(x 1 1)4
10x 2 1 6
b 5 (10x 2 1)6 (3x 1 5)26
d. y 5 a
3x 1 5
yr 5 (10x 2 1)6 326(3x 1 5)27 (3)4
1 6(10x 2 1)5 (10)(3x 1 5)26
5 (10x 2 1)5 (3x 1 5)27 3x 2 18(10x 2 1)4
1 60(3x 1 5)
5 (10x 2 1)5 (3x 1 5)27
3 (2180x 1 18 1 180x 1 300)
318(10x 2 1)5
5
(3x 1 5)7
e. y 5 (x 2 2)3 Ax2 1 9B 4
yr 5 (x 2 2)3 C4Ax2 1 9B 3 (2x)D
1 3(x 2 2)2 (1)Ax2 1 9B 4
5 (x 2 2)2 Ax2 1 9B 3 C8x(x 2 2) 1 3Ax2 1 9B D
5 (x 2 2)2 Ax2 1 9B 3 A11x2 2 16x 1 27B
f. y 5 A1 2 x2 B 3 (6 1 2x)23
1 2 x2 3
5a
b
6 1 2x
1 2 x2 2
b
yr 5 3a
6 1 2x
c. y 5
3 c
(6 1 2x)(22x) 2 A1 2 x2 B (2)
d
(6 1 2x)2
3A1 2 x2 B 2 A212x 2 4x2 2 2 1 2x2 B
(6 1 2x)4
2 2
3A1 2 x B A2x2 1 12x 1 2B
52
(6 1 2x)4
3A1 2 x2 B 2 Ax2 1 6x 1 1B
52
8(3 2 x)4
2
6. a. g(x) 5 f(x )
gr(x) 5 f(x2 ) 3 2x
b. h(x) 5 2xf(x)
hr(x) 5 2xf r(x) 1 2f(x)
18
7. a. y 5 5u2 1 3u 2 1, u 5 2
x 15
x52
u52
5
Chapter 2: Derivatives
dy
5 10u 1 3
du
du
36x
52 2
dx
Ax 1 5B 2
When x 5 2,
du
72
8
52 52
dx
81
9
When u 5 2,
dy
5 20 1 3
du
5 23
dy
8
5 23a2 b
dx
9
184
52
9
!x 1 x
u14
,u5
,
b. y 5
u24
10
x54
3
u5
5
dy
(u 2 4) 2 (u 1 4)
5
du
(u 2 4)2
du
1 1 1
5 a x22 1 1b
dx
10 2
When x 5 4,
du
8
1 5
52
5 a b
(u 2 4)2 dx
10 4
1
5
8
3
When u 5 ,
5
dy
8
52
du
3
20 2
a 2 b
5
5
8(25)
52
(217)2
When x 5 4,
dy
8(25)
1
5
3
dx
172
8
25
5
289
c. y 5 f("x2 1 9), f r(5) 5 22, x 5 4
dy
1
1
5 f r("x2 1 9) 3 Ax2 1 9B 22 (2x)
dx
2
dy
1 1
5 f r(5) ? ? ? 8
dx
2 5
Calculus and Vectors Solutions Manual
5 22 ?
4
5
8
5
2
8. f(x) 5 A9 2 x2 B 3
2
1
f r(x) 5 A9 2 x2 B 23 (22x)
3
24x
5
1
3(9 2 x2 )3
2
f r(1) 5 2
3
The slope of the tangent line at (1, 4) is 2 23.
9. y 5 2x3 1 6x2
yr 5 23x2 1 12x
23x2 1 12x 5 215
23x2 1 12x 5 212
2
x2 2 4x 2 5 5 0
x 2 4x 2 4 5 0
52
4 6 !16 1 16
2
4 6 4!2
5
2
x 5 2 6 2 !2
x5
(x 2 5)(x 1 1) 5 0
x 5 5, x 5 21
10. a. i. y 5 Ax2 2 4B 5
yr 5 5Ax2 2 4B 4 (2x)
Horizontal tangent,
10xAx2 2 4B 4 5 0
x 5 0, x 5 62
ii. y 5 Ax3 2 xB 2
yr 5 2Ax3 2 xB A3x2 2 1B
Horizontal tangent,
2x(x2 2 1)(3x2 2 1) 5 0
!3
x 5 0, x 5 61, x 5 6
.
3
2-35
b. i.
ii.
11. a. y 5 Ax2 1 5x 1 2B 4 at (0, 16)
yr 5 4Ax2 1 5x 1 2B 3 (2x 1 5)
At x 5 0,
yr 5 4(2)3 (5)
5 160
Equation of the tangent at (0, 16) is
y 2 16 5 160(x 2 0)
y 5 160x 1 16
or 160x 2 y 1 16 5 0
b. y 5 A3x22 2 2x3 B 5 at (1, 1)
yr 5 5A3x22 2 2x3 B 4 A26x23 2 6x2 B
At x 5 1,
yr 5 5(1)4 (26 2 6)
5 260
Equation of the tangent at (1, 1) is
y 2 1 5 260(x 2 1)
60x 1 y 2 61 5 0.
12. y 5 3x2 2 7x 1 5
dy
5 6x 2 7
dx
2-36
Slope of x 1 5y 2 10 5 0 is 2 15.
Since perpendicular, 6x 2 7 5 5
x52
y 5 3(4) 2 14 1 5
5 3.
Equation of the tangent at (2, 3) is
y 2 3 5 5(x 2 2)
5x 2 y 2 7 5 0.
13. y 5 8x 1 b is tangent to y 5 2x2
dy
5 4x
dx
Slope of the tangent is 8, therefore 4x 5 8, x 5 2.
Point of tangency is (2, 8).
Therefore, 8 5 16 1 b, b 5 28.
Or 8x 1 b 5 2x2
2x2 2 8x 2 b 5 0
8 6 !64 1 8b
x5
.
2(2)
For tangents, the roots are equal, therefore
64 1 8b 5 0, b 5 28.
Point of tangency is (2, 8), b 5 28.
14. a.
b.
The equation of the tangent is y 5 0.
The equation of the tangent is y 5 6.36.
The equation of the tangent is y 5 26.36.
Chapter 2: Derivatives
c.
Ax2 2 6B A3x2 B 2 x3 (2x)
Ax2 2 6B 2
x4 2 18x2
5 2
Ax 2 6B 2
f r(x) 5
x4 2 18x2
50
Ax2 2 6B 2
x2 Ax2 2 18B 5 0
x2 5 0 or x2 2 18 5 0
x50
x 5 63 !2
The coordinates of the points where the slope is 0
9 !2
are (0, 0), Q 3 !2, 9 !2
2 R , and Q 23!2, 2 2 R .
d. Substitute into the expression for f r(x) from
part b.
16 2 72
f r(2) 5
(22)2
256
5
4
5 214
5
2
15. a. f(x) 5 2x3 2 5x3
5 2
2 1
f r(x) 5 2 3 x3 2 5 3 x3
3
3
10 23
10
5 x 2 13
3
3x
2
f(x) 5 0 6 x3 32x 2 54 5 0
5
x 5 0 or x 5
2
y 5 f(x) crosses the x-axis at x 5 52, and
10 x 2 1
f r(x) 5 a 13 b
3
x
5
10
3
1
f ra b 5
3 3 5 13
2
3
2
Q2 R
553
3
!
2
2
1
5 53 3 23
3
!5
1
5 (25 3 2)3
3
5!
50
b. To find a, let f(x) 5 0.
10 23
10
x 2 13 5 0
3
3x
Calculus and Vectors Solutions Manual
30x 5 30
x51
Therefore a 5 1.
16. M 5 0.1t2 2 0.001t3
a. When t 5 10,
M 5 0.1(100) 2 0.001(1000)
59
When t 5 15,
M 5 0.1(225) 2 0.001(3375)
5 19.125
One cannot memorize partial words, so 19 words
are memorized after 15 minutes.
b. Mr 5 0.2t 2 0.003t2
When t 5 10,
Mr 5 0.2(10) 2 0.003(100)
5 1.7
The number of words memorized is increasing by
1.7 words> min.
When t 5 15,
Mr 5 0.2(15) 2 0.003(225)
5 2.325
The number of words memorized is increasing by
2.325 words> min.
30
17. a. N(t) 5 20 2
"9 1 t2
30t
3
A9 1 t2 B 2
b. No, according to this model, the cashier never
stops improving. Since t . 0, the derivative is always
positive, meaning that the rate of change in the
cashier’s productivity is always increasing. However,
these increases must be small, since, according to the
model, the cashier’s productivity can never exceed 20.
1
18. C(x) 5 x3 1 40x 1 700
3
a. Cr(x) 5 x2 1 40
b. Cr(x) 5 76
x2 1 40 5 76
x2 5 36
x56
Production level is 6 gloves> week.
x2
2
19. R(x) 5 750x 2 2 x3
6
3
a. Marginal Revenue
x
Rr(x) 5 750 2 2 2x2
3
Nr(t) 5
2-37
10
2 2(100)
3
5 $546.67
20
20. D(p) 5
,p.1
!p 2 1
1
3
Dr(p) 5 20a2 b (p 2 1)22
2
10
52
3
(p 2 1)2
10
10
Dr(5) 5
52
3
8
"4
5
52
4
Slope of demand curve at (5, 10) is 2 54.
21. B(x) 5 20.2x2 1 500, 0 # x # 40
a. B(0) 5 20.2(0)2 1 500 5 500
B(30) 5 20.2(30)2 1 500 5 320
b. Br(x) 5 20.4x
Br(0) 5 20.4(0) 5 0
Br(30) 5 20.4(30) 5 212
c. B(0) 5 blood sugar level with no insulin
B(30) 5 blood sugar level with 30 mg of insulin
Br(0) 5 rate of change in blood sugar level
with no insulin
Br(30) 5 rate of change in blood sugar level
with 30 mg of insulin
d. Br(50) 5 20.4(50) 5 220
B(50) 5 20.2(50)2 1 500 5 0
Br(50) 5 220 means that the patient’s blood sugar
level is decreasing at 20 units per mg of insulin 1 h
after 50 mg of insulin is injected.
B(50) 5 0 means that the patient’s blood sugar level
is zero 1 h after 50 mg of insulin is injected. These
values are not logical because a person’s blood sugar
level can never reach zero and continue to decrease.
3x
22. a. f(x) 5
1 2 x2
3x
5
(1 2 x)(1 1 x)
f(x) is not differentiable at x 5 1 because it is not
defined there (vertical asymptote at x 5 1).
x21
b. g(x) 5 2
x 1 5x 2 6
x21
5
(x 1 6)(x 2 1)
1
5
for x 2 1
(x 1 6)
g(x) is not differentiable at x 5 1 because it is not
defined there (hole at x 5 1).
b. Rr(10) 5 750 2
2-38
3
c. h(x) 5 "
(x 2 2)2
The graph has a cusp at (2, 0) but it is differentiable
at x 5 1.
d. m(x) 5 Z3x 2 3 Z 2 1.
The graph has a corner at x 5 1, so m(x) is not
differentiable at x 5 1.
3
23. a. f(x) 5 2
4x 2 x
3
5
x(4x 2 1)
f(x) is not defined at x 5 0 and x 5 0.25. The
graph has vertical asymptotes at x 5 0 and
x 5 0.25. Therefore, f(x) is not differentiable at
x 5 0 and x 5 0.25.
x2 2 x 2 6
b. f(x) 5
x2 2 9
(x 2 3)(x 1 2)
5
(x 2 3)(x 1 3)
(x 1 2)
5
for x 2 3
(x 1 3)
f(x) is not defined at x 5 3 and x 5 23. At
x 5 23, the graph as a vertical symptote and at
x 5 3 it has a hole. Therefore, f(x) is not
differentiable at x 5 3 and x 5 23.
c. f(x) 5 "x2 2 7x 1 6
5 !(x 2 6)(x 2 1)
f(x) is not defined for 1 , x , 6. Therefore,
f(x) is not differentiable for 1 , x , 6.
(t 1 1)(25) 2 (25t)(t)
24. pr(t) 5
(t 1 1)2
25t 1 25 2 25t
5
(t 1 1)2
25
5
(t 1 1)2
25. Answers may vary. For example,
f(x) 5 2x 1 3
1
y5
2x 1 3
(2x 1 3)(0) 2 (1)(2)
yr 5
(2x 1 3)2
2
52
(2x 1 3)2
f(x) 5 5x 1 10
1
y5
5x 1 10
(5x 1 10)(0) 2 (1)(5)
yr 5
(5x 1 10)2
Chapter 2: Derivatives
52
5
(5x 1 10)2
1
Rule: If f(x) 5 ax 1 b and y 5 f (x), then
yr 5
2a
(ax 1 b)2
1
1
1
yr 5 lim c
2
d
hS0 h a(x 1 h) 1 b
ax 1 b
1 ax 1 b 2 3a(x 1 h)b4
d
5 lim c
hS0 h 3a(x 1 h) 1 b4 (ax 1 h)
1 ax 1 b 2 ax 2 ah 2 b
5 lim c
d
hS0 h 3a(x 1 h) 1 b4 (ax 1 b)
1
2ah
5 lim c
d
hS0 h 3a(x 1 h) 1 b4 (ax 1 b)
2a
5 lim
hS0 3a(x 1 h) 1 b4 (ax 1 b)
2a
5
(ax 1 b)2
26. a. Let y 5 f(x)
(2x 2 3)2 1 5
y5
2x 2 3
Let u 5 2x 2 3.
u2 1 5
Then y 5
.
u
y 5 u 1 5u21
dy
b. f r(x) 5
dx
dy
dy
du
5
3
dx
du
dx
5 (1 2 5u22 )(2)
5 2(1 2 5(2x 2 3)22 )
27. g(x) 5 !2x 2 3 1 5(2x 2 3)
a. Let y 5 g(x).
y 5 !2x 2 3 1 5(2x 2 3)
Let u 5 2x 2 3.
Then y 5 !u 1 5u.
dy
dy
du
b. gr(x) 5
5
3
dx
du
dx
1 212
5 a u 1 5b (2)
2
212
5 u 1 10
1
5 (2x 2 3)22 1 10
Calculus and Vectors Solutions Manual
28. a. f(x) 5 (2x 2 5)3 (3x2 1 4)5
f r(x) 5 (2x 2 5)3 (5)A3x2 1 4B 4 (6x)
1 A3x2 1 4B 5 (3)(2x 2 5)2 (2)
5 30x(2x 2 5)3 A3x2 1 4B 4
1 6(3x2 1 4)5 (2x 2 5)2
5 6(2x 2 5)2 A3x2 1 4B 4
3 C5x(2x 2 5) 1 A3x2 1 4B D
5 6(2x 2 5)2 A3x2 1 4B 4
3 A10x2 2 25x 1 3x2 1 4B
5 6(2x 2 5)2 (3x2 1 4)4
3 (13x2 2 25x 1 4)
3
b. g(x) 5 (8x )(4x2 1 2x 2 3)5
gr(x) 5 (8x3 )(5)(4x2 1 2x 2 3)4 (8x 1 2)
1 (4x2 1 2x 2 3)5 (24x2 )
5 40x3 (4x2 1 2x 2 3)4 (8x 1 2)
1 24x2 (4x2 1 2x 2 3)5
5 8x2 (4x2 1 2x 2 3)4 35x(8x 1 2)
1 3(4x2 1 2x 2 3)4
5 8x2 (4x2 1 2x 2 3)4
(40x2 1 10x 1 12x2 1 6x 2 9)
5 8x2 (4x2 1 2x 2 3)4 (52x2 1 16x 2 9)
c. y 5 (5 1 x)2 (4 2 7x3 )6
yr 5 (5 1 x)2 (6)(4 2 7x3 )5 (221x2 )
1 (4 2 7x3 )6 (2)(5 1 x)
5 2126x2 (5 1 x)2 (4 2 7x3 )5
1 2(5 1 x)(4 2 7x3 )6
5 2(5 1 x)(4 2 7x3 )5 3263x2 (5 1 x)
1 4 2 7x34
5 2(5 1 x)(4 2 7x3 )5 (4 2 315x2 2 70x3 )
6x 2 1
d. h(x) 5
(3x 1 5)4
(3x 1 5)4 (6) 2 (6x 2 1)(4)(3x 1 5)3 (3)
hr(x) 5
((3x 1 5)4 )2
3
6(3x 1 5) 3(3x 1 5) 2 2(6x 2 1)4
5
(3x 1 5)8
6(29x 1 7)
5
(3x 1 5)5
(2x2 2 5)3
e. y 5
(x 1 8)2
dy
(x 1 8)2 (3)(2x2 2 5)2 (4x)
5
dx
A(x 1 8)2 B 2
(2x2 2 5)3 (2)(x 1 8)
2
A(x 1 8)2 B 2
2(x 1 8)(2x2 2 5)2 36x(x 1 8) 2 (2x2 2 5)4
(x 1 8)4
2(2x2 2 5)2 (4x2 1 48x 1 5)
5
(x 1 8)3
5
2-39
f. f(x) 5
23x4
"4x 2 8
23x4
5
1
(4x 2 8)2
1
(4x 2 8)2 (212x3 )
f r(x) 5
1
A(4x 2 8)2 B 2
1
1
(23x4 )a b (4x 2 8) 2 2 (4)
2
2
1
A(4x 2 8)2 B 2
26x3 (4x 2 8)22 32(4x 2 8) 2 x4
5
4x 2 8
26x3 (7x 2 16)
5
3
(4x 2 8)2
23x3 (7x 2 16)
5
3
(4x 2 8)2
2x 1 5 4
g. g(x) 5 a
b
6 2 x2
1
2x 1 5 3
b
6 2 x2
(6 2 x2 )(2) 2 (2x 1 5)(22x)
b
3a
(6 2 x2 )2
2x 1 5 3 2(6 1 x2 1 5x)
5 4a
b a
b
6 2 x2
(6 2 x2 )2
2x 1 5 3 (x 1 2)(x 1 3)
5 8a
b a
b
6 2 x2
(6 2 x2 )2
3
1
h. y 5 c
2 3d
(4x 1 x )
In (1),
4a 2 8a 5 16
24a 5 16
a 5 24
Using (1),
b 5 28(24) 5 32
a 5 24, b 5 32, c 5 0, f(x) 5 24x2 1 32x
30. a. A(t) 5 2t3 1 5t 1 750
Ar(t) 5 23t2 1 5
b. Ar(5) 5 23(25) 1 5
5 270
At 5 h, the number of ants living in the colony is
decreasing by 7000 ants> h.
c. A(0) 5 750, so there were 750 (100) or
75 000 ants living in the colony before it was
treated with insecticide.
d. Determine t so that A(t) 5 0. 2t3 1 5t 1 750
cannot easily be factored, so find the zeros by using
a graphing calculator.
gr(x) 5 4a
5 (4x 1 x2 )29
dy
5 29(4x 1 x2 )210 (4 1 2x)
dx
29. f(x) 5 ax2 1 bx 1 c,
It is given that (0, 0) and (8, 0) are on the curve,
and f r(2) 5 16.
Calculate f r(x) 5 2ax 1 b.
Then,
16 5 2a(2) 1 b
(1)
4a 1 b 5 16
Since (0, 0) is on the curve,
0 5 a(0)2 1 b(0) 1 c
c50
Since (8, 0) is on the curve,
0 5 a(8)2 1 b(8) 1 c
0 5 64a 1 8b 1 0
(2)
8a 1 b 5 0
Solve (1) and (2):
From (2), b 5 28a
(1)
2-40
All of the ants have been killed after about 9.27 h.
Chapter 2 Test, p. 114
1. You need to use the chain rule when the derivative
for a given function cannot be found using the sum,
difference, product, or quotient rules or when writing
the function in a form that would allow the use of
these rules is tedious. The chain rule is used when
a given function is a composition of two or more
functions.
2. f is the blue graph (it's a cubic). f' is the red graph
(it is quadratic). The derivative of a polynomial
function has degree one less than the derivative of
the function. Since the red graph is a quadratic
(degree 2) and the blue graph is cubic (degree 3),
the blue graph is f and the red graph is f r.
f(x 1 h) 2 f(x)
3. f(x) 5 lim
hS0
h
x 1 h 2 (x 1 h)2 2 (x 2 x2 )
5 lim
hS0
h
x 1 h 2 (x2 1 2hx 1 h2 ) 2 x 1 x2
5 lim
hS0
h
h 2 2hx 2 h2
5 lim
hS0
h
Chapter 2: Derivatives
h(1 2 2x 2 h)
hS0
h
5 lim (1 2 2x 2 h)
5 lim
hS0
5 1 2 2x
d
Therefore,
(x 2 x2 ) 5 1 2 2x.
dx
1
4. a. y 5 x3 2 3x25 1 4p
3
dy
5 x2 1 15x26
dx
b. y 5 6(2x 2 9)5
dy
5 30(2x 2 9)4 (2)
dx
5 60(2x 2 9)4
2
x
3
1
1 6"
x
c. y 5
"x
"3
1
1
1
5 2x22 1
x 1 6x3
"3
dy
1
3
2
5 2x22 1
1 2x23
dx
"3
x2 1 6 5
b
d. y 5 a
3x 1 4
dy
x2 1 6 4 2x(3x 1 4) 2 (x2 1 6)3
5 5a
b
dx
3x 1 4
(3x 1 4)2
5(x2 1 6)4 (3x2 1 8x 2 18)
5
(3x 1 4)6
3
6x2 2 7
e. y 5 x2 "
dy
1
1
2
5 2x(6x2 2 7)3 1 x2 (6x2 2 7)23 (12x)
dx
3
2
5 2x(6x2 2 7)23 ( (6x2 2 7) 1 2x2 )
2
5 2x(6x2 2 7)23 (8x2 2 7)
4x5 2 5x4 1 6x 2 2
f. y 5
x4
5 4x 2 5 1 6x23 2 2x24
dy
5 4 2 18x24 1 8x25
dx
4x5 2 18x 1 8
5
x5
5. y 5 (x2 1 3x 2 2)(7 2 3x)
dy
5 (2x 1 3)(7 2 3x) 1 (x2 1 3x 2 2)(23)
dx
At (1, 8),
dy
5 (5)(4) 1 (2)(23)
dx
5 14.
The slope of the tangent to
y 5 (x2 1 3x 2 2)(7 2 3x) at (1, 8) is 14.
Calculus and Vectors Solutions Manual
6. y 5 3u2 1 2u
dy
5 6u 1 2
du
u 5 "x2 1 5
du
1
1
5 (x2 1 5)22 2x
dy
2
dy
x
5 (6u 1 2)a
b
2
dx
"x 1 5
At x 5 22, u 5 3.
dy
2
5 (20)a2 b
dx
3
40
52
3
7. y 5 (3x22 2 2x3 )5
dy
5 5(3x22 2 2x3 )4 (26x23 2 6x2 )
dx
At (1, 1),
dy
5 5(1)4 (26 2 6)
dx
5 260.
Equation of tangent line at (1, 1) is y 2 1 5 60(x 2 1)
y 2 1 5 260x 1 60
60x 1 y 2 61 5 0.
1
8. P(t) 5 (t 4 1 3)3
1 3
1
Pr(t) 5 3(t 4 1 3)2 a t24 b
4
1
1
3
Pr(16) 5 3(16 4 1 3)2 a 3 1624 b
4
1
1
5 3(2 1 3)2 a 3 b
4
8
75
5
32
The amount of pollution is increasing at a rate of
75
32 ppm>year.
9. y 5 x4
dy
5 4x3
dx
1
2 5 4x3
16
Normal line has a slope of 16. Therefore,
dy
1
52 .
dx
16
1
x3 5 2
64
2-41
1
4
1
y5
256
Therefore, y 5 x4 has a normal line with a slope of
1
16 at Q 2 14, 256
R.
x52
10. y 5 x3 2 x2 2 x 1 1
dy
5 3x2 2 2x 2 1
dx
dy
For a horizontal tangent line, dx 5 0.
3x2 2 2x 2 1 5 0
(3x 1 1)(x 2 1) 5 0
1
or
x52
x51
3
1
1
1
y52 2 1 11
y51212111
27
9
3
50
21 2 3 1 9 1 27
5
27
32
5
27
The required points are Q 2 13, 32
27 ), (1, 0 R .
2-42
11. y 5 x2 1 ax 1 b
dy
5 2x 1 a
dx
y 5 x3
dy
5 3x2
dx
Since the parabola and cubic function are tangent at
(1, 1), then 2x 1 a 5 3x2.
At (1, 1) 2(1) 1 a 5 3(1)2
a 5 1.
Since (1, 1) is on the graph of
y 5 x2 1 x 1 b, 1 5 12 1 1 1 b
b 5 21.
The required values are 1 and 21 for a and b,
respectively.
Chapter 2: Derivatives
CHAPTER 2
Derivatives
Review of Prerequisite Skills,
pp. 62–63
1. a. a5 3 a3 5 a513
5 a8
b. A22a2 B 3 5 (22)3 Aa2 B 3
5 28Aa233 B
5 28a6
7
4p 3 6p9
24p719
c.
5
12p15
12p15
5 2p16215
5 2p
4 25
26 22
d. Aa b B Aa b B 5 Aa426 B Ab2522 B
5 a22b27
1
5 2 7
ab
e. A3e6 B A2e3 B 4 5 (3)Ae6 B A24 B Ae3 B 4
5 (3)A24 BAe6 B Ae334 B
5 (3)(16)Ae6112 B
5 48e18
24
3
A3a B C2a (2b)3 D
(3)(2)A21B 3 Aa2413 B Ab3 B
f.
5
12a5b2
12a5b2
2125
B Ab322 B
26Aa
5
12
26
21Aa B AbB
5
2
b
52 6
2a
1
2
1
2
2. a. Ax2 B Ax3 B 5 x2 1 3
7
5 x6
2
2
2
b. A8x6 B 3 5 83x63 3
5 4x4
1
1
3
( a2)( a3)
"a "
a
c.
5
1
a2
"a
1
5 a3
3. A perpendicular line will have a slope that is the
negative reciprocal of the slope of the given line:
21
a. slope 5 2
3
52
3
2
Calculus and Vectors Solutions Manual
21
2 12
52
21
c. slope 5 5
b. slope 5
3
3
5
21
d. slope 5
21
51
52
24 2 (22)
23 2 9
22
5
212
1
5
6
The equation of the desired line is therefore
y 1 4 5 16 (x 1 3) or x 2 6y 2 21 5 0.
b. The equation 3x 2 2y 5 5 can be rewritten as
2y 5 3x 2 5 or y 5 32x 2 52, which has slope 32.
The equation of the desired line is therefore
y 1 5 5 32 (x 1 2) or 3x 2 2y 2 4 5 0.
c. The line perpendicular to y 5 34 x 2 6 will have
21
slope m 5 3 5 2 43. The equation of the desired line
4. a. This line has slope m 5
4
is therefore y 1 3 5 2 43 (x 2 4) or 4x 1 3y 2 7 5 0.
5. a. (x 2 3y)(2x 1 y) 5 2x2 1 xy 2 6xy 2 3y2
5 2x2 2 5xy 2 3y2
2
b. (x 2 2)(x 2 3x 1 4)
5 x3 2 3x2 1 4x 2 2x2 1 6x 2 8
5 x3 2 5x2 1 10x 2 8
c. (6x 2 3)(2x 1 7) 5 12x2 1 42x 2 6x 2 21
5 12x2 1 36x 2 21
d. 2(x 1 y) 2 5(3x 2 8y) 5 2x 1 2y 2 15x 1 40y
5 213x 1 42y
2
2
e. (2x 2 3y) 1 (5x 1 y)
5 4x2 2 12xy 1 9y2 1 25x2 1 10xy 1 y2
5 29x2 2 2xy 1 10y2
f. 3x(2x 2 y)2 2 x(5x 2 y)(5x 1 y)
5 3x(4x2 2 4xy 1 y2 ) 2 x(25x2 2 y2 )
5 12x3 2 12x2y 1 3xy2 2 25x3 1 xy2
5 213x3 2 12x2y 1 4xy2
2-1
3x(x 1 2)
5x3
15x4 (x 1 2)
3
5
x2
2x(x 1 2)
2x3 (x 1 2)
15
5 x423
2
15
5 x
2
x 2 0, 22
y
(y 2 5)2
3
b.
(y 1 2)(y 2 5)
4y3
y(y 2 5)(y 2 5)
5 3
4y (y 1 2)(y 2 5)
y25
5 2
4y (y 1 2)
y 2 22, 0, 5
4
9
4
2(h 1 k)
c.
4
5
3
h 1 k 2(h 1 k)
h1k
9
8(h 1 k)
5
9(h 1 k)
8
5
9
h 2 2k
(x 1 y)(x 2 y) (x 1 y)3
d.
4
5(x 2 y)
10
(x 1 y)(x 2 y)
10
5
3
5(x 2 y)
(x 1 y)3
10(x 1 y)(x 2 y)
5
5(x 2 y)(x 1 y)3
2
5
(x 1 y)2
x 2 2y, 1y
x27
5x
(x 2 7)(x 2 1)
(5x)(2x)
1
5
1
e.
2x
x21
2x(x 2 1)
2x(x 2 1)
x2 2 7x 2 x 1 7 1 10x2
5
2x(x 2 1)
11x2 2 8x 1 7
5
2x(x 2 1)
x 2 0, 1
x12
x11
2
f.
x22
x13
(x 1 2)(x 2 2)
(x 1 1)(x 1 3)
2
5
(x 2 2)(x 1 3)
(x 1 3)(x 2 2)
x2 1 x 1 3x 1 3 2 x2 1 4
5
(x 1 3)(x 2 2)
4x 1 7
5
(x 1 3)(x 2 2)
x 2 23, 2
7. a. 4k2 2 9 5 (2k 1 3)(2k 2 3)
6. a.
2-2
b. x2 1 4x 2 32 5 x2 1 8x 2 4x 2 32
5 x(x 1 8) 2 4(x 1 8)
5 (x 2 4)(x 1 8)
c. 3a2 2 4a 2 7 5 3a2 2 7a 1 3a 2 7
5 a(3a 2 7) 1 1(3a 2 7)
5 (a 1 1)(3a 2 7)
d. x4 2 1 5 (x2 1 1)(x2 2 1)
5 (x2 1 1)(x 1 1)(x 2 1)
3
3
e. x 2 y 5 (x 2 y)(x2 1 xy 1 y2 )
f. r4 2 5r 2 1 4 5 r4 2 4r 2 2 r 2 1 4
5 r 2 (r 2 2 4) 2 1(r 2 2 4)
5 (r2 2 1)(r2 2 4)
5 (r 1 1)(r 2 1)(r 1 2)(r 2 2)
8. a. Letting f(a) 5 a3 2 b3, f(b) 5 b3 2 b3
50
So b is a root of f (a), and so by the factor theorem,
a 2 b is a factor of a3 2 b3. Polynomial long
division provides the other factor:
a2 1 ab 1 b2
a 2 bq a3 1 0a2 1 0a 2 b3
a3 2 a2b
a2b 1 0a 2 b3
a2b 2 ab2
ab2 2 b3
ab2 2 b3
0
So a3 2 b3 5 (a 2 b)(a2 1 ab 1 b2 ).
b. Using long division or recognizing a pattern from
the work in part a.:
a5 2 b5 5 (a 2 b)(a4 1 a3b 1 a2b2 1 ab3 1 b4 ).
c. Using long division or recognizing a pattern from
the work in part a.: a7 2 b7
5 (a 2 b)(a6 1 a5b 1 a4b2 1 a3b3
1 a2b4 1 ab5 1 b6 ).
d. Using the pattern from the previous parts:
an 2 bn 5 (a 2 b)(an21 1 an22b 1 an23b2 1 c
1 a2bn23 1 abn22 1 bn21 ).
9. a. f(2) 5 22(24 ) 1 3(22 ) 1 7 2 2(2)
5 232 1 12 1 7 2 4
5 217
b. f(21) 5 22(21)4 1 3(21)2 1 7 2 2(21)
5 22 1 3 1 7 1 2
5 10
1 4
1 2
1
1
c. f a b 5 22a b 1 3a b 1 7 2 2a b
2
2
2
2
3
1
52 1 1721
8
4
53
5
8
Chapter 2: Derivatives
1
d. f(20.25) 5 f a2 b
4
1 4
1 2
1
5 2a2 b 1 3a2 b 1 7 2 2a2 b
4
4
4
1
3
1
52
1
171
128
16
2
983
5
128
8 7.68
3
3 !2
5
10. a.
!2
( !2)( !2)
3 !2
5
2
4 2 !2
(4 2 !2)( !3)
5
b.
!3
( !3)(!3)
4!3 2 !6
5
3
2 1 3 !2
(2 1 3 !2)(3 1 4 !2)
5
c.
3 2 4 !2
(3 2 4 !2)(3 1 4 !2)
6 1 9!2 1 8!2 1 12(2)
5
32 2 A4 !2B 2
30 1 17 !2
5
9 2 16(2)
30 1 17!2
52
23
3!2 2 4!3
(3!2 2 4 !3)(3 !2 2 4 !3)
5
d.
3!2 1 4!3
(3 !2 1 4 !3)(3"2 2 4 !3)
(3!2)2 2 24!6 1 (4 !3)2
5
(3!2)2 2 (4!3)2
9(2) 2 24!6 1 16(3)
5
9(2) 2 16(3)
66 2 24 !6
52
30
11 2 4 !6
52
5
11. a. f(x) 5 3x2 2 2x
When a 5 2,
f(a 1 h) 2 f(a)
f(2 1 h) 2 f(2)
5
h
h
2
3(2 1 h) 2 2(2 1 h) 2 C3(2)2 2 2(2)D
5
h
3(4 1 4h 1 h2 ) 2 4 2 2h 2 8
5
h
12 1 12h 1 3h2 2 2h 2 12
5
h
Calculus and Vectors Solutions Manual
3h2 1 10h
h
5 3h 1 10
This expression can be used to determine the slope of
the secant line between (2, 8) and (2 1 h, f(2 1 h)).
b. For h 5 0.01: 3(0.01) 1 10 5 10.03
c. The value in part b. represents the slope of the
secant line through (2, 8) and (2.01, 8.1003).
5
2.1 The Derivative Function, pp. 73–75
1. A function is not differentiable at a point where its
graph has a cusp, a discontinuity, or a vertical tangent:
a. The graph has a cusp at x 5 22, so f is
differentiable on 5xPR Z x 2 226.
b. The graph is discontinuous at x 5 2, so f is
differentiable on 5xPR Z x 2 26.
c. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
d. The graph has a cusp at x 5 1, so f is
differentiable on 5xPR Z x 2 16.
e. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
f. The function does not exist for x , 2, but has
no cusps, discontinuities, or vertical tangents
elsewhere, so f is differentiable on 5xPR Z x . 26.
2. The derivative of a function represents the slope of
the tangent line at a given value of the independent
variable or the instantaneous rate of change of the
function at a given value of the independent variable.
3.
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
f(x) 5 5x 2 2
f(a 1 h) 5 5(a 1 h) 2 2
5 5a 1 5h 2 2
f(a 1 h) 2 f(a) 5 5a 1 5h 2 2 2 (5a 2 2)
5 5h
b.
f(x) 5 x2 1 3x 2 1
f(a 1 h) 5 (a 1 h)2 1 3(a 1 h) 2 1
5 a2 1 2ah 1 h2 1 3a
1 3h 2 1
f(a 1 h) 2 f(a) 5 a2 1 2ah 1 h2 1 3a 1 3h
2 1 2 (a2 1 3a 2 1)
5 2ah 1 h2 1 3h
4. a.
2-3
f(x) 5 x3 2 4x 1 1
f(a 1 h) 5 (a 1 h)3 2 4(a 1 h) 1 1
5 a3 1 3a2h 1 3ah2 1 h3
2 4a 2 4h 1 1
f(a 1 h) 2 f(a) 5 a3 1 3a2h 1 3ah2 1 h3 2 4a
2 4h 1 1 2 (a3 2 4a 1 1)
5 3a2h 1 3ah2 1 h3 2 4h
f(x) 5 x2 1 x 2 6
d.
f(a 1 h) 5 (a 1 h)2 1 (a 1 h) 2 6
5 a2 1 2ah 1 h2 1 a 1 h 2 6
f(a 1 h) 2 f(a) 5 a2 1 2ah 1 h2 1 a 1 h 2 6
2 (a2 1 a 2 6)
5 2ah 1 h2 1 h
e.
f(x) 5 27x 1 4
f(a 1 h) 5 27(a 1 h) 1 4
5 27a 2 7h 1 4
f(a 1 h) 2 f(a) 5 27a 2 7h 1 4 2 (27a 1 4)
5 27h
f(x) 5 4 2 2x 2 x2
f.
f(a 1 h) 5 4 2 2(a 1 h) 2 (a 1 h)2
5 4 2 2a 2 2h 2 a2 2 2ah 2 h2
f(a 1 h) 2 f(a) 5 4 2 2a 2 2h 2 a2 2 2ah
2 h2 2 4 1 2a 1 a2
5 22h 2 h2 2 2ah
f(1 1 h) 2 f(1)
5. a. f r (1) 5 lim
hS0
h
(1 1 h)2 2 12
5 lim
hS0
h
1 1 2h 1 h2 2 1
5 lim
hS0
h
2h 1 h2
5 lim
hS0
h
5 lim (2 1 h)
c.
hS0
52
f(3 1 h) 2 f(3)
b. f r(3) 5 lim
hS0
h
(3 1 h)2 1 3(3 1 h) 1 1
5 lim c
hS0
h
2
(3 1 3(3) 1 1)
d
2
h
9 1 6h 1 h2 1 9 1 3h 1 1 2 19
5 lim
hS0
h
9h 1 h2
5 lim
hS0
h
5 lim (9 1 h)
hS0
c.
f(0 1 h) 2 f(0)
hS0
h
!h 1 1 2 !0 1 1
5 lim
hS0
h
!h 1 1 2 1
5 lim
hS0
h
( ! h 1 1 2 1)( ! h 1 1 1 1)
5 lim
hS0
h( ! h 1 1 1 1)
f r (0) 5 lim
( "h 1 1) 2 2 1
hS0 h( !h 1 1 1 1)
5 lim
h1121
hS0 h( !h 1 1 1 1)
1
5 lim
hS0 ( !h 1 1 1 1)
1
5 lim
hS0 ( !1 1 1)
1
5
2
f(21 1 h) 2 f(21)
d. f r(21) 5 lim
hS0
h
5 lim
5 lim
hS0
5 lim
hS0
5
5
2 21
21 1 h
h
5
21 1 h 1 5
h
5
5(21 1 h)
1 21 1 h
21
1
h
5 lim
h
5 2 5 1 5h
5 lim
hS0 h(21 1 h)
5h
5 lim
hS0 h(21 1 h)
5
5 lim
hS0 (21 1 h)
5
5
21
5 25
f(x 1 h) 2 f(x)
6. a. f r (x) 5 lim
hS0
h
25(x 1 h) 2 8 2 (25x 2 8)
5 lim
hS0
h
25x 2 5h 2 8 1 5x 1 8
5 lim
hS0
h
hS0
59
2-4
Chapter 2: Derivatives
25h
hS0 h
5 lim 25
5 lim
hS0
5 25
f(x 1 h) 2 f(x)
b. f r(x) 5 lim
hS0
h
2(x 1 h)2 1 4(x 1 h)
5 lim c
hS0
h
A2x2 1 4xB
d
h
2x2 1 4xh 1 2h2 1 4x
5 lim c
hS0
h
4h 2 2x2 2 4x
1
d
h
4xh 1 2h2 1 4h
5 lim
hS0
h
5 lim (4x 1 2h 1 4)
2
hS0
5 4x 1 4
f(x 1 h) 2 f(x)
c. f r(x) 5 lim
hS0
h
6(x 1 h)3 2 7(x 1 h)
5 lim c
hS0
h
(6x3 2 7x)
2
d
h
6x3 1 18x2h 1 18xh2 1 6h3
5 lim c
hS0
h
27x 2 7h 2 6x3 1 7x
1
d
h
18x2h 1 18xh2 1 6h3 2 7h
5 lim
hS0
h
2
5 lim (18x 1 18xh 1 6h2 2 7)
hS0
5 18x2 2 7
f(x 1 h) 2 f(x)
d. f r(x) 5 lim
hS0
h
5 lim
hS0
5 lim
hS0
A !3x 1 3h 1 2 B 2 2 A !3x 1 2 B 2
hA !3x 1 3h 1 2 1 !3x 1 2 B
3x 1 3h 1 2 2 3x 2 2
hS0 hA !3x 1 3h 1 2 1 !3x 1 2B
3
5 lim
hS0 !3x 1 3h 1 2 1 !3x 1 2
3
5
2 !3x 1 2
7. a. Let y 5 f(x), then
f(x 1 h) 2 f(x)
dy
5 f r(x) 5 lim
dx
hS0
h
6 2 7(x 1 h) 2 (6 2 7x)
5 lim
hS0
h
6 2 7x 2 7h 2 6 1 7x
5 lim
hS0
h
27h
5 lim
hS0
h
5 lim 27
5 lim
hS0
5 27
b. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
5 lim
!3x 1 3h 1 2 2 !3x 1 2
hS0
h
A !3x 1 3h 1 2 2 !3x 1 2 B
5 lim c
hS0
h
3
A !3x 1 3h 1 2 1 !3x 1 2B
d
A !3x 1 3h 1 2 1 !3x 1 2B
Calculus and Vectors Solutions Manual
h
hS0
5 lim
hS0
(x 1 h 1 1)(x 2 1)
£ (x 1 h 2 1)(x 2 1)
h
(x 1 1)(x 1 h 2 1)
(x 2 1)(x 1 h 2 1) §
2
h
x2
5 lim £
2
5 lim
1 hx 1 x 2 x 2 h 2 1
(x 1 h 2 1)(x 2 1)
h
hS0
!3(x 1 h) 1 2 2 !3x 1 2
h
5 lim
x1h11
x11
2x21
x1h21
x2 1 hx 2 x 1 x 1 h 2 1
(x 1 h 2 1)(x 2 1)
§
h
22h
(x 1 h 2 1)(x 2 1)
h
22
5 lim
hS0 (x 1 h 2 1)(x 2 1)
2
52
(x 2 1)2
hS0
2-5
c. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
3(x 1 h)2 2 3x2
5 lim
hS0
h
3x2 1 6xh 1 3h2 2 3x2
5 lim
hS0
h
6xh 1 3h2
5 lim
hS0
h
5 lim 6x 1 3h
9. a.
5 6x
8. Let y 5 f(x), then the slope of the tangent at
each point x can be found by calculating f r(x)
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
2(x 1 h)2 2 4(x 1 h) 2 2x2 1 4x
5 lim
hS0
h
2
2x 1 4xh 1 2h2 2 4x 2 4h
5 lim c
hS0
h
2
22x 1 4x
1
d
h
4xh 1 h2 2 4h
5 lim
hS0
h
5 lim 4x 1 h 2 4
b. Let y 5 f(x), then the slope of the tangent at
each point x can be found by calculating f(x)
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
(x 1 h)3 2 x3
5 lim
hS0
h
x3 1 3x2h 1 3xh2 1 h3 2 x3
5 lim
hS0
h
3x2h 1 3xh2 1 h3
5 lim
hS0
h
5 lim 3x2 1 3xh 1 h2
hS0
hS0
5 4x 2 4
So the slope of the tangent at x 5 0 is
f r(0) 5 4(0) 2 4
5 24
At x 5 1, the slope of the tangent is
f r(1) 5 4(1) 2 4
50
At x 5 2, the slope of the tangent is
f r(2) 5 4(2) 2 4
54
y
4
3
2
1
x
0
–2 –1
1 2 3 4
–1
–2
2-6
12
10
8
6
4
2
–6 –4 –2 0
–2
–4
y
x
2 4 6
hS0
5 3x2
So the slope of the tangent at x 5 22 is
f r(22) 5 3(22)2
5 12
At x 5 21, the slope of the tangent is
f r(21) 5 3(21)2
53
At x 5 0, the slope of the tangent is
f r(0) 5 3(0)2
50
At x 5 1, the slope of the tangent is
f r(1) 5 3(1)2
53
At x 5 2, the slope of the tangent is
f r(2) 5 3(2)2
5 12
c.
y
12
10
8
6
4
2
x
0
–6 –4 –2
2 4 6
–2
Chapter 2: Derivatives
d. The graph of f(x) is a cubic. The graph of f r(x)
seems to be a parabola.
10. The velocity the particle at time t is given by sr(t)
s(t 1 h) 2 s(t)
sr(t) 5 lim
hS0
h
2 (t 1 h)2 1 8(t 1 h) 2 (2t2 1 8t)
5 lim
hS0
h
2t2 2 2th 2 h2 1 8t 1 8h 1 t2 2 8t
5 lim
hS0
h
22th 2 h2 1 8h
5 lim
hS0
h
5 lim 2 2t 2 h 1 8
hS0
5 22t 1 8
So the velocity at t 5 0 is
sr(0) 5 22(0) 1 8
5 8 m>s
At t 5 4, the velocity is
sr(4) 5 22(4) 1 8
5 0 m>s
At t 5 6, the velocity is
sr(6) 5 22(6) 1 8
5 24 m>s
f(x 1 h) 2 f(x)
11. f r(x) 5 lim
hS0
h
!x 1 h 1 1 2 !x 1 1
hS0
h
5 lim
5 lim c
hS0
A !x 1 h 1 1 2 !x 1 1 B
h
3
A !x 1 h 1 1 1 !x 1 1 B
d
A !x 1 h 1 1 1 !x 1 1 B
A !x 1 h 1 1 B 2 2 A !x 1 1 B 2
hS0 hA !x 1 h 1 1 1 !x 1 1 B
5 lim
x1h112x21
hS0 hA !x 1 h 1 1 1 !x 1 1B
5 lim
h
hS0 hA !x 1 h 1 1 1 !x 1 1B
5 lim
5 lim
hS0
5
1
A !x 1 h 1 1 1 !x 1 1 B
1
2 !x 1 1
The equation x 2 6y 1 4 5 0 can be rewritten as
y 5 16 x 1 23, so this line has slope 16. The value of x
where the tangent to f(x) has slope 16 will satisfy
f r(x) 5 16.
Calculus and Vectors Solutions Manual
1
1
5
2 !x 1 1
6
6 5 2 !x 1 1
32 5 A !x 1 1B 2
95x11
85x
f(8) 5 !8 1 1
5 !9
53
So the tangent passes through the point (8, 3), and its
equation is y 2 3 5 16 (x 2 8) or x 2 6y 1 10 5 0.
12. a. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
c2c
5 lim
hS0
h
0
5 lim
hS0 h
50
b. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
(x 1 h) 2 x
5 lim
hS0
h
h
5 lim
hS0 h
5 lim 1
hS0
51
c. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
m(x 1 h) 1 b 2 mx 2 b
5 lim
hS0
h
mx 1 mh 1 b 2 mx 2 b
5 lim
hS0
h
mh
5 lim
hS0 h
5 lim m
hS0
5m
d. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
a(x 1 h)2 1 b(x 1 h) 1 c
5 lim c
hS0
h
(ax2 1 bx 1 c)
2
d
h
2-7
5 lim c
ax2 1 2axh 1 ah2 1 bx 1 bh
hS0
h
2ax2 2 bx 2 c
1
d
h
2axh 1 ah2 1 bh
5 lim
hS0
h
5 lim (2ax 1 ah 1 b)
hS0
5 2ax 1 b
13. The slope of the function at a point x is given by
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
(x 1 h)3 2 x3
5 lim
hS0
h
3
x 1 3x2h 1 3xh2 1 h3 2 x3
5 lim
hS0
h
2
2
3x h 1 3xh 1 h3
5 lim
hS0
h
2
5 lim 3x 1 3xh 1 h2
hS0
2
5 3x
Since 3x2 is nonnegative for all x, the original
function never has a negative slope.
14. h(t) 5 18t 2 4.9t2
h(t 1 k) 2 h(t)
a. hr(t) 5 lim
kS0
k
18(t 1 k) 2 4.9(t 1 k)2
5 lim
kS0
k
A18t 2 4.9t2 B
2
k
18t 1 18k 2 4.9t 2 2 9.8tk 2 4.9k2
5 lim
kS0
k
18t 1 4.9t2
2
k
18k 2 9.8tk 2 4.9k2
5 lim
kS0
k
5 lim (18 2 9.8t 2 4.9k)
kS0
5 18 2 9.8t 2 4.9(0)
5 18 2 9.8t
Then hr(2) 5 18 2 9.8(2) 5 21.6 m>s.
b. hr(2) measures the rate of change in the height
of the ball with respect to time when t 5 2.
15. a. This graph has positive slope for x , 0, zero
slope at x 5 0, and negative slope for x . 0, which
corresponds to graph e.
b. This graph has positive slope for x , 0, zero
slope at x 5 0, and positive slope for x . 0, which
corresponds to graph f.
2-8
c. This graph has negative slope for x , 22,
positive slope for 22 , x , 0, negative slope for
0 , x , 2, positive slope for x . 2, and zero slope
at x 5 22, x 5 0, and x 5 2, which corresponds to
graph d.
16. This function is defined piecewise as f(x) 5 2x2
for x , 0, and f(x) 5 x2 for x $ 0. The derivative
will exist if the left-side and right-side derivatives are
the same at x 5 0:
f(0 1 h) 2 f(0)
2 (0 1 h)2 2 A20 2 B
lim2
5 lim2
hS0
h
hS0
h
2h2
5 lim2
hS0
h
5 lim2 (2h)
hS0
50
lim1
hS0
f(0 1 h) 2 f(0)
(0 1 h)2 2 A02 B
5 lim1
h
hS0
h
h2
5 lim1
hS0 h
5 lim1 (h)
hS0
50
Since the limits are equal for both sides, the derivative
exists and f r(0) 5 0.
17. Since f r(a) 5 6 and f(a) 5 0,
f(a 1 h) 2 f(a)
6 5 lim
hS0
h
f(a 1 h) 2 0
6 5 lim
hS0
h
f(a 1 h)
3 5 lim
hS0
2h
18.
y
6
4
2
x
–1
–2
1 2 3 4 5
f(x) is continuous.
f(3) 5 2
But f r(3) 5 `.
(Vertical tangent)
19. y 5 x2 2 4x 2 5 has a tangent parallel to
2x 2 y 5 1.
Let f(x) 5 x2 2 4x 2 5. First, calculate
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
Chapter 2: Derivatives
5 lim c
(x 1 h)2 2 4(x 1 h) 2 5
hS0
h
2
Ax 2 4x 2 5B
2
d
h
x2 1 2xh 1 h2 2 4x 2 4h 2 5
5 lim c
hS0
h
2x2 1 4x 1 5
1
d
h
2xh 1 h2 2 4h
5 lim
hS0
h
5 lim (2x 1 h 2 4)
hS0
5 2x 1 0 2 4
5 2x 2 4
Thus, 2x 2 4 is the slope of the tangent to the curve
at x. We want the tangent parallel to 2x 2 y 5 1.
Rearranging, y 5 2x 2 1.
If the tangent is parallel to this line,
2x 2 4 5 2
x53
When x 5 3, y 5 (3)2 2 4(3) 2 5 5 28.
The point is (3, 28).
20. f(x) 5 x2
The slope of the tangent at any point Ax, x2 B is
f(x 1 h) 2 f(x)
f r 5 lim
hS0
h
(x 1 h)2 2 x2
5 lim
hS0
h
(x 1 h 1 x)(x 1 h 2 x)
5 lim
hS0
h
h(2x 1 h)
5 lim
hS0
h
5 lim (2x 1 h)
hS0
5 2x 1 0
5 2x
Let (a, a2 ) be a point of tangency. The equation of
the tangent is
y 2 a2 5 (2a)(x 2 a)
y 5 (2a)x 2 a2
Suppose the tangent passes through (1, 23).
Substitute x 5 1 and y 5 23 into the equation of
the tangent:
23 5 (2a)(1) 2 a2
2
a 2 2a 2 3 5 0
(a 2 3)(a 1 1) 5 0
a 5 21, 3
So the two tangents are y 5 22x 2 1 or
2x 1 y 1 1 5 0 and y 5 6x 2 9 or 6x 2 y 2 9 5 0.
Calculus and Vectors Solutions Manual
2.2 The Derivatives of Polynomial
Functions, pp. 82–84
1. Answers may vary. For example:
d
constant function rule:
(5) 5 0
dx
d 3
Ax B 5 3x2
power rule:
dx
d
A4x3 B 5 12x2
constant multiple rule:
dx
d 2
Ax 1 xB 5 2x 1 1
sum rule:
dx
d 3
Ax 2 x2 1 3xB 5 3x2 2 2x 1 3
difference rule:
dx
d
d
2. a. f r(x) 5 (4x) 2 (7)
dx
dx
d
d
5 4 (x) 2
(7)
dx
dx
5 4Ax0 B 2 0
54
d
d
b. f r(x) 5 Ax3 B 2 Ax2 B
dx
dx
5 3x2 2 2x
d
d
d
c. f r(x) 5 A2x2 B 1 (5x) 1 (8)
dx
dx
dx
d
d
d
5 2 Ax2 B 1 5 (x) 1
(8)
dx
dx
dx
5 2 (2x) 1 5 1 0
5 22x 1 5
d 3
x)
d. f r(x) 5 ("
dx
d 1
5 ( x3 )
dx
1 1
5 ( x(3 21))
3
1 2
5 (x23)
3
1
5 3 2
3"x
d x 4
e. f r(x) 5 aa b b
dx 2
1 4d 4
5a b
Ax B
2 dx
1
5 A4x3 B
16
x3
5
4
2-9
d 23
(x )
dx
5 (23)( x2321)
5 23x24
d
3. a. hr(x) 5 ((2x 1 3)(x 1 4))
dx
d
5
A2x2 1 8x 1 3x 1 12B
dx
d
d
d
5
A2x2 B 1
(11x) 1
(12)
dx
dx
dx
d
d
d
5 2 Ax2 B 1 11 (x) 1
(12)
dx
dx
dx
5 2(2x) 1 11(1) 1 0
5 4x 1 11
d
b. f r(x) 5 A2x3 1 5x2 2 4x 2 3.75B
dx
d
d
d
5
(2x3 ) 1
A5x2 B 2
(4x)
dx
dx
dx
d
2
(3.75)
dx
d
d
d
5 2 Ax3 B 1 5 Ax2 B 2 4 (x)
dx
dx
dx
d
2
(3.75)
dx
5 2A3x2 B 1 5(2x) 2 4(1) 2 0
5 6x2 1 10x 2 4
ds
d
c.
5 At 2 At 2 2 2tB B
dt
dt
d
5 At 4 2 2t 3 B
dt
d
d
5 At 4 B 2 A2t 3 B
dt
dt
d 4
d
5 At B 2 2 At 3 B
dt
dt
5 4t 3 2 2A3t 2 B
5 4t 3 2 6t 2
dy
d 1 5 1 3 1 2
d.
5
a x 1 x 2 x 1 1b
dx
dx 5
3
2
d 1 5
d 1 3
d 1 2
d
5
a xb1
a xb2
a xb1
(1)
dx 5
dx 3
dx 2
dx
1 d
1 d
1 d
5 a b Ax5 B 1 a b Ax3 B 2 a b Ax2 B
5 dx
3 dx
2 dx
d
1
(1)
dx
1
1
1
5 A5x4 B 1 A3x2 B 2 (2x) 1 0
5
3
2
5 x4 1 x2 2 x
f. f r(x) 5
2-10
d
A5Ax2 B 4 B
dx
d
5 5 Ax234 B
dx
d
5 5 Ax8 B
dx
5 5A8x7 B
5 40x7
d t5 2 3t2
f. sr(t) 5 a
b
dt
2t
1 d
5 a b At 4 2 3tB
2 dt
1 d
d
5 a b a At4 B 2 (3t)b
2 dt
dt
1 d
d
5 a b a At4 B 2 3 (t)b
2 dt
dt
1
5 a b A4t 3 2 3(1)B
2
3
5 2t 3 2
2
dy
d
5
4. a.
5
A3x3 B
dx
dx
d 5
5 3 ( x3)
dx
5
5
5 a b3( xA3 21))
3
2
5 5x3
dy
6
d
1
b.
5
a4x22 2 b
x
dx
dx
d
d
1
5 4 ( x22) 2 6 Ax21 B
dx
dx
21
1
5 4a
b ( x22 21) 2 6(21)Ax2121 B
2
3
5 22x22 1 6x22
dy
d 6
2
5
a 3 1 2 2 3b
c.
dx
dx x
x
d
d
d
5 6 Ax23 B 1 2 Ax22 B 2
(3)
dx
dx
dx
5 6(23)Ax2321 B 1 2(22)Ax2221 B 2 0
e. gr(x) 5
5 218x24 2 4x23
218
4
5 4 2 3
x
x
Chapter 2: Derivatives
d.
dy
d
5
9x22 1 3"x
dx
dx
d
d 1
5 9 Ax22 B 1 3 ( x2)
dx
dx
(
)
(
1
1
5 9(22)Ax2221 B 1 3a b ( x2 21)
2
3
1
5 218x23 1 x22
2
dy
d
e.
5
"x 1 6"x3 1 "2
dx
dx
d 1
d 3
d
5 ( x2) 1 6 ( x2) 1
"2
dx
dx
dx
1 1
3
3
5 ( x2 21) 1 6a b ( x2 21) 1 0
2
2
1
1
1
5 ( x22) 1 9x2
2
(
5 2t 2 6(1) 1 0
5 2t 2 6
d 3
6. a. f r(x) 5
x 2 "x
dx
d 3
d 1
5
Ax B 2 ( x2)
dx
dx
1 1
5 3x2 2 ( x2 21)
2
1 1
5 3x2 2 x22
2
1
1
so f r(a) 5 f r(4) 5 3(4)2 2 (4)22
2
1 1
5 3(16) 2
2 "4
1 1
5 48 2 a b a b
2 2
5 47.75
d
2
7 2 6"x 1 5x3
b. f r(x) 5
dx
d
d 1
d 2
5
(7) 2 6 ( x2) 1 5 ( x3)
dx
dx
dx
1
2
1
2
5 0 2 6a b ( x2 21) 1 5a b ( x3 21)
2
3
10
1
1
5 23x22 1 a b ( x23)
3
10
1
1
so f r(a) 5 f r(64) 5 23( 6422) 1 a b ( 6423)
3
1
10 1
5 23a b 1 a b
8
3 4
11
5
24
dy
d
5
A3x4 B
7. a.
dx
dx
d
5 3 Ax4 B
dx
5 3A4x3 B
5 12x3
The slope at (1, 3) is found by substituting x 5 1 into
)
( )
dy
d 1 1 "x
5
a
b
x
dx
dx
1
d 1
d x2
5
a b1
a b
dx x
dx x
d 21
d
1
5
(x ) 1 ( x22)
dx
dx
21 212 21
5 (21)x2121 1
(x )
2
1 3
5 2x22 2 x22
2
ds
d
5. a.
5 A22t 2 1 7tB
dt
dt
d
d
5 (22)a At2 B b 1 7a (t)b
dt
dt
5 (22)(2t) 1 7(1)
5 24t 1 7
ds
d
1
b.
5 a18 1 5t 2 t 3 b
dt
dt
3
d
d
1 d
5 (18) 1 5 (t) 2 a b At 3 B
dt
dt
3 dt
1
5 0 1 5(1) 2 a b A3t 2 B
3
5 5 2 t2
d
ds
c.
5 A(t 2 3)2B
dt
dt
d
5 At 2 2 6t 1 9B
dt
d
d
d
5 At 2 B 2 (6) (t) 1 (9)
dt
dt
dt
f.
Calculus and Vectors Solutions Manual
(
)
)
dy
the equation for dx. So the slope 5 12(1)3
5 12
dy
d 1
b
5
a
b.
dx
dx x25
d 5
5
Ax B
dx
5 5x4
2-11
The slope at (21, 21) is found by substituting x 5 21
The slope at x 5 4 is found by substituting x 5 4
dy
into the equation for dx. So the
into the equation for dx. So the slope is (4) 2 5 12.
16
c. y 5 2
x
dy
d 16
5
a b
dx
dx x2
d
5 16 Ax22 B
dx
5 16(22)x2221
5 232x23
The slope at x 5 22 is found by substituting
dy
slope 5 5(21)4
55
dy
d 2
c.
5
a b
dx
dx x
d
5 2 Ax21 B
dx
5 2(21)x2121
5 22x22
The slope at (22 , 21) is found by substituting
dy
x 5 22 into the equation for dx. So the
slope 5 22(22)22
1
52
2
dy
d
d.
5 ("16x3)
dx
dx
d 3
5 "16 ( x2)
dx
3 3
5 4a bx2 21
2
1
2
5 6x
The slope at (4, 32) is found by substituting x 5 4
into the equation for
1
dy
. So the
dx
slope 5 6(4)2
5 12
8. a. y 5 2x3 1 3x
dy
d
5
A2x3 1 3xB
dx
dx
d
d
5 2 Ax3 B 1 3 (x)
dx
dx
5 2A3x2 B 1 3(1)
5 6x2 1 3
The slope at x 5 1 is found by substituting x 5 1
dy
into the equation for dx. So the slope is
6(1)2 1 3 5 9.
b. y 5 2"x 1 5
dy
d
5 (2"x 1 5)
dx
dx
d 1
d
5 2 ( x2 ) 1
(5)
dx
dx
1
1
5 2a b ( x2 21) 1 0
2
21
5 x2
2-12
21
x 5 22 into the equation for
23
232(22)
5
(232)
(22)3
dy
.
dx
So the slope is
5 4.
d. y 5 x23 (x21 1 1)
5 x24 1 x23
dy
d 24
5
(x 1 x23 )
dx
dx
5 24x25 2 3x24
3
4
52 52 4
x
x
The slope at x 5 1 is found by substituting
dy
x 5 1 into the equation for dx. So the slope is
2 145 2 134 5 27.
dy
d
1
5
a2x 2 b
x
dx
dx
d
d 21
5 2 (x) 2
Ax B
dx
dx
5 2(1) 2 (21)x2121
5 2 1 x22
The slope at x 5 0.5 is found by substituting
9. a.
dy
x 5 0.5 into the equation for dx.
So the slope is 2 1 (0.5)22 5 6.
The equation of the tangent line is therefore
y 1 1 5 6(x 2 0.5) or 6x 2 y 2 4 5 0.
dy
d 3
4
b.
5
a 2 3b
dx
dx x2
x
d
d
5 3 Ax22 B 2 4 Ax23 B
dx
dx
5 3(22)x2221 2 4(23)x2321
5 12x24 2 6x23
The slope at x 5 21 is found by substituting
dy
x 5 21 into the equation for dx. So the slope is
12(21)24 2 6(21)23 5 18.
Chapter 2: Derivatives
The equation of the tangent line is therefore
y 2 7 5 18(x 1 1) or 18x 2 y 1 25 5 0.
d
dy
c.
5 ("3x3)
dx
dx
d 3
5 "3 ( x2)
dx
3 32 21
5 "3a bx
2
1
3"3x2
5
2
The slope at x 5 3 is found by substituting x 5 3
dy
into the equation for dx.
1
3"3(3)2
9
So the slope is
5 .
2
2
The equation of the tangent line is therefore
y 2 9 5 92 (x 2 3) or 9x 2 2y 2 9 5 0.
dy
d 1 2 1
d.
5
a ax 1 bb
x
dx
dx x
d
1
5
ax 1 2 b
dx
x
d
d 22
5
(x) 1
Ax B
dx
dx
5 1 1 (22)x2221
5 1 2 2x23
The slope at x 5 1 is found by substituting
dy
into the equation for dx.
So the slope is 1 2 2(1)23 5 21.
The equation of the tangent line is therefore
y 2 2 5 2 (x 2 1) or x 1 y 2 3 5 0.
dy
d
e.
5 (("x 2 2)(3"x 1 8))
dx
dx
d
5 (3("x)2 1 8"x 2 6"x 2 16)
dx
d
5 (3x 1 2"x 2 16)
dx
d
d 1
d
5
(3x) 1 2 ( x2) 2
(16)
dx
dx
dx
1 1
5 3(1) 1 2a bx2 21 2 0
2
1
5 3 1 x22
The slope at x 5 4 is found by substituting x 5 4
The equation of the tangent line is therefore
y 5 3.5(x 2 4) or 7x 2 2y 2 28 5 0.
dy
d "x 2 2
f.
5
a 3
b
dx
dx
"x
1
d x2 2 2
5
a
b
1
dx
x3
d 1 1
1
5 ( x2 2 3 2 2x23)
dx
d 1
d
1
5 ( x6) 2 2 Ax23 B
dx
dx
1 1
1
1
5 ( x6 21) 2 2a2 bx23 21 2 0
6
3
2 243
1 256
5 (x ) 1 x
6
3
The slope at x 5 1 is found by substituting x 5 1
dy
into the equation for dx.
5
4
So the slope is 16 (1)26 1 23 (1)23 5 56.
The equation of the tangent line is therefore
y 1 1 5 56 (x 2 1) or 5x 2 6y 2 11 5 0.
10. A normal to the graph of a function at a point is
a line that is perpendicular to the tangent at the
given point.
3
4
y 5 2 2 3 at P(21, 7)
x
x
Slope of the tangent is 18, therefore, the slope of
the normal is 2 181 .
1
Equation is y 2 7 5 2 (x 1 1).
18
x 1 18y 2 125 5 0
3
1
11. y 5 3 5 3x23
"x
Parallel to x 1 16y 1 3 2 0
Slope of the line is 2 161 .
dy
4
5 2x23
dx
1
4
x23 5
16
1
1
4 5
3
16
x
4
x3 5 16
3
x 5 (16)4 5 8
dy
into the equation for dx.
1
So the slope is 3 1 (4)22 5 3.5.
Calculus and Vectors Solutions Manual
2-13
1
5 x21 : y 5 x3
x
dy
1 dy
52 2:
5 3x2
dx
x dx
1
Now, 2 2 5 3x2
x
1
x4 5 2 .
3
No real solution. They never have the same slope.
dy
5 2x
13. y 5 x2,
dx
The slope of the tangent at A(2, 4) is 4 and at
B Q 2 18 , 641 R is 2 14.
Since the product of the slopes is 21, the tangents
at A(2, 4) and B Q 2 18 , 641 R will be perpendicular.
12. y 5
4 y
3
2
1
–3 –2 –1 0
–1
x
1 2 3
14. y 5 2x2 1 3x 1 4
dy
5 22x 1 3
dx
dy
55
For
dx
5 5 22x 1 3
x 5 21.
The point is (21, 0).
6
5
4
3
2
1
–2 –1 0
–1
–2
y
x
1
2 3 4
15. y 5 x3 1 2
dy
5 3x2, slope is 12
dx
x2 5 4
x 5 2 or x 5 22
Points are (2, 10) and (22, 26).
2-14
16. y 5 15x5 2 10x, slope is 6
dy
5 x4 2 10 5 6
dx
x4 5 16
x2 5 4 or x2 5 24
x 5 62 non-real
Tangents with slope 6 are at the points Q 2, 2 685 R
and Q2 2, 685 R .
17. y 5 2x2 1 3
a. Equation of tangent from A(2, 3):
If x 5 a, y 5 2x2 1 3.
Let the point of tangency be PAa, 2a2 1 3B.
dy
dy
Now, dx 5 4x and when x 5 a, dx 5 4a.
The slope of the tangent is the slope of AP.
2a2
5 4a.
a22
2a2 5 4a2 2 8a
2
2a 2 8a 5 0
2a(a 2 4) 5 0
a 5 0 or a 5 4.
Point (2, 3):
Slope is 0.
Slope is 16.
Equation of tangent is
Equation of tangent is
y 2 3 5 0.
y 2 3 5 16(x 2 2) or
16x 2 y 2 29 5 0.
b. From the point B(2, 27):
2a2 1 10
Slope of BP:
5 4a
a22
2a2 1 10 5 4a2 2 8a
2
2a 2 8a 2 10 5 0
a2 2 4a 2 5 5 0
(a 2 5)(a 1 1) 5 0
a55
a 5 21
Slope is 4a 5 20.
Slope is 4a 5 24.
Equation is
Equation is
y 1 7 5 20(x 2 2)
y 1 7 5 24(x 2 2)
or 20x 2 y 2 47 5 0.
or 4x 1 y 2 1 5 0.
a
18. ax 2 4y 1 21 5 0 is tangent to y 5 x2 at x 5 22.
Therefore, the point of tangency is a22, 4 b,
This point lies on the line, therefore,
a
a(22) 2 4a b 1 21 5 0
4
23a 1 21 5 0
a 5 7.
a
Chapter 2: Derivatives
19. a. When h 5 200,
d 5 3.53"200
8 49.9
Passengers can see about 49.9 km.
1
b. d 5 3.53!h 5 3.53h2
1 1
dr 5 3.53a h 2 2 b
2
3.53
5
2!h
When h 5 200,
3.53
dr 5
2 !200
8 0.12
The rate of change is about 0.12 km>m.
20. d(t) 5 4.9t2
a. d(2) 5 4.9(2)2 5 19.6 m
d(5) 5 4.9(5)2 5 122.5 m
The average rate of change of distance with respect
to time from 2 s to 5 s is
Dd
122.5 2 19.6
5
Dt
522
5 34.3 m>s
b. dr(t) 5 9.8t
Thus, dr(4) 5 9.8(4) 5 39.2 m>s.
c. When the object hits the ground, d 5 150.
Set d(t) 5 150:
4.9t2 5 150
1500
t2 5
49
10
t 5 6 "15
7
10
Since t $ 0, t 5 "15
7
Then,
10
10
dra "15b 5 9.8a "15b
7
7
8 54.2 m>s
21. v(t) 5 sr(t) 5 2t 2 t2
0.5 5 2t 2 t2
t2 2 2t 1 0.5 5 0
2t2 2 4t 1 1 5 0
4 6 "8
t5
4
t 8 1.71, 0.29
The train has a velocity of 0.5 km>min at about
0.29 min and 1.71 min.
Calculus and Vectors Solutions Manual
22. v(t) 5 Rr(t) 5 210t
v(2) 5 220
The velocity of the bolt at t 5 2 is 220 m>s.
23.
y
3 (0, 3)
2
1
–3 –2 –1 0
–1
–2
–3
1
x
2 3
Let the coordinates of the points of tangency be
AAa,23a2 B.
dy
5 26x, slope of the tangent at A is 26a
dx
23a2 2 3
5 26a
a
23a2 2 3 5 26a2
3a2 5 3
a 5 1 or a 5 21
Coordinates of the points at which the tangents
touch the curve are (1, 23) and (21, 23).
24. y 5 x3 2 6x2 1 8x, tangent at A(3, 23)
dy
5 3x2 2 12x 1 8
dx
When x 5 3,
dy
5 27 2 36 1 8 5 21
dx
The slope of the tangent at A(3, 23) is 21.
Equation will be
y 1 3 5 21(x 2 3)
y 5 2x.
2x 5 x3 2 6x2 1 8x
x3 2 6x2 1 9x 5 0
xAx2 2 6x 1 9B 5 0
x(x 2 3)2 5 0
x 5 0 or x 5 3
Coordinates are B(0, 0).
Slope of PA:
3
2
1
–1 0
–11
–2
–3
y
x
1
2 3 4
2-15
25. a. i. f(x) 5 2x 2 5x2
f r(x) 5 2 2 10x
Set f r(x) 5 0:
2 2 10x 5 0
10x 5 2
1
x5
5
Then,
1
1
1 2
f a b 5 2a b 2 5a b
5
5
5
2
1
5 2
5
5
1
5
5
Thus the point is Q 15, 15 R .
ii. f(x) 5 4x2 1 2x 2 3
f r(x) 5 8x 1 2
Set f r(x) 5 0:
8x 1 2 5 0
8x 5 22
1
x52
4
Then,
1
1 2
1
f a2 b 5 4a2 b 1 2a2 b 2 3
4
4
4
1
2
12
5 2 2
4
4
4
13
52
4
Thus the point is Q 2 14, 2 134 R .
iii. f(x) 5 x3 2 8x2 1 5x 1 3
f r(x) 5 3x2 2 16x 1 5
Set f r(x) 5 0:
3x2 2 16x 1 5 5 0
2
3x 2 15x 2 x 1 5 5 0
3x(x 2 5) 2 (x 2 5) 5 0
(3x 2 1)(x 2 5) 5 0
1
x5 ,5
3
1 3
1 2
1
1
f a b 5 a b 2 8a b 1 5a b 1 3
3
3
3
3
1
24
45
81
5
2
1
1
27
27
27
27
103
5
27
f(5) 5 (5)3 2 8(5)2 1 5(5) 1 3
5 25 2 200 1 25 1 3
5 247
Thus the two points are Q 13, 103
27 R and (5, 247).
b. At these points, the slopes of the tangents are
zero, meaning that the rate of change of the value
of the function with respect to the domain is zero.
These points are also local maximum or minimum
points.
26. "x 1 "y 5 1
P(a, b) is on the curve, therefore a $ 0, b $ 0.
!y 5 1 2 !x
y 5 1 2 2 !x 1 x
1
dy
1
5 2 ? 2x22 1 1
dx
2
21 1 !a
1
At x 5 a, slope is 2
115
.
!a
!a
But !a 1 !b 5 1
2 !b 5 !a 2 1.
"b
b
Therefore, slope is 2
52
.
Åa
"a
27. f(x) 5 xn, f r(x) 5 nxn21
Slope of the tangent at x 5 1 is f r(1) 5 n,
The equation of the tangent at (1, 1) is:
y 2 1 5 n(x 2 1)
nx 2 y 2 n 1 1 5 0
Let y 5 0, nx 5 n 2 1
n21
1
x5
512 .
n
n
1
1
The x-intercept is 1 2 ; as n S `, S 0, and
n
n
the x-intercept approaches 1. As n S `, the slope
of the tangent at (1, 1) increases without bound, and
the tangent approaches a vertical line having equation
x 2 1 5 0.
28. a.
y
9
8
7
6
5
4
3
2
1
–2 –1 0
2-16
x
1
2 3 4
Chapter 2: Derivatives
f(x) 5 e
x2, if x , 3
x 1 6, if x $ 3
f r(3) does not exist.
b.
7
6
5
4
3
2
1
2x, if x , 3
1, if x $ 3
y
x
–3 –2 –1 0
–1
f(x) 5 e
f '(x) 5 e
1 2 3
3x2 2 6, if x , 2"2 or x . "2
6 2 3x2, if 2"2 , x , "2
f r(x) 5 e
6x, if x , 2"2 or x . "2
26x, if 2"2 # x # "2
f r "2 and f r 2"2 do not exist.
( )
(
c.
3
2
1
–3 –2 –1 0
–1
)
y
c. h(x) 5 (3x 1 2)(2x 2 7)
hr(x) 5 (3x 1 2)(2) 1 (3)(2x 2 7)
5 12x 2 17
d. h(x) 5 A5x7 1 1B Ax2 2 2xB
hr(x) 5 A5x7 1 1B (2x 2 2) 1 A35x6 BAx2 2 2xB
5 45x8 2 80x7 1 2x 2 2
e. s(t) 5 At2 1 1B A3 2 2t2 B
sr(t) 5 At2 1 1B (24t) 1 (2t)A3 2 2t2 B
5 28t3 1 2t
x23
f. f(x) 5
x13
f(x) 5 (x 2 3)(x 1 3)21
f r(x) 5 (x 2 3)(21)(x 1 3)22 1 (1)(x 1 3)21
5 (x 1 3)22 (2x 1 3 1 x 1 3)
6
5
(x 1 3)2
2. a. y 5 (5x 1 1)3 (x 2 4)
dy
5 (5x 1 1)3 (1) 1 3(5x 1 1)2 (5)(x 2 4)
dx
5 (5x 1 1)3 1 15(5x 1 1)2 (x 2 4)
b. y 5 A3x2 1 4BA3 1 x3 B 5
dy
5 A3x2 1 4B (5)A3 1 x3 B 4 A3x2 B
dx
1 (6x)A3 1 x3 B 5
5 15x2 (3x2 1 4)(3 1 x3 )4 1 6x(3 1 x3 )5
x
1 2 3
x 2 1, if x $ 1
1 2 x, if 0 # x , 1
f(x) 5 μ
x 1 1, if 21 , x , 0
2x 2 1, if x # 21
since
since
since
since
1, if x . 1
21, if 0 , x , 1
f'(x) 5 μ
1, if 21 , x , 0
21, if x , 21
f r(0), f r(21), and f r(1) do not exist.
2.3 The Product Rule, pp. 90–91
1. a. h(x) 5 x(x 2 4)
hr(x) 5 x(1) 1 (1)(x 2 4)
5 2x 2 4
b. h(x) 5 x2 (2x 2 1)
hr(x) 5 x2 (2) 1 (2x)(2x 2 1)
5 6x2 2 2x
Calculus and Vectors Solutions Manual
Zx 2 1 Z 5 x 2 1
Zx 2 1 Z 5 1 2 x
Z2x 2 1 Z 5 x 1 1
Z2x21 Z 5 2x 2 1
y 5 A1 2 x2 B 4 (2x 1 6)3
dy
5 4A1 2 x2 B 3 (22x)(2x 1 6)3
dx
1 A1 2 x2 B 4 3(2x 1 6)2 (2)
5 28xA1 2 x2 B 3 (2x 1 6)3
1 6A1 2 x2 B 4 (2x 1 6)2
d. y 5 Ax2 2 9B 4 (2x 2 1)3
dy
5 Ax2 2 9B 4 (3)(2x 2 1)2 (2)
dx
1 4Ax2 2 9B 3 (2x)(2x 2 1)3
5 6(x2 2 9)4 (2x 2 1)2
1 8x(x2 2 9)3 (2x 2 1)3
c.
2-17
3. It is not appropriate or necessary to use the product
rule when one of the factors is a constant or when it
would be easier to first determine the product of the
factors and then use other rules to determine the
derivative. For example, it would not be best to
use the product rule for f(x) 5 3Ax2 1 1B or
g(x) 5 (x 1 1) (x 2 1).
4. F(x) 5 3b(x)43c(x)4
F r(x) 5 3b(x)43cr(x)4 1 3br(x)43c(x)4
5. a. y 5 (2 1 7x)(x 2 3)
dy
5 (2 1 7x)(1) 1 7(x 2 3)
dx
At x 5 2,
dy
5 (2 1 14) 1 7(21)
dx
5 16 2 7
59
b.
y 5 (1 2 2x)(1 1 2x)
dy
5 (1 2 2x)(2) 1 (22)(1 1 2x)
dx
1
At x 5 ,
2
dy
5 (0)(2) 2 2(2)
dx
5 24
c.
y 5 A3 2 2x 2 x2 B Ax2 1 x 2 2B
dy
5 A3 2 2x 2 x2 B A2x 1 1B
dx
1 (22 2 2x)Ax2 1 x 2 2B
At x 5 22,
dy
5 (3 1 4 2 4)(24 1 1)
dx
1 (22 1 4)(4 2 2 2 2)
5 (3)(23) 1 (2)(0)
5 29
d. y 5 x3 (3x 1 7)2
dy
5 3x2 (3x 1 7)2 1 x36(3x 1 7)
dx
At x 5 22,
dy
5 12(1)2 1 (28)(6)(1)
dx
5 12 2 48
5 236
e.
y 5 (2x 1 1)5 (3x 1 2)4, x 5 21
dy
5 5(2x 1 1)4 (2)(3x 1 2)4
dx
1 (2x 1 1)54(3x 1 2)3 (3)
At x 5 21,
2-18
dy
5 5(21)4 (2)(21)4
dx
1 (21)5 (4)(21)3 (3)
5 10 1 12
5 22
f.
y 5 x(5x 2 2)(5x 1 2)
5 xA25x2 2 4B
dy
5 x(50x) 1 (25x2 2 4)(1)
dx
At x 5 3,
dy
5 3(150) 1 (25 ? 9 2 4)
dx
5 450 1 221
5 671
6. Tangent to y 5 Ax3 2 5x 1 2B A3x2 2 2xB
at (1, 22)
dy
5 A3x2 2 5B A3x2 2 2xB
dx
1 Ax3 2 5x 1 2B (6x 2 2)
when x 5 1,
dy
5 (22)(1) 1 (22)(4)
dx
5 22 1 28
5 210
Slope of the tangent at (1, 22) is 210.
The equation is y 1 2 5 210(x 2 1);
10x 1 y 2 8 5 0.
7. a. y 5 2(x 2 29)(x 1 1)
dy
5 2(x 2 29)(1) 1 2(1)(x 1 1)
dx
2x 2 58 1 2x 1 2 5 0
4x 2 56 5 0
4x 5 56
x 5 14
Point of horizontal tangency is (14, 2450).
b. y 5 Ax2 1 2x 1 1B Ax2 1 2x 1 1B
5 Ax2 1 2x 1 1B 2
dy
5 2Ax2 1 2x 1 1B (2x 1 2)
dx
Ax2 1 2x 1 1B (2x 1 2) 5 0
2(x 1 1)(x 1 1)(x 1 1) 5 0
x 5 21
Point of horizontal tangency is (21, 0).
8. a. y 5 (x 1 1)3 (x 1 4)(x 2 3)2
dy
5 3(x 1 1)2 (x 1 4)(x 2 3)2
dx
1 (x 1 1)3 (1)(x 2 3)2
1 (x 1 1)3 (x 1 4)32(x 2 3)4
Chapter 2: Derivatives
b. y 5 x2 A3x2 1 4B 2 A3 2 x3 B 4
dy
5 2xA3x2 1 4B 2 A3 2 x3 B 4
dx
1 x2 32A3x2 1 4B (6x)4 A3 2 x3 B 4
1 x2 A3x2 1 4B 2 34A3 2 x3 B 3 A23x2 B4
t 2
9. V(t) 5 75a1 2 b , 0 # t # 24
24
75 L 3 60% 5 45 L
t 2
45
5 a1 2 b
Set
75
24
t
3
512
6
24
Å5
3
t5a6
2 1b (224)
Å5
t 8 42.590 (inadmissable) or t 8 5.4097
t 2
V(t) 5 75a1 2 b
24
t
t
V(t) 5 75a1 2 b a1 2 b
24
24
t
1
Vr(t) 5 75 c a1 2 b a2 b
24
24
1
t
1 a2 b a1 2 b d
24
24
t
1
5 (75)(2)a1 2 b a2 b
24
24
Vr(5.4097) 5 24.84 L>h
10. Determine the point of tangency, and then find the
negative reciprocal of the slope of the tangent. Use
this information to find the equation of the normal.
h(x) 5 2x(x 1 1)3 Ax2 1 2x 1 1B 2
hr(x) 5 2(x 1 1)3 Ax2 1 2x 1 1B 2
1 (2x)(3)(x 1 1)2 Ax2 1 2x 1 1B 2
1 2x(x 1 1)3 2Ax2 1 2x 1 1B (2x 1 2)
hr(22) 5 2(21)3 (1)2
1 2(22)(3)(21)2 (1)2
1 2(22)(21)3 (2)(1)(22)
5 22 2 12 2 16
5 230
11.
a. f(x) 5 g1 (x)g2 (x)g3 (x) c gn21 (x)gn (x)
f r(x) 5 g1r(x)g2 (x)g3 (x) c gn21 (x)gn (x)
1 g1 (x)g2r(x)g3 (x) c gn21 (x)gn (x)
1 g1 (x)g2 (x)g3r(x) c gn21 (x)gn (x)
1 c 1 g1 (x)g2 (x)g3 (x) cgn21 (x)gnr(x)
Calculus and Vectors Solutions Manual
f(x) 5 (1 1 x)(1 1 2x)(1 1 3x) c
(1 1 nx)
f r(x) 5 1(1 1 2x)(1 1 3x) c(1 1 nx)
1 (1 1 x)(2)(1 1 3x) c(1 1 nx)
1 (1 1 x)(1 1 2x)(3) c(1 1 nx)
1 c 1 (1 1 x)(1 1 2x)(1 1 3x)
c (n)
f r(0) 5 1(1)(1)(1) c (1)
1 1(2)(1)(1) c (1)
1 1(1)(3)(1) c (1)
1 c 1 (1)(1)(1) c (n)
5112131c1n
n(n 1 1)
f r(0) 5
2
12. f(x) 5 ax2 1 bx 1 c
f r(x) 5 2ax 1 b
(1)
Horizontal tangent at (21, 28)
f r(x) 5 0 at x 5 21
22a 1 b 5 0
Since (2, 19) lies on the curve,
4a 1 2b 1 c 5 19
(2)
Since (21, 28) lies on the curve,
a 2 b 1 c 5 28
(3)
4a 1 2b 1 c 5 19
23a 2 3b 5 227
a1b59
22a 1 b 5 0
3a 5 9
a 5 3, b 5 6
3 2 6 1 c 5 28
c 5 25
The equation is y 5 3x2 1 6x 2 5.
13.
y
3
2
1
x
0
–3 –2 –1
1 2 3
–1
b.
a. x 5 1 or x 5 21
b. f r(x) 5 2x, x , 21 or x . 1
f r(x) 5 22x, 21 , x , 1
6
4
2
–3 –2 –1 0
–2
–4
–6
y
x
1 2 3
2-19
c. f r(22) 5 2(22) 5 24
f r(0) 5 22(0) 5 0
f r(3) 5 2(3) 5 6
16
14. y 5 2 2 1
x
dy
32
52 3
dx
x
Slope of the line is 4.
32
2 3 54
x
4x3 5 232
x3 5 28
x 5 22
16
y5
21
4
53
Point is at (22, 3).
Find intersection of line and curve:
4x 2 y 1 11 5 0
y 5 4x 1 11
Substitute,
16
4x 1 11 5 2 2 1
x
4x3 1 11x2 5 16 2 x2 or 4x3 1 12x2 2 16 5 0.
Let x 5 22
RS 5 4(22)3 1 12(22)2 2 16
50
Since x 5 22 satisfies the equation, therefore it is
a solution.
When x 5 22, y 5 4(22) 1 11 5 3.
Intersection point is (22, 3). Therefore, the line is
tangent to the curve.
Mid-Chapter Review, pp. 92–93
1. a.
6
4
2
y
x
–2 0
–2
–4
–6
2 4 6
˛
x
Slope of Tangent
f 9(x)
0
25
1
23
2
21
3
1
4
3
5
5
y
c.
6
4
2
–2 0
–2
–4
–6
x
2 4 6
d. f(x) is quadratic; f r(x) is linear.
(6(x 1 h) 1 15) 2 (6x 1 15)
2. a. f r(x) 5 lim
hS0
h
6h
5 lim
hS0 h
5 lim 6
˛
hS0
56
A2(x 1 h)2 2 4B 2 A2x2 2 4B
b. f r(x) 5 lim
hS0
h
(x 1 h) 2 2 x2
5 lim2
hS0
h
((x 1 h) 2 x)((x 1 h) 1 x)
5 lim2
hS0
h
h(2x 1 h)
5 lim2
hS0
h
5 lim2(2x 1 h)
hS0
5 4x
((x 1 h)2 2 5(x 1 h)) 2 (x2 2 5x)
b. f r(x) 5 lim
hS0
h
x2 1 2hx 1 h2 2 5x 2 5h 2 x2 1 5x
5 lim
hS0
h
h2 1 2hx 2 5h
5 lim
hS0
h
˛
˛
˛
2-20
h(h 1 2x 2 5)
hS0
h
5 2x 2 5
Use the derivative function to calculate the slopes of
the tangents.
5 lim
c. f r(x) 5 lim
5
5
2x15
(x 1 h) 1 5
h
5(x 1 5) 2 5((x 1 h) 1 5)
5 lim
hS0
((x 1 h) 1 5)(x 1 5)h
25h
5 lim
h S 0 ((x 1 h) 1 5) (x 1 5)h
hS0
Chapter 2: Derivatives
25
h S 0 ((x 1 h) 1 5) (x 1 5)
25
5
(x 1 5) 2
!(x 1 h) 2 2 2 !x 2 2
d. f r(x) 5 lim
hS0
h
5 lim
5 lim c
hS0
!(x 1 h) 2 2 2 !x 2 2
h
3
!(x 1 h) 2 2 1 !x 2 2
d
!(x 1 h) 2 2 1 !x 2 2
((x 1 h) 2 2) 2 (x 2 2)
1 h) 2 2 1 !x 2 2B
hS0 hA !(x
5 lim
h
hS0 hA !(x 1 h) 2 2 1 !x 2 2B
5 lim
1
hS0 !(x 1 h) 2 2 1 !x 2 2
1
5
2 !x 2 2
3. a. yr 5 2x 2 4
When x 5 1,
yr 5 2(1) 2 4
5 22.
When x 5 1,
y 5 (1)2 2 4(1) 1 3
5 0.
Equation of the tangent line:
y 2 0 5 22(x 2 1), or y 5 22x 1 2
b.
y
6
4
2
x
0
–4 –2
2 4 6
–2
–4
–6
5 lim
dy
5 24x3
dx
dy
1
b.
5 5x22
dx
5
5
!x
c. gr(x) 5 26x24
6
52 4
x
4. a.
Calculus and Vectors Solutions Manual
dy
5 5 2 6x23
dx
6
552 3
x
dy
5 2(11t 1 1)(11)
e.
dt
5 242t 1 22
1
f. y 5 1 2
x
5 1 2 x21
dy
5 x22
dx
1
5 2
x
5. f r(x) 5 8x3
8x3 5 1
1
x3 5
8
1
x5
2
1
1 4
f a b 5 2a b
2
2
1
5
8
Equation of the tangent line:
3
1
1
y 2 5 1ax 2 b, or y 5 x 2
8
2
8
6. a. f r(x) 5 8x 2 7
b. f r(x) 5 26x2 1 8x 1 5
c. f(x) 5 5x22 2 3x23
f r(x) 5 210x23 1 9x24
10
9
52 3 1 4
x
x
1
1
d. f(x) 5 x2 1 x3
1 1
1 2
f r(x) 5 x22 1 x23
2
3
1
1
5 12 1 23
2x
3x
1
e. f(x) 5 7x22 2 3x2
3 1
f r(x) 5 214x23 2 x22
2
14
3
5 2 3 2 12
x
2x
f. f r(x) 5 4x22 1 5
4
5 215
x
d.
2-21
7. a. yr 5 26x 1 6
When x 5 1,
yr 5 26(1) 1 6
5 0.
When x 5 1,
y 5 23A12 B 1 6(1) 1 4
5 7.
Equation of the tangent line:
y 2 7 5 0(x 2 1), or
y57
1
b. y 5 3 2 2x2
1
yr 5 2x22
21
5
!x
When x 5 9,
21
yr 5
!9
1
52 .
3
When x 5 9,
y 5 3 2 2 !9
5 23.
Equation of the tangent line:
1
1
y 2 (23) 5 2 (x 2 9), or y 5 2 x
3
3
c. f r(x) 5 28x3 1 12x2 2 4x 2 8
f r(3) 5 28(3)3 1 12(3)2 2 4(3) 2 8
5 2216 1 108 2 12 2 8
5 2218
f(3) 5 22(3)4 1 4(3)3 2 2(3)2 2 8(3) 1 9
5 2162 1 108 2 18 2 24 1 9
5 287
Equation of the tangent line:
y 2 (287) 5 2128(x 2 3), or
y 5 2128x 1 297
d
8. a. f r(x) 5 A4x2 2 9xB A3x2 1 5B
dx
d
1 A4x2 2 9xB A3x2 1 5B
dx
5 (8x 2 9)A3x2 1 5B 1 A4x2 2 9xB (6x)
5 24x3 2 27x2 1 40x 2 45
1 24x3 2 54x2
5 48x3 2 81x2 1 40x 2 45
d
b. f r(t) 5 A23t2 2 7t 1 8B (4t 2 1)
dt
d
1 A23t2 2 7t 1 8B (4t 2 1)
dt
5 (26t 2 7)(4t 2 1)
1 A23t2 2 7t 1 8B (4)
2-22
5 224t2 2 28t 1 6t 1 7 2 12t2 2 28t 1 32
5 236t2 2 50t 1 39
dy
d
c.
5
A3x2 1 4x 2 6B (2x2 2 9)
dx
dx
d
1 A3x2 1 4x 2 6B A2x2 2 9B
dx
5 (6x 1 4)A2x2 2 9B 1 A3x2 1 4x 2 6B (4x)
5 12x3 2 54x 1 8x2 2 36 1 12x3
1 16x2 2 24x
5 24x3 1 24x2 2 78x 2 36
dy
d
d.
5
A3 2 2x3 B 2 A3 2 2x3 B
dx
dx
d
1 A3 2 2x3 B 2 A3 2 2x3 B
dx
d
5 c A3 2 2x3 B A3 2 2x3 B
dx
d
1 A3 2 2x3 B A3 2 2x3 B d A3 2 2x3 B
dx
1 A3 2 2x3 B 2 A26x2 B
5 S 2A26x2 B A3 2 2x3 B T A3 2 2x3 B
1 A3 2 2x3 B 2 A26x2 B
5 3A3 2 2x3 B 2 A26x2 B
5 A3 2 2x3 B 2 A218x2 B
5 A9 2 12x3 1 4x6 BA218x2 B
5 2162x2 1 216x5 2 72x8
d
9. yr 5 A5x2 1 9x 2 2B A2x2 1 2x 1 3B
dx
d
1 A5x2 1 9x 2 2B A2x2 1 2x 1 3B
dx
5 (10x 1 9)A2x2 1 2x 1 3B
1 A5x2 1 9x 2 2B (2 2 2x)
yr(1) 5 (10(1) 1 9)(2 (1)2 1 2(1) 1 3)
1 (5(1)2 1 9(1) 2 2)(2 2 2(1))
5 (19)(4)
5 76
Equation of the tangent line:
y 2 48 5 76(x 2 1), or 76x 2 y 2 28 5 0
dy
d
10.
5 2 (x 2 1)(5 2 x)
dx
dx
d
1 2(x 2 1) (5 2 x)
dx
5 2(5 2 x) 2 2(x 2 1)
5 12 2 4x
dy
The tangent line is horizontal when dx 5 0.
12 2 4x 5 0
12 5 4x
x53
Chapter 2: Derivatives
When x 5 3,
y 5 2((3) 2 1)(5 2 (3))
5 8.
Point where tangent line is horizontal: (3, 8)
dy
(5(x 1 h)2 2 8(x 1 h) 1 4)
11.
5 lim c
dx hS0
h
A5x2 2 8x 1 4B
2
d
h
5(x 1 h)2 2 5x2 2 8h
5 lim
hS0
h
5((x 1 h) 2 x)((x 1 h) 1 x) 2 8h
5 lim
hS0
h
5h(2x 1 h) 2 8h
5 lim
hS0
h
5 lim (5(2x 1 h) 2 8)
t1h
5 lim
3
5 lim
30 2
b
90
1 2
5 500a b
3
500
5
L
9
b. V(0) 5 500(1 2 0)2 5 500 L
500
V(60) 5
L
9
The average rate of change of volume with respect
to time from 0 min to 60 min is
500
DV
2 500
5 9
Dt
60 2 0
28
(500)
5 9
60
200
52
L>min
27
c. Calculate Vr(t):
V(t 1 h) 2 V(t)
Vr(t) 5 lim
hS0
h
5 lim
hS0
500a1 2 90 b 2 500a21 1 90 b
t
h
Calculus and Vectors Solutions Manual
h
500a2 b a2 2
h
90
2t 1 h
b
90
h
500
2t 1 h
5 lim 2
a2 2
b
hS0
90
90
250
2t
5
a2 2 b
9
90
2900 1 10t
5
81
hS0
2900 1 10(30)
81
200
52
L>min
27
4
13. V(r) 5 pr3
3
4
4
V(15) 5 p(15)3
a. V(10) 5 p(10)3
3
3
4
4
5 p(3375)
5 p(1000)
3
3
4000
p
5
5 4500p
3
Then, the average rate of change of volume with
respect to radius is
DV
4500p 2 4000
3 p
5
Dr
15 2 10
500p Q 9 2 83 R
5
5
19
5 100pa b
3
1900
p cm3>cm
5
3
b. First calculate Vr(r):
V(r 1 h) 2 V(r)
Vr(r) 5 lim
hS0
h
4
p3(r
1
h)3 2 r34
5 lim 3
hS0
h
4
3
p
r
1
3r2h 1 3rh2 1 h3 2 r3 R
Q
3
5 lim
hS0
h
4
2
2
3
3 p Q 3r h 1 3rh 1 h R
5 lim
hS0
h
Vr(30) 5
5 500a
2
t1h
t
a1 2 90 1 1 2 90 b
Then,
t 2
12. V(t) 5 500a1 2 b . 0 # t # 90
90
a. After 1 h, t 5 60, and the volume is
2
V(60) 5 500 Q 1 2 60
90 R
t1h
t
h
hS0
hS0
5 10x 2 8
500a1 2 90 2 1 1 90 b
2
2-23
4
5 lim pA3r2 1 3rh 1 h2 B
hS0 3
4
5 pA3r2 1 3r(0) 1 (0)2 B
3
5 4pr2
Then, Vr(8) 5 4p(8)2
5 4p(64)
5 256p cm3>cm
14. This statement is always true. A cubic polynomial
function will have the form f(x) 5 ax3 1 bx2 1
cx 1 d, a 2 0. So the derivative of this cubic is
f r(x) 5 3ax2 1 2bx 1 c, and since 3a 2 0, this
derivative is a quadratic polynomial function. For
example, if f(x) 5 x3 1 x2 1 1,
we get
f r(x) 5 3x2 1 2x,
and if
f(x) 5 2x3 1 3x2 1 6x 1 2,
we get
f r(x) 5 6x2 1 6x 1 6
x2a13b
15. y 5 a2b , a, bPI
x
Simplifying,
y 5 x2a13b2 (a2b) 5 xa14b
Then,
yr 5 (a 1 4b)a14b21
16. a. f(x) 5 26x3 1 4x 2 5x2 1 10
f r(x) 5 218x2 1 4 2 10x
Then, f r(x) 5 218(3)2 1 4 2 10(3)
5 2188
b. f r(3) is the slope of the tangent line to f(x) at
x 5 3 and the rate of change in the value of f(x)
with respect to x at x 5 3.
17. a. P(t) 5 100 1 120t 1 10t2 1 2t3
P(t) 5 100 1 120t 1 10t2 1 2t3
P(0) 5 100 1 120(0) 1 10(0)2 1 2(0)3
5 100 bacteria
b. At 5 h, the population is
P(5) 5 100 1 120(5) 1 10(5)2 1 2(5)3
5 1200 bacteria
c. Pr(t) 5 120 1 20t 1 6t2
At 5 h, the colony is growing at
Pr(5) 5 120 1 20(5) 1 6(5)2
5 370 bacteria> h
100
18. C(t) 5
,t.2
t
Simplifying, C(t) 5 100t21.
100
Then, Cr(t) 5 2100t22 5 2 2 .
t
2-24
Cr(5)
Cr(50)
Cr(100)
100
100
100
52
52
52
2
2
(5)
(50)
(100)2
1
100
100
52
52
52
25
2500
100
5 24
5 20.04
5 20.01
These are the rates of change of the percentage with
respect to time at 5, 50, and 100 min. The percentage
of carbon dioxide that is released per unit time from
the pop is decreasing. The pop is getting flat.
2.4 The Quotient Rule, pp. 97–98
1. For x, a, b real numbers,
xaxb 5 xa1b
For example,
x9x26 5 x3
Also,
Axa B b 5 xab
For example,
Ax2 B 3 5 x6
Also,
xa
5 xa2b, x 2 0
xb
For example,
x5
5 x2
x3
2.
Function
x2 1 3x
,
x
x20
f(x) 5
Rewrite
Differentiate
and Simplify,
If Necessary
f(x) 5 x 1 3
f r(x) 5 1
5
g(x) 5
3x3
,x20
x
1
,
10x5
x20
h(x) 5
y5
8x3 1 6x
,
2x
h(x) 5
1 25
x
10
hr(x) 5
21 26
x
2
y 5 4x2 1 3
dy
5 8x
dx
s5t13
ds
51
dt
x20
t2 2 9
s5
,t23
t23
1
gr(x) 5 2x23
2
g(x) 5 3x3
Chapter 2: Derivatives
3. In the previous problem, all of these rational
examples could be differentiated via the power rule
after a minor algebraic simplification.
A second approach would be to rewrite a rational
example
f(x)
h(x) 5
g(x)
using the exponent rules as
h(x) 5 f(x)(g(x))21,
and then apply the product rule for differentiation
(together with the power of a function rule to find
hr(x).
A third (and perhaps easiest) approach would be to
just apply the quotient rule to find hr(x).
(x 1 1)(1) 2 x(1)
4. a. hr(x) 5
Ax 1 1B 2
1
5
Ax 1 1B 2
(t 1 5)(2) 2 (2t 2 3)(1)
b. hr(t) 5
At 1 5B 2
13
5
At 1 5B 2
A2x2 2 1B A3x2 B 2 x3 A4xB
c. hr(x) 5
A2x2 2 1B 2
4
2x 2 3x2
5
A2x2 2 1B 2
Ax2 1 3B (0) 2 1(2x)
d. hr(x) 5
Ax2 1 3B 2
22x
5 2
Ax 1 3B 2
x(3x 1 5)
3x2 1 5x
e. y 5
5
2
A1 2 x B
1 2 x2
dy
(6x 1 5)A1 2 x2 B 2 A3x2 1 5xB (22x)
5
dx
A1 2 x2 B 2
6x 1 5 2 6x3 2 5x2 1 6x3 1 10x2
5
A1 2 x2 B 2
5x2 1 6x 1 5
5
A1 2 x2 B 2
dy
Ax2 1 3B A2x 2 1B 2 Ax2 2 x 1 1B (2x)
f.
5
dx
Ax2 1 3B 2
2x3 1 6x 2 x2 2 3 2 2x3 1 2x2 2 2x
5
Ax2 1 3B 2
x2 1 4x 2 3
5
Ax2 1 3B 2
Calculus and Vectors Solutions Manual
3x 1 2
, x 5 23
x15
dy
(x 1 5)(3) 2 (3x 1 2)(1)
5
dx
(x 1 5)2
At x 5 23:
dy
(2)(3) 2 (27)(1)
5
dx
A2B 2
13
5
4
x3
b. y 5 2
,x51
x 19
dy
Ax2 1 9B A3x2 B 2 Ax3 B (2x)
5
dx
Ax2 1 9B 2
At x 5 1:
dy
(10)(3) 2 (1)(2)
5
dx
A10B 2
28
5
100
7
5
25
x2 2 25
c. y 5 2
,x52
x 1 25
dy
2xAx2 1 25B 2 Ax2 2 25B (2x)
5
dx
Ax2 1 25B 2
At x 5 2:
dy
4(29) 2 (221)(4)
5
dx
A29B 2
116 1 84
5
292
200
5
841
(x 1 1)(x 1 2)
,x54
d. y 5
(x 2 1)(x 2 2)
x2 1 3x 1 2
5 2
x 2 3x 1 2
dy
(2x 1 3)Ax2 2 3x 1 2B
5
dx
Ax 2 1B 2 Ax 2 2B 2
Ax2 1 3x 1 2B (2x 2 3)
2
Ax 2 1B 2 Ax 2 2B 2
At x 5 4:
dy
(11)(6) 2 (30)(5)
5
dx
(9)(4)
84
52
36
7
52
3
5. a. y 5
2-25
x3
x2 2 6
dy
3x2 Ax2 2 6B 2 x3 A2xB
5
dx
Ax2 2 6B 2
At (3, 9):
3(9)(3) 2 (27)(6)
dy
5
dx
A3B 2
5 9 2 18
5 29
The slope of the tangent to the curve at (3, 9) is 29.
3x
7. y 5
x24
dy
3(x 2 4) 2 3x
12
5
52
2
dx
Ax 2 4B
Ax 2 4B 2
12
Slope of the tangent is 2 25.
6. y 5
Therefore,
12
2 5
(x 2 4)
dy
12
25
x 2 4 5 5 or x 2 4 5 25
x 5 9 or x 5 21
Points are Q 9, 275 R and Q 21, 35 R .
5x 1 2
8. f(x) 5
x12
(x 1 2)(5) 2 (5x 1 2)(1)
f r(x) 5
Ax 1 2B 2
8
f r(x) 5
Ax 1 2B 2
Since Ax 1 2B 2 is positive or zero for all xPR,
8
.0
(x 1 2)2
for x 2 22. Therefore, tangents to
5x 1 2
the graph of f(x) 5 x 1 2 do not have a negative
slope.
2x2
9. a. y 5
,x24
x24
dy
(x 2 4)(4x) 2 A2x2 B (1)
5
dx
Ax 2 4B 2
2
4x 2 16x 2 2x2
5
(x 2 4)2
2
2x 2 16x
5
(x 2 4)2
2x(x 2 8)
5
(x 2 4)2
dy
Curve has horizontal tangents when dx 5 0, or
when x 5 0 or 8. At x 5 0:
0
y5
24
50
At x 5 8:
2-26
2(8)2
4
5 32
So the curve has horizontal tangents at the points
(0, 0) and (8, 32).
x2 2 1
b. y 5 2
x 1x22
(x 2 1)(x 1 1)
5
(x 1 2)(x 2 1)
x11
5
,x21
x12
dy
(x 1 2) 2 (x 1 1)
5
dx
Ax 1 2B 2
1
5
Ax 1 2B 2
y5
Curve has horizontal tangents when dx 5 0.
No value of x will produce a slope of 0, so there
are no horizontal tangents.
4t
10. p(t) 5 1000a1 1 2
b
t 1 50
4At 2 1 50B 2 4t(2t)
pr(t) 5 1000a
b
At 2 1 50B 2
1000A200 2 4t 2 B
5
At 2 1 50B 2
1000(196)
5 75.36
pr(1) 5
A51B 2
1000(184)
5 63.10
pr(2) 5
A54B 2
Population is growing at a rate of 75.4 bacteria per
hour at t 5 1 and at 63.1 bacteria per hour at t 5 2.
x2 2 1
11. y 5
3x
1
1
5 x 2 x21
3
3
dy
1
1 22
5 1 x
dx
3
3
1
1
5 1 2
3
3x
At x 5 2:
A2B 2 2 1
y5
3(2)
1
5
2
and
dy
1
1
5 1
dx
3
3A2B 2
Chapter 2: Derivatives
1
1
1
3
12
5
5
12
So the equation of the tangent to the curve at x 5 2 is:
1
5
y 2 5 (x 2 2), or 5x 2 12y 2 4 5 0.
2
12
10(6 2 t)
, 0 # t # 6, t 5 0,
12. a. s(t) 5
t13
s(0) 5 20
The boat is initially 20 m from the dock.
(t 1 3)(21) 2 (6 2 t)(1)
b. v(t) 5 sr(t) 5 10 c
d
At 1 3B 2
290
v(t) 5
At 1 3B 2
At t 5 0, v(0) 5 210, the boat is moving towards
the dock at a speed of 10 m> s. When s(t) 5 0, the
boat will be at the dock.
10(6 2 t)
5 0, t 5 6.
t13
290
10
v(6) 5 2 5 2
9
9
The speed of the boat when it bumps into the dock
is 109 m> s.
13. a. i. t 5 0
1 1 2(0)
r(0) 5
110
5 1 cm
1 1 2t
ii.
5 1.5
11t
1 1 2t 5 1.5(1 1 t)
1 1 2t 5 1.5 1 1.5t
0.5t 5 0.5
t 5 1s
(1 1 t)(2) 2 (1 1 2t)(1)
iii. rr(t) 5
A1 1 tB 2
2 1 2t 2 1 2 2t
5
A1 1 tB 2
1
5
A1 1 tB 2
1
rr(1.5) 5
A1 1 1B 2
1
5
4
5 0.25 cm> s
b. No, the radius will never reach 2 cm, because
y 5 2 is a horizontal asymptote of the graph of the
function. Therefore, the radius approaches but never
equals 2 cm.
5
Calculus and Vectors Solutions Manual
ax 1 b
(x 2 1)(x 2 4)
(x 2 1)(x 2 4)(a)
f r(x) 5
Ax 2 1B 2 Ax 2 4B 2
d
(ax 1 b) 3(x 2 1)(x 2 4)4
dx
2
Ax 2 1B 2 Ax 2 4B 2
(x 2 1)(x 2 4)(a)
5
Ax 2 1B 2 Ax 2 4B 2
(ax 1 b)3(x 2 1) 1 (x 2 4)4
2
Ax 2 1B 2 Ax 2 4B 2
2
(x 2 5x 1 4)(a) 2 (ax 1 b)(2x 2 5)
5
(x 2 1)2 (x 2 4)2
2
2ax 2 2bx 1 4a 1 5b
5
(x 2 1)2 (x 2 4)2
Since the point (2, 21) is on the graph (as it’s on
the tangent line) we know that
21 5 f(2)
2a 1 b
5
(1)(22)
2 5 2a 1 b
b 5 2 2 2a
Also, since the tangent line is horizontal at (2, 21),
we know that
0 5 f r(2)
2aA2B 2 2 2b(2) 1 4a 1 5b
5
A1B 2 A22B 2
b50
0 5 2 2 2a
a51
So we get
x
f(x) 5
(x 2 1)(x 2 4)
Since the tangent line is horizontal at the point
(2, 21), the equation of this tangent line is
y 2 (21) 5 0(x 2 2), or y 5 21
Here are the graphs of both f(x) and this horizontal
tangent line:
x
f (x) =
(x – 1) (x –4)
y
8
6
4
2
x
0
–2 –1
1 2 3 4 5
–2
y =–1
–4
–6
14. f(x) 5
2-27
A2t 2 1 7B (5) 2 (5t)(4t)
A2t 2 1 7B 2
10t 2 1 35 2 20t 2
5
A2t 2 1 7B 2
210t 2 1 35
5
A2t 2 1 7B 2
Set cr(t) 5 0 and solve for t.
210t 2 1 35
50
(2t 2 1 7)2
210t 2 1 35 5 0
10t 2 5 35
t 2 5 3.5
15. cr(t) 5
t 5 6"3.5
t 8 61.87
To two decimal places, t 5 21.87 or t 5 1.87,
because sr(t) 5 0 for these values. Reject the
negative root in this case because time is positive
(t $ 0). Therefore, the concentration reaches its
maximum value at t 5 1.87 hours.
16. When the object changes direction, its velocity
changes sign.
At 2 1 8B (1) 2 t(2t)
sr(t) 5
At 2 1 8B 2
2
t 1 8 2 2t 2
5
(t 2 1 8)2
2t 2 1 8
5 2
(t 1 8)2
solve for t when sr(t) 5 0.
2t 2 1 8
50
At 2 1 8B 2
2t 2 1 8 5 0
t2 5 8
t 5 6"8
t 8 62.83
To two decimal places, t 5 2.83 or t 5 22.83,
because sr(t) 5 0 for these values. Reject the
negative root because time is positive (t $ 0).
The object changes direction when t 5 2.83 s.
d
ax 1 b
17. f(x) 5
,x22
c
cx 1 d
(cx 1 d)(a) 2 (ax 1 b)(c)
f r(x) 5
Acx 1 dB 2
ad 2 bc
f r(x) 5
Acx 1 dB 2
For the tangents to the graph of y 5 f(x) to have
positive slopes, f r(x) . 0. (cx 1 d)2 is positive for
all xPR. ad 2 bc . 0 will ensure each tangent has
a positive slope.
2-28
2.5 The Derivatives of Composite
Functions, pp. 105–106
1. f(x) 5 !x, g(x) 5 x2 2 1
a. f(g(1)) 5 f(1 2 1)
5 f(0)
50
b. g(f(1)) 5 g(1)
50
c. g(f(0)) 5 g(0)
5021
5 21
d. f(g(24)) 5 f(16 2 1)
5 f(15)
5 !15
e. f(g(x)) 5 f Ax2 2 1B
5 "x2 2 1
f. g(f(x)) 5 gA !xB
5 A !xB 2 2 1
5x21
2. a. f(x) 5 x2, g(x) 5 !x
(f + g)(x) 5 f(g(x))
5 f A !xB
5 A !xB 2
5x
Domain 5 5x $ 06
(g + f)(x) 5 g(f(x))
5 gAx2 B
5 "x2
5 Zx Z
Domain 5 5xPR6
The composite functions are not equal for negative
x-values (as (f + g) is not defined for these x), but
are equal for non-negative x-values.
1
b.
f(x) 5 , g(x) 5 x2 1 1
x
(f + g)(x) 5 f(g(x))
5 f Ax2 1 1B
1
5 2
x 11
Domain 5 5xPR6
(g + f)(x) 5 g(f(x))
1
5 ga b
x
1 2
5a b 11
x
Chapter 2: Derivatives
1
11
x2
Domain 5 5x 2 06
The composite functions are not equal here. For
instance, (f + g)(1) 5 12 and (g + f )(1) 5 2.
1
c.
f(x) 5 , g(x) 5 !x 1 2
x
( f + g)(x) 5 f(g(x))
5
5 f( !x 1 2)
1
5
!x 1 2
Domain 5 5x . 226
(g + f)(x) 5 g(f(x))
1
5 ga b
x
1
12
Åx
The domain is all x such that
5
1
12$0
x
and x 2 0, or equivalently
Domain 5 5x # 2 12 or x . 06
The composite functions are not equal here. For
instance, ( f + g)(2) 5 12 and (g + f )(2) 5 # 52.
3. If f(x) and g(x) are two differentiable functions
of x, and
h(x) 5 (f + g)(x)
5 f(g(x))
is the composition of these two functions, then
hr(x) 5 f r(g(x)) ? gr(x)
This is known as the “chain rule” for differentiation of
composite functions. For example, if f(x) 5 x10 and
g(x) 5 x2 1 3x 1 5, then h(x) 5 Ax2 1 3x 1 5B 10,
and so
hr(x) 5 f r(g(x)) ? gr(x)
5 10Ax2 1 3x 1 5B 9 (2x 1 3)
2
As another example, if f(x) 5 x3 and
2
g(x) 5 x2 1 1, then h(x) 5 Ax2 1 1B 3,
and so
2
1
hr(x) 5 Ax2 1 1B 23 (2x)
3
4. a. f(x) 5 (2x 1 3)4
f r(x) 5 4A2x 1 3B 3 (2)
5 8A2x 1 3B 3
b. g(x) 5 Ax2 2 4B 3
gr(x) 5 3Ax2 2 4B 2 (2x)
5 6xAx2 2 4B 2
Calculus and Vectors Solutions Manual
c. h(x) 5 A2x2 1 3x 2 5B 4
hr(x) 5 4A2x2 1 3x 2 5B 3 (4x 1 3)
d. f(x) 5 Ap2 2 x2 B 3
f r(x) 5 3Ap2 2 x2 B 2 (22x)
5 26xAp2 2 x2 B 2
e. y 5 "x2 2 3
1
5 Ax2 2 3B 2
1
1
yr 5 Ax2 2 3B 2 (2x)
2
x
5
2
"x 2 3
1
f. f(x) 5 2
Ax 2 16B 5
5 Ax2 2 16B 25
f r(x) 5 25Ax2 2 16B 26 (2x)
210x
5 2
Ax 2 16B 6
2
5. a. y 5 2 3
x
5 22x23
dy
5 (22)(23)x24
dx
6
5 4
x
1
b. y 5
x11
5 (x 1 1)21
dy
5 (21)(x 1 1)22 (1)
dx
21
5
Ax 1 1B 2
1
c. y 5 2
x 24
5 Ax2 2 4B 21
dy
5 (21)Ax2 2 4B 22 (2x)
dx
22x
5 2
Ax 2 4B 2
3
5 3A9 2 x2 B 21
d. y 5
9 2 x2
dy
6x
5
dx
A9 2 x2 B 2
2-29
1
5x 1 x
5 A5x2 1 xB 21
e. y 5
2
dy
5 (21)A5x2 1 xB 22 (10x 1 1)
dx
10x 1 1
52 2
A5x 1 xB 2
1
f. y 5 2
Ax 1 x 1 1B 4
5 Ax2 1 x 1 1B 24
dy
5 (24)Ax2 1 x 1 1B 25 (2x 1 1)
dx
8x 1 4
52 2
Ax 1 x 1 1B 5
h5g+f
6.
5 g(f(x))
h(21) 5 g(f(21))
5 g(1)
5 24
h(x) 5 g(f(x))
hr(x) 5 gr(f(x))f r(x)
hr(21) 5 gr(f(21))f r(21)
5 gr(1)(25)
5 (27)(25)
5 35
1
7. f(x) 5 (x 2 3)2, g(x) 5 , h(x) 5 f(g(x)),
x
1
f r(x) 5 2(x 2 3), gr(x) 5 2 2
x
hr(x) 5 f r(g(x))gr(x)
1
1
5 f ra b a2 2 b
x
x
1
1
5 2a 2 3b a2 2 b
x
x
2 1
5 2 2 a 2 3b
x x
8. a. f(x) 5 Ax 1 4B 3 Ax 2 3B 6
d
3Ax 1 4B 34 ? Ax 2 3B 6
f r(x) 5
dx
d
1 Ax 1 4B 3 3Ax 2 3B 64
dx
5 3Ax 1 4B 2 Ax 2 3B 6
1 Ax 1 4B 3 (6)Ax 2 3B 5
5 Ax 1 4B 2 Ax 2 3B 5
3 33(x 2 3) 1 6(x 1 4)4
5 Ax 1 4B 2 Ax 2 3B 5 (9x 1 15)
2-30
b. y 5 Ax2 1 3B 3 Ax3 1 3B 2
dy
d 2
5
3Ax 1 3B 34 ? Ax3 1 3B 2
dx
dx
d 3
1 Ax2 1 3B 3 ?
3Ax 1 3B 24
dx
5 3Ax2 1 3B 2 (2x)Ax3 1 3B 2
1 Ax2 1 3B 3 (2)Ax3 1 3BA3x2 B
5 6xAx2 1 3B 2 Ax3 1 3B 3Ax3 1 3B 1 xAx2 1 3B4
5 6xAx2 1 3B 2 Ax3 1 3B A2x3 1 3x 1 3B
3x2 1 2x
c. y 5 2
x 11
dy
A6x 1 2B Ax2 1 1B 2 A3x2 1 2xBA2xB
5
dx
Ax2 1 1B 2
3
2
6x 1 2x 1 6x 1 2 2 6x3 2 4x2
5
Ax2 1 1B 2
2
22x 1 6x 1 2
5
Ax2 1 1B 2
d. h(x) 5 x3 A3x 2 5B 2
d 3
d
hr(x) 5
3x 4 ? A3x 2 5B 2 1 x3 3A3x 2 5B 24
dx
dx
5 3x2 A3x 2 5B 2 1 x3 (2)(3x 2 5)(3)
5 3x2 (3x 2 5)3(3x 2 5) 1 2x4
5 3x2 (3x 2 5)(5x 2 5)
5 15x2 (3x 2 5)(x 2 1)
4
e. y 5 x A1 2 4x2 B 3
dy
d 4
d
5
3x 4A1 2 4x2 B 3 1 x4 ?
3A1 2 4x2 B 34
dx
dx
dx
5 4x3 A1 2 4x2 B 3 1 x4 (3)A1 2 4x2 B 2 A28xB
5 4x3 A1 2 4x2 B 2 3A1 2 4x2 B 2 6x24
5 4x3 A1 2 4x2 B 2 A1 2 10x2 B
x2 2 3 4
b
f. y 5 a 2
x 13
dy
x2 2 3 3 d x2 2 3
5 4a 2
b
c
d
dx
x 1 2 dx x2 1 3
x2 2 3 3 Ax2 1 3B (2x) 2 Ax2 2 3B (2x)
5 4a 2
b ?
x 13
Ax2 1 3B 2
x2 2 3 3
12x
5 4a 2
b ? 2
x 13
Ax 1 3B 2
48xAx2 2 3B 3
5
Ax2 1 3B 5
1
2
9. a. s(t) 5 t3 (4t 2 5)3
1
1
5 t3 3(4t 2 5)24 3
1
5 3t(4t 2 5)24 3
1
5 3t A16t2 2 40t 1 25B4 3
1
5 A16t3 2 40t2 1 25tB 3, t 5 8
Chapter 2: Derivatives
1
2
sr(t) 5 A16t3 2 40t2 1 25tB 23
3
3 A48t2 2 80t 1 25B
A48t2 2 80t 1 25B
5
2
3A16t3 2 40t2 1 25tB 3
Rate of change at t 5 8:
(48(8)2 2 80(8) 1 25)
sr(8) 5
2
3(16(8)3 2 40(8)2 1 25(8))3
2457
5
972
91
5
36
1
t2p 3
b. s(t) 5 a
b , t 5 2p
t 2 6p
2
1 t 2 p 23 d t 2 p
sr(t) 5 a
b ? c
d
3 t 2 6p
dt t 2 6p
2
1 t 2 6p 3 (t 2 6p) 2 (t 2 p)
5 a
b ?
3 t2p
(t 2 6p)2
2
1 t 2 6p 3
25p
5 a
b ?
3 t2p
(t 2 6p)2
Rate of change at t 5 2p:
1
25p
2
sr(2p) 5 (24)3 ?
3
16p2
3
5"2
52
24p
y 5 2x6
10. y 5 A1 1 x3 B 2
dy
dy
5 2A1 1 x3 B (3x2 )
5 12x5
dx
dx
For the same slope,
6x2 A1 1 x3 B 5 12x5
6x2 1 6x5 5 12x5
6x2 2 6x5 5 0
6x2 Ax3 2 1B 5 0
x 5 0 or x 5 1.
Curves have the same slope at x 5 0 and x 5 1.
11. y 5 A3x 2 x2 B 22
dy
5 22A3x 2 x2 B 23 (3 2 2x)
dx
At x 5 2,
dy
5 2236 2 44 23 (3 2 4)
dx
5 2(2)23
1
5
4
The slope of the tangent line at x 5 2 is 41.
Calculus and Vectors Solutions Manual
y 5 Ax3 2 7B 5 at x 5 2
dy
5 5Ax3 2 7B 4 A3x2 B
dx
When x 5 2,
dy
5 5(1)4 (12)
dx
5 60
Slope of the tangent is 60.
Equation of the tangent at (2, 1) is
y 2 1 5 60(x 2 2)
60x 2 y 2 119 5 0.
13. a. y 5 3u2 2 5u 1 2
u 5 x2 2 1, x 5 2
u53
du
dy
5 6u 2 5,
5 2x
du
dx
dy
dy
du
5
3
dx
du
dx
5 (6u 2 5)(2x)
5 (18 2 5)(4)
5 13(4)
5 52
1
b. y 5 2u3 1 3u2, u 5 x 1 x2, x 5 1
dy
dy du
5
?
dx
du dx
1
5 (6u2 1 6u)a1 1
b
2!x
At x 5 1:
1
u 5 1 1 12
52
dy
1
b
5 (6(2)2 1 6(2))a1 1
dx
2 !1
3
5 36 3
2
5 54
c. y 5 uAu2 1 3B 3, u 5 (x 1 3)2, x 5 22
du
dy
5 Au2 1 3B 3 1 6u2 Au2 1 3B 2,
5 2(x 1 3)
du
dx
dy
dy du
5
5 373 1 6(4)2432(1)4
dx
du dx
5 439 3 2
5 878
d. y 5 u3 2 5Au3 2 7uB 2,
12.
u 5 "x
1
5 x2, x 5 4
dy
dy du
5
?
dx
du dx
2-31
1 1
5 333u2 2 10Au3 2 7uB A3u2 2 7B4 ? a x2 b
2
1
5 33u2 2 10Au3 2 7uB A3u2 2 7B4 ?
2"x
At x 5 4:
u 5 "4
52
dy
1
5 33(2)2 2 10( (2)3 2 7(2))(3(2)2 2 7)4
dx
2(2)
5 78
14. h(x) 5 f(g(x)), therefore
hr(x) 5 f r(g(x)) 3 gr(x)
f(u) 5 u2 2 1, g(2) 5 3, gr(2) 5 21
Now, hr(2) 5 f r(g(2)) 3 gr(2)
5 f r(3) 3 gr(2).
Since f(u) 5 u2 2 1, f r(u) 5 2u, and f r(3) 5 6,
hr(2) 5 6(21)
5 26.
t 2
15. V(t) 5 50 000a1 2 b
30
1
t
Vr(t) 5 50 000 c2a1 2 b a2 b d
30
30
10
1
Vr(10) 5 50 000 c2a1 2 b a2 b d
30
30
2
1
5 50 000 c2a b a2 b d
3
30
8 2222
At t 5 10 minutes, the water is flowing out of the
tank at a rate of 2222 L> min.
16. The velocity function is the derivative of the
position function.
1
s(t) 5 At 3 1 t 2 B 2
1
1
v(t) 5 sr(t) 5 At 3 1 t 2 B 22 A3t 2 1 2tB
2
3t 2 1 2t
5
2"t 3 1 t 2
3(3)2 1 2(3)
v(3) 5
2"33 1 32
27 1 6
5
2"36
33
5
12
5 2.75
The particle is moving at 2.75 m/s.
17. a. h(x) 5 p(x)q(x)r(x)
hr(x) 5 pr(x)q(x)r(x) 1 p(x)qr(x)r(x)
1 p(x)q(x)rr(x)
2-32
b. h(x) 5 x(2x 1 7)4 (x 2 1)2
Using the result from part a.,
hr(x) 5 (1)(2x 1 7)4 (x 2 1)2
1 x34(2x 1 7)3 (2)4 (x 2 1)2
1 x(2x 1 7)4 32(x 2 1)4
hr(23) 5 1(16) 1 (23)34(1)(2)4 (16)
1 (23)(1)32(24)4
5 16 2 384 1 24
5 2344
18. y 5 Ax2 1 x 2 2B 3 1 3
dy
5 3Ax2 1 x 2 2B 2 (2x 1 1)
dx
At the point (1, 3), x 5 1 and the slope of the
tangent will be 3(1 1 1 2 2)2 (2 1 1) 5 0.
Equation of the tangent at (1, 3) is y 2 3 5 0.
Solving this equation with the function, we have
Ax2 1 x 2 2B 3 1 3 5 3
(x 1 2)3 (x 2 1)3 5 0
x 5 22 or x 5 1
Since 22 and 1 are both triple roots, the line with
equation y 2 3 5 0 will be a tangent at both x 5 1
and x 5 22. Therefore, y 2 3 5 0 is also a tangent
at (22, 3).
x2 (1 2 x)3
19. y 5
(1 1 x)3
12x 3
5 x2 c a
bd
11x
dy
12x 3
12x 2
5 2xa
b 1 3x2 a
b
dx
11x
11x
2 (1 1 x) 2 (1 2 x)(1)
3 c
d
(1 1 x)2
12x 3
12x 2
22
5 2xa
b 1 3x2 a
b c
d
11x
11x
(1 1 x)2
12x 2 12x
3x
5 2xa
b c
2
d
11x
11x
(1 1 x)2
1 2 x 2 1 2 x2 2 3x
5 2xa
b c
d
11x
(1 1 x)2
2xAx2 1 3x 2 1B (1 2 x)2
52
(1 1 x)4
Review Exercise, pp. 110–113
1. To find the derivative f r(x), the limit
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
must be computed, provided it exists. If this limit
does not exist, then the derivative of f (x) does not
Chapter 2: Derivatives
exist at this particular value of x. As an alternative
to this limit, we could also find f r(x) from the
definition by computing the equivalent limit
f(z) 2 f(x)
.
z2x
zSx
f r(x) 5 lim
These two limits are seen to be equivalent by
substituting z 5 x 1 h.
2. a. y 5 2x2 2 5x
dy
(2(x 1 h)2 2 5(x 1 h)) 2 A2x2 2 5xB
5 lim
dx hS0
h
2A(x 1 h)2 2 x2 B 2 5h
5 lim
hS0
h
2((x 1 h) 2 x)((x 1 h) 1 x) 2 5h
5 lim
hS0
h
2h(2x 1 h) 2 5h
5 lim
hS0
h
5 lim (2(2x 1 h) 2 5)
hS0
5 4x 2 5
b. y 5 !x 2 6
!(x 1 h) 2 6 2 !x 2 6
dy
5 lim
dx hS0
h
!(x 1 h) 2 6 2 !x 2 6
5 lim c
hS0
h
!(x 1 h) 2 6 1 !x 2 6
d
3
!(x 1 h) 2 6 1 !x 2 6
( (x 1 h) 2 6) 2 (x 2 6)
5 lim
hS0 h( !(x 1 h) 2 6 1 !x 2 6)
1
5 lim
hS0 !(x 1 h) 2 6 1 !x 2 6
1
5
2 !x 2 6
x
c. y 5
42x
x1h
x
2
dy
4 2 (x 1 h)
42x
5 lim
dx hS0
h
(x 1 h)(4 2 x) 2 x(4 2 (x 1 h))
(4 2 (x 1 h))(4 2 x)
5 lim
hS0
h
4h
5 lim
hS0 h(4 2 (x 1 h))(4 2 x)
4
5 lim
hS0 (4 2 (x 1 h))(4 2 x)
4
5
(4 2 x)2
Calculus and Vectors Solutions Manual
3. a. y 5 x2 2 5x 1 4
dy
5 2x 2 5
dx
3
b. f(x) 5 x4
3 1
f r(x) 5 x24
4
3
5 14
4x
7
c. y 5 4
3x
7
5 x24
3
dy
228 25
5
x
dx
3
28
52 5
3x
1
d. y 5 2
x 15
5 Ax2 1 5B 21
dy
5 (21)Ax2 1 5B 22 ? (2x)
dx
2x
52 2
Ax 1 5B 2
3
e. y 5
A3 2 x2 B 2
5 3A3 2 x2 B 22
dy
5 (26)A3 2 x2 B 23 ? (22x)
dx
12x
5
A3 2 x2 B 3
f. y 5 "7x2 1 4x 1 1
1
5 A7x2 1 4x 1 1B 2
dy
1
1
5 A7x2 1 4x 1 1B 22 A14x 1 4B
dx
2
7x 1 2
5
"7x2 1 4x 1 1
2x3 2 1
4. a. f(x) 5
x2
1
5 2x 2 2
x
5 2x 2 x22
f r(x) 5 2 1 2x23
2
521 3
x
2-33
b. g(x) 5 !xAx3 2 xB
1
5 x2 Ax3 2 xB
7
3
5 x2 2 x2
7 5
3 1
gr(x) 5 x2 2 x2
2
2
!x 2
5
A7x 2 3B
2
x
c. y 5
3x 2 5
dy
(3x 2 5)(1) 2 (x)(3)
5
dx
(3x 2 5)2
5
52
(3x 2 5)2
1
d. y 5 (x 2 1)2 (x 1 1)
1
1
1
yr 5 (x 2 1)2 1 (x 1 1)a b (x 2 1)22
2
x11
5 !x 2 1 1
2!x 2 1
2x 2 2 1 x 1 1
5
2!x 2 1
3x 2 1
5
2!x 2 1
2
e. f(x) 5 A !x 1 2B 23
1
2
5 ( x2 1 2) 23
1
22 12
5 1
f r(x) 5
x 1 2) 23 # x2 2
(
3
2
1
52
5
3 !xA !x 1 2B 3
x2 1 5x 1 4
f. y 5
x14
(x 1 4)(x 1 1)
5
x14
5 x 1 1, x 2 24
dy
51
dx
5. a. y 5 x4 (2x 2 5)6
yr 5 x4 36(2x 2 5)5 (2)4 1 4x3 (2x 2 5)6
5 4x3 (2x 2 5)5 33x 1 (2x 2 5)4
5 4x3 (2x 2 5)5 (5x 2 5)
5 20x3 (2x 2 5)5 (x 2 1)
b. y 5 x"x2 1 1
1
1
yr 5 x c Ax2 1 1B 22 (2x)d 1 (1)"x2 1 1
2
x2
5
1 "x2 1 1
"x2 1 1
2-34
(2x 2 5)4
(x 1 1)3
(x 1 1)34(2x 2 5)3 (2)
yr 5
(x 1 1)6
3(2x 2 5)4 (x 1 1)2
2
(x 1 1)6
(x 1 1)2 (2x 2 5)3 38x 1 8 2 6x 1 154
5
(x 1 1)6
3
(2x 2 5) (2x 1 23)
yr 5
(x 1 1)4
10x 2 1 6
b 5 (10x 2 1)6 (3x 1 5)26
d. y 5 a
3x 1 5
yr 5 (10x 2 1)6 326(3x 1 5)27 (3)4
1 6(10x 2 1)5 (10)(3x 1 5)26
5 (10x 2 1)5 (3x 1 5)27 3x 2 18(10x 2 1)4
1 60(3x 1 5)
5 (10x 2 1)5 (3x 1 5)27
3 (2180x 1 18 1 180x 1 300)
318(10x 2 1)5
5
(3x 1 5)7
e. y 5 (x 2 2)3 Ax2 1 9B 4
yr 5 (x 2 2)3 C4Ax2 1 9B 3 (2x)D
1 3(x 2 2)2 (1)Ax2 1 9B 4
5 (x 2 2)2 Ax2 1 9B 3 C8x(x 2 2) 1 3Ax2 1 9B D
5 (x 2 2)2 Ax2 1 9B 3 A11x2 2 16x 1 27B
f. y 5 A1 2 x2 B 3 (6 1 2x)23
1 2 x2 3
5a
b
6 1 2x
1 2 x2 2
b
yr 5 3a
6 1 2x
c. y 5
3 c
(6 1 2x)(22x) 2 A1 2 x2 B (2)
d
(6 1 2x)2
3A1 2 x2 B 2 A212x 2 4x2 2 2 1 2x2 B
(6 1 2x)4
2 2
3A1 2 x B A2x2 1 12x 1 2B
52
(6 1 2x)4
3A1 2 x2 B 2 Ax2 1 6x 1 1B
52
8(3 2 x)4
2
6. a. g(x) 5 f(x )
gr(x) 5 f(x2 ) 3 2x
b. h(x) 5 2xf(x)
hr(x) 5 2xf r(x) 1 2f(x)
18
7. a. y 5 5u2 1 3u 2 1, u 5 2
x 15
x52
u52
5
Chapter 2: Derivatives
dy
5 10u 1 3
du
du
36x
52 2
dx
Ax 1 5B 2
When x 5 2,
du
72
8
52 52
dx
81
9
When u 5 2,
dy
5 20 1 3
du
5 23
dy
8
5 23a2 b
dx
9
184
52
9
!x 1 x
u14
,u5
,
b. y 5
u24
10
x54
3
u5
5
dy
(u 2 4) 2 (u 1 4)
5
du
(u 2 4)2
du
1 1 1
5 a x22 1 1b
dx
10 2
When x 5 4,
du
8
1 5
52
5 a b
(u 2 4)2 dx
10 4
1
5
8
3
When u 5 ,
5
dy
8
52
du
3
20 2
a 2 b
5
5
8(25)
52
(217)2
When x 5 4,
dy
8(25)
1
5
3
dx
172
8
25
5
289
c. y 5 f("x2 1 9), f r(5) 5 22, x 5 4
dy
1
1
5 f r("x2 1 9) 3 Ax2 1 9B 22 (2x)
dx
2
dy
1 1
5 f r(5) ? ? ? 8
dx
2 5
Calculus and Vectors Solutions Manual
5 22 ?
4
5
8
5
2
8. f(x) 5 A9 2 x2 B 3
2
1
f r(x) 5 A9 2 x2 B 23 (22x)
3
24x
5
1
3(9 2 x2 )3
2
f r(1) 5 2
3
The slope of the tangent line at (1, 4) is 2 23.
9. y 5 2x3 1 6x2
yr 5 23x2 1 12x
23x2 1 12x 5 215
23x2 1 12x 5 212
2
x2 2 4x 2 5 5 0
x 2 4x 2 4 5 0
52
4 6 !16 1 16
2
4 6 4!2
5
2
x 5 2 6 2 !2
x5
(x 2 5)(x 1 1) 5 0
x 5 5, x 5 21
10. a. i. y 5 Ax2 2 4B 5
yr 5 5Ax2 2 4B 4 (2x)
Horizontal tangent,
10xAx2 2 4B 4 5 0
x 5 0, x 5 62
ii. y 5 Ax3 2 xB 2
yr 5 2Ax3 2 xB A3x2 2 1B
Horizontal tangent,
2x(x2 2 1)(3x2 2 1) 5 0
!3
x 5 0, x 5 61, x 5 6
.
3
2-35
b. i.
ii.
11. a. y 5 Ax2 1 5x 1 2B 4 at (0, 16)
yr 5 4Ax2 1 5x 1 2B 3 (2x 1 5)
At x 5 0,
yr 5 4(2)3 (5)
5 160
Equation of the tangent at (0, 16) is
y 2 16 5 160(x 2 0)
y 5 160x 1 16
or 160x 2 y 1 16 5 0
b. y 5 A3x22 2 2x3 B 5 at (1, 1)
yr 5 5A3x22 2 2x3 B 4 A26x23 2 6x2 B
At x 5 1,
yr 5 5(1)4 (26 2 6)
5 260
Equation of the tangent at (1, 1) is
y 2 1 5 260(x 2 1)
60x 1 y 2 61 5 0.
12. y 5 3x2 2 7x 1 5
dy
5 6x 2 7
dx
2-36
Slope of x 1 5y 2 10 5 0 is 2 15.
Since perpendicular, 6x 2 7 5 5
x52
y 5 3(4) 2 14 1 5
5 3.
Equation of the tangent at (2, 3) is
y 2 3 5 5(x 2 2)
5x 2 y 2 7 5 0.
13. y 5 8x 1 b is tangent to y 5 2x2
dy
5 4x
dx
Slope of the tangent is 8, therefore 4x 5 8, x 5 2.
Point of tangency is (2, 8).
Therefore, 8 5 16 1 b, b 5 28.
Or 8x 1 b 5 2x2
2x2 2 8x 2 b 5 0
8 6 !64 1 8b
x5
.
2(2)
For tangents, the roots are equal, therefore
64 1 8b 5 0, b 5 28.
Point of tangency is (2, 8), b 5 28.
14. a.
b.
The equation of the tangent is y 5 0.
The equation of the tangent is y 5 6.36.
The equation of the tangent is y 5 26.36.
Chapter 2: Derivatives
c.
Ax2 2 6B A3x2 B 2 x3 (2x)
Ax2 2 6B 2
x4 2 18x2
5 2
Ax 2 6B 2
f r(x) 5
x4 2 18x2
50
Ax2 2 6B 2
x2 Ax2 2 18B 5 0
x2 5 0 or x2 2 18 5 0
x50
x 5 63 !2
The coordinates of the points where the slope is 0
9 !2
are (0, 0), Q 3 !2, 9 !2
2 R , and Q 23!2, 2 2 R .
d. Substitute into the expression for f r(x) from
part b.
16 2 72
f r(2) 5
(22)2
256
5
4
5 214
5
2
15. a. f(x) 5 2x3 2 5x3
5 2
2 1
f r(x) 5 2 3 x3 2 5 3 x3
3
3
10 23
10
5 x 2 13
3
3x
2
f(x) 5 0 6 x3 32x 2 54 5 0
5
x 5 0 or x 5
2
y 5 f(x) crosses the x-axis at x 5 52, and
10 x 2 1
f r(x) 5 a 13 b
3
x
5
10
3
1
f ra b 5
3 3 5 13
2
3
2
Q2 R
553
3
!
2
2
1
5 53 3 23
3
!5
1
5 (25 3 2)3
3
5!
50
b. To find a, let f(x) 5 0.
10 23
10
x 2 13 5 0
3
3x
Calculus and Vectors Solutions Manual
30x 5 30
x51
Therefore a 5 1.
16. M 5 0.1t2 2 0.001t3
a. When t 5 10,
M 5 0.1(100) 2 0.001(1000)
59
When t 5 15,
M 5 0.1(225) 2 0.001(3375)
5 19.125
One cannot memorize partial words, so 19 words
are memorized after 15 minutes.
b. Mr 5 0.2t 2 0.003t2
When t 5 10,
Mr 5 0.2(10) 2 0.003(100)
5 1.7
The number of words memorized is increasing by
1.7 words> min.
When t 5 15,
Mr 5 0.2(15) 2 0.003(225)
5 2.325
The number of words memorized is increasing by
2.325 words> min.
30
17. a. N(t) 5 20 2
"9 1 t2
30t
3
A9 1 t2 B 2
b. No, according to this model, the cashier never
stops improving. Since t . 0, the derivative is always
positive, meaning that the rate of change in the
cashier’s productivity is always increasing. However,
these increases must be small, since, according to the
model, the cashier’s productivity can never exceed 20.
1
18. C(x) 5 x3 1 40x 1 700
3
a. Cr(x) 5 x2 1 40
b. Cr(x) 5 76
x2 1 40 5 76
x2 5 36
x56
Production level is 6 gloves> week.
x2
2
19. R(x) 5 750x 2 2 x3
6
3
a. Marginal Revenue
x
Rr(x) 5 750 2 2 2x2
3
Nr(t) 5
2-37
10
2 2(100)
3
5 $546.67
20
20. D(p) 5
,p.1
!p 2 1
1
3
Dr(p) 5 20a2 b (p 2 1)22
2
10
52
3
(p 2 1)2
10
10
Dr(5) 5
52
3
8
"4
5
52
4
Slope of demand curve at (5, 10) is 2 54.
21. B(x) 5 20.2x2 1 500, 0 # x # 40
a. B(0) 5 20.2(0)2 1 500 5 500
B(30) 5 20.2(30)2 1 500 5 320
b. Br(x) 5 20.4x
Br(0) 5 20.4(0) 5 0
Br(30) 5 20.4(30) 5 212
c. B(0) 5 blood sugar level with no insulin
B(30) 5 blood sugar level with 30 mg of insulin
Br(0) 5 rate of change in blood sugar level
with no insulin
Br(30) 5 rate of change in blood sugar level
with 30 mg of insulin
d. Br(50) 5 20.4(50) 5 220
B(50) 5 20.2(50)2 1 500 5 0
Br(50) 5 220 means that the patient’s blood sugar
level is decreasing at 20 units per mg of insulin 1 h
after 50 mg of insulin is injected.
B(50) 5 0 means that the patient’s blood sugar level
is zero 1 h after 50 mg of insulin is injected. These
values are not logical because a person’s blood sugar
level can never reach zero and continue to decrease.
3x
22. a. f(x) 5
1 2 x2
3x
5
(1 2 x)(1 1 x)
f(x) is not differentiable at x 5 1 because it is not
defined there (vertical asymptote at x 5 1).
x21
b. g(x) 5 2
x 1 5x 2 6
x21
5
(x 1 6)(x 2 1)
1
5
for x 2 1
(x 1 6)
g(x) is not differentiable at x 5 1 because it is not
defined there (hole at x 5 1).
b. Rr(10) 5 750 2
2-38
3
c. h(x) 5 "
(x 2 2)2
The graph has a cusp at (2, 0) but it is differentiable
at x 5 1.
d. m(x) 5 Z3x 2 3 Z 2 1.
The graph has a corner at x 5 1, so m(x) is not
differentiable at x 5 1.
3
23. a. f(x) 5 2
4x 2 x
3
5
x(4x 2 1)
f(x) is not defined at x 5 0 and x 5 0.25. The
graph has vertical asymptotes at x 5 0 and
x 5 0.25. Therefore, f(x) is not differentiable at
x 5 0 and x 5 0.25.
x2 2 x 2 6
b. f(x) 5
x2 2 9
(x 2 3)(x 1 2)
5
(x 2 3)(x 1 3)
(x 1 2)
5
for x 2 3
(x 1 3)
f(x) is not defined at x 5 3 and x 5 23. At
x 5 23, the graph as a vertical symptote and at
x 5 3 it has a hole. Therefore, f(x) is not
differentiable at x 5 3 and x 5 23.
c. f(x) 5 "x2 2 7x 1 6
5 !(x 2 6)(x 2 1)
f(x) is not defined for 1 , x , 6. Therefore,
f(x) is not differentiable for 1 , x , 6.
(t 1 1)(25) 2 (25t)(t)
24. pr(t) 5
(t 1 1)2
25t 1 25 2 25t
5
(t 1 1)2
25
5
(t 1 1)2
25. Answers may vary. For example,
f(x) 5 2x 1 3
1
y5
2x 1 3
(2x 1 3)(0) 2 (1)(2)
yr 5
(2x 1 3)2
2
52
(2x 1 3)2
f(x) 5 5x 1 10
1
y5
5x 1 10
(5x 1 10)(0) 2 (1)(5)
yr 5
(5x 1 10)2
Chapter 2: Derivatives
52
5
(5x 1 10)2
1
Rule: If f(x) 5 ax 1 b and y 5 f (x), then
yr 5
2a
(ax 1 b)2
1
1
1
yr 5 lim c
2
d
hS0 h a(x 1 h) 1 b
ax 1 b
1 ax 1 b 2 3a(x 1 h)b4
d
5 lim c
hS0 h 3a(x 1 h) 1 b4 (ax 1 h)
1 ax 1 b 2 ax 2 ah 2 b
5 lim c
d
hS0 h 3a(x 1 h) 1 b4 (ax 1 b)
1
2ah
5 lim c
d
hS0 h 3a(x 1 h) 1 b4 (ax 1 b)
2a
5 lim
hS0 3a(x 1 h) 1 b4 (ax 1 b)
2a
5
(ax 1 b)2
26. a. Let y 5 f(x)
(2x 2 3)2 1 5
y5
2x 2 3
Let u 5 2x 2 3.
u2 1 5
Then y 5
.
u
y 5 u 1 5u21
dy
b. f r(x) 5
dx
dy
dy
du
5
3
dx
du
dx
5 (1 2 5u22 )(2)
5 2(1 2 5(2x 2 3)22 )
27. g(x) 5 !2x 2 3 1 5(2x 2 3)
a. Let y 5 g(x).
y 5 !2x 2 3 1 5(2x 2 3)
Let u 5 2x 2 3.
Then y 5 !u 1 5u.
dy
dy
du
b. gr(x) 5
5
3
dx
du
dx
1 212
5 a u 1 5b (2)
2
212
5 u 1 10
1
5 (2x 2 3)22 1 10
Calculus and Vectors Solutions Manual
28. a. f(x) 5 (2x 2 5)3 (3x2 1 4)5
f r(x) 5 (2x 2 5)3 (5)A3x2 1 4B 4 (6x)
1 A3x2 1 4B 5 (3)(2x 2 5)2 (2)
5 30x(2x 2 5)3 A3x2 1 4B 4
1 6(3x2 1 4)5 (2x 2 5)2
5 6(2x 2 5)2 A3x2 1 4B 4
3 C5x(2x 2 5) 1 A3x2 1 4B D
5 6(2x 2 5)2 A3x2 1 4B 4
3 A10x2 2 25x 1 3x2 1 4B
5 6(2x 2 5)2 (3x2 1 4)4
3 (13x2 2 25x 1 4)
3
b. g(x) 5 (8x )(4x2 1 2x 2 3)5
gr(x) 5 (8x3 )(5)(4x2 1 2x 2 3)4 (8x 1 2)
1 (4x2 1 2x 2 3)5 (24x2 )
5 40x3 (4x2 1 2x 2 3)4 (8x 1 2)
1 24x2 (4x2 1 2x 2 3)5
5 8x2 (4x2 1 2x 2 3)4 35x(8x 1 2)
1 3(4x2 1 2x 2 3)4
5 8x2 (4x2 1 2x 2 3)4
(40x2 1 10x 1 12x2 1 6x 2 9)
5 8x2 (4x2 1 2x 2 3)4 (52x2 1 16x 2 9)
c. y 5 (5 1 x)2 (4 2 7x3 )6
yr 5 (5 1 x)2 (6)(4 2 7x3 )5 (221x2 )
1 (4 2 7x3 )6 (2)(5 1 x)
5 2126x2 (5 1 x)2 (4 2 7x3 )5
1 2(5 1 x)(4 2 7x3 )6
5 2(5 1 x)(4 2 7x3 )5 3263x2 (5 1 x)
1 4 2 7x34
5 2(5 1 x)(4 2 7x3 )5 (4 2 315x2 2 70x3 )
6x 2 1
d. h(x) 5
(3x 1 5)4
(3x 1 5)4 (6) 2 (6x 2 1)(4)(3x 1 5)3 (3)
hr(x) 5
((3x 1 5)4 )2
3
6(3x 1 5) 3(3x 1 5) 2 2(6x 2 1)4
5
(3x 1 5)8
6(29x 1 7)
5
(3x 1 5)5
(2x2 2 5)3
e. y 5
(x 1 8)2
dy
(x 1 8)2 (3)(2x2 2 5)2 (4x)
5
dx
A(x 1 8)2 B 2
(2x2 2 5)3 (2)(x 1 8)
2
A(x 1 8)2 B 2
2(x 1 8)(2x2 2 5)2 36x(x 1 8) 2 (2x2 2 5)4
(x 1 8)4
2(2x2 2 5)2 (4x2 1 48x 1 5)
5
(x 1 8)3
5
2-39
f. f(x) 5
23x4
"4x 2 8
23x4
5
1
(4x 2 8)2
1
(4x 2 8)2 (212x3 )
f r(x) 5
1
A(4x 2 8)2 B 2
1
1
(23x4 )a b (4x 2 8) 2 2 (4)
2
2
1
A(4x 2 8)2 B 2
26x3 (4x 2 8)22 32(4x 2 8) 2 x4
5
4x 2 8
26x3 (7x 2 16)
5
3
(4x 2 8)2
23x3 (7x 2 16)
5
3
(4x 2 8)2
2x 1 5 4
g. g(x) 5 a
b
6 2 x2
1
2x 1 5 3
b
6 2 x2
(6 2 x2 )(2) 2 (2x 1 5)(22x)
b
3a
(6 2 x2 )2
2x 1 5 3 2(6 1 x2 1 5x)
5 4a
b a
b
6 2 x2
(6 2 x2 )2
2x 1 5 3 (x 1 2)(x 1 3)
5 8a
b a
b
6 2 x2
(6 2 x2 )2
3
1
h. y 5 c
2 3d
(4x 1 x )
In (1),
4a 2 8a 5 16
24a 5 16
a 5 24
Using (1),
b 5 28(24) 5 32
a 5 24, b 5 32, c 5 0, f(x) 5 24x2 1 32x
30. a. A(t) 5 2t3 1 5t 1 750
Ar(t) 5 23t2 1 5
b. Ar(5) 5 23(25) 1 5
5 270
At 5 h, the number of ants living in the colony is
decreasing by 7000 ants> h.
c. A(0) 5 750, so there were 750 (100) or
75 000 ants living in the colony before it was
treated with insecticide.
d. Determine t so that A(t) 5 0. 2t3 1 5t 1 750
cannot easily be factored, so find the zeros by using
a graphing calculator.
gr(x) 5 4a
5 (4x 1 x2 )29
dy
5 29(4x 1 x2 )210 (4 1 2x)
dx
29. f(x) 5 ax2 1 bx 1 c,
It is given that (0, 0) and (8, 0) are on the curve,
and f r(2) 5 16.
Calculate f r(x) 5 2ax 1 b.
Then,
16 5 2a(2) 1 b
(1)
4a 1 b 5 16
Since (0, 0) is on the curve,
0 5 a(0)2 1 b(0) 1 c
c50
Since (8, 0) is on the curve,
0 5 a(8)2 1 b(8) 1 c
0 5 64a 1 8b 1 0
(2)
8a 1 b 5 0
Solve (1) and (2):
From (2), b 5 28a
(1)
2-40
All of the ants have been killed after about 9.27 h.
Chapter 2 Test, p. 114
1. You need to use the chain rule when the derivative
for a given function cannot be found using the sum,
difference, product, or quotient rules or when writing
the function in a form that would allow the use of
these rules is tedious. The chain rule is used when
a given function is a composition of two or more
functions.
2. f is the blue graph (it's a cubic). f' is the red graph
(it is quadratic). The derivative of a polynomial
function has degree one less than the derivative of
the function. Since the red graph is a quadratic
(degree 2) and the blue graph is cubic (degree 3),
the blue graph is f and the red graph is f r.
f(x 1 h) 2 f(x)
3. f(x) 5 lim
hS0
h
x 1 h 2 (x 1 h)2 2 (x 2 x2 )
5 lim
hS0
h
x 1 h 2 (x2 1 2hx 1 h2 ) 2 x 1 x2
5 lim
hS0
h
h 2 2hx 2 h2
5 lim
hS0
h
Chapter 2: Derivatives
h(1 2 2x 2 h)
hS0
h
5 lim (1 2 2x 2 h)
5 lim
hS0
5 1 2 2x
d
Therefore,
(x 2 x2 ) 5 1 2 2x.
dx
1
4. a. y 5 x3 2 3x25 1 4p
3
dy
5 x2 1 15x26
dx
b. y 5 6(2x 2 9)5
dy
5 30(2x 2 9)4 (2)
dx
5 60(2x 2 9)4
2
x
3
1
1 6"
x
c. y 5
"x
"3
1
1
1
5 2x22 1
x 1 6x3
"3
dy
1
3
2
5 2x22 1
1 2x23
dx
"3
x2 1 6 5
b
d. y 5 a
3x 1 4
dy
x2 1 6 4 2x(3x 1 4) 2 (x2 1 6)3
5 5a
b
dx
3x 1 4
(3x 1 4)2
5(x2 1 6)4 (3x2 1 8x 2 18)
5
(3x 1 4)6
3
6x2 2 7
e. y 5 x2 "
dy
1
1
2
5 2x(6x2 2 7)3 1 x2 (6x2 2 7)23 (12x)
dx
3
2
5 2x(6x2 2 7)23 ( (6x2 2 7) 1 2x2 )
2
5 2x(6x2 2 7)23 (8x2 2 7)
4x5 2 5x4 1 6x 2 2
f. y 5
x4
5 4x 2 5 1 6x23 2 2x24
dy
5 4 2 18x24 1 8x25
dx
4x5 2 18x 1 8
5
x5
5. y 5 (x2 1 3x 2 2)(7 2 3x)
dy
5 (2x 1 3)(7 2 3x) 1 (x2 1 3x 2 2)(23)
dx
At (1, 8),
dy
5 (5)(4) 1 (2)(23)
dx
5 14.
The slope of the tangent to
y 5 (x2 1 3x 2 2)(7 2 3x) at (1, 8) is 14.
Calculus and Vectors Solutions Manual
6. y 5 3u2 1 2u
dy
5 6u 1 2
du
u 5 "x2 1 5
du
1
1
5 (x2 1 5)22 2x
dy
2
dy
x
5 (6u 1 2)a
b
2
dx
"x 1 5
At x 5 22, u 5 3.
dy
2
5 (20)a2 b
dx
3
40
52
3
7. y 5 (3x22 2 2x3 )5
dy
5 5(3x22 2 2x3 )4 (26x23 2 6x2 )
dx
At (1, 1),
dy
5 5(1)4 (26 2 6)
dx
5 260.
Equation of tangent line at (1, 1) is y 2 1 5 60(x 2 1)
y 2 1 5 260x 1 60
60x 1 y 2 61 5 0.
1
8. P(t) 5 (t 4 1 3)3
1 3
1
Pr(t) 5 3(t 4 1 3)2 a t24 b
4
1
1
3
Pr(16) 5 3(16 4 1 3)2 a 3 1624 b
4
1
1
5 3(2 1 3)2 a 3 b
4
8
75
5
32
The amount of pollution is increasing at a rate of
75
32 ppm>year.
9. y 5 x4
dy
5 4x3
dx
1
2 5 4x3
16
Normal line has a slope of 16. Therefore,
dy
1
52 .
dx
16
1
x3 5 2
64
2-41
1
4
1
y5
256
Therefore, y 5 x4 has a normal line with a slope of
1
16 at Q 2 14, 256
R.
x52
10. y 5 x3 2 x2 2 x 1 1
dy
5 3x2 2 2x 2 1
dx
dy
For a horizontal tangent line, dx 5 0.
3x2 2 2x 2 1 5 0
(3x 1 1)(x 2 1) 5 0
1
or
x52
x51
3
1
1
1
y52 2 1 11
y51212111
27
9
3
50
21 2 3 1 9 1 27
5
27
32
5
27
The required points are Q 2 13, 32
27 ), (1, 0 R .
2-42
11. y 5 x2 1 ax 1 b
dy
5 2x 1 a
dx
y 5 x3
dy
5 3x2
dx
Since the parabola and cubic function are tangent at
(1, 1), then 2x 1 a 5 3x2.
At (1, 1) 2(1) 1 a 5 3(1)2
a 5 1.
Since (1, 1) is on the graph of
y 5 x2 1 x 1 b, 1 5 12 1 1 1 b
b 5 21.
The required values are 1 and 21 for a and b,
respectively.
Chapter 2: Derivatives
CHAPTER 2
Derivatives
Review of Prerequisite Skills,
pp. 62–63
1. a. a5 3 a3 5 a513
5 a8
b. A22a2 B 3 5 (22)3 Aa2 B 3
5 28Aa233 B
5 28a6
7
4p 3 6p9
24p719
c.
5
12p15
12p15
5 2p16215
5 2p
4 25
26 22
d. Aa b B Aa b B 5 Aa426 B Ab2522 B
5 a22b27
1
5 2 7
ab
e. A3e6 B A2e3 B 4 5 (3)Ae6 B A24 B Ae3 B 4
5 (3)A24 BAe6 B Ae334 B
5 (3)(16)Ae6112 B
5 48e18
24
3
A3a B C2a (2b)3 D
(3)(2)A21B 3 Aa2413 B Ab3 B
f.
5
12a5b2
12a5b2
2125
B Ab322 B
26Aa
5
12
26
21Aa B AbB
5
2
b
52 6
2a
1
2
1
2
2. a. Ax2 B Ax3 B 5 x2 1 3
7
5 x6
2
2
2
b. A8x6 B 3 5 83x63 3
5 4x4
1
1
3
( a2)( a3)
"a "
a
c.
5
1
a2
"a
1
5 a3
3. A perpendicular line will have a slope that is the
negative reciprocal of the slope of the given line:
21
a. slope 5 2
3
52
3
2
Calculus and Vectors Solutions Manual
21
2 12
52
21
c. slope 5 5
b. slope 5
3
3
5
21
d. slope 5
21
51
52
24 2 (22)
23 2 9
22
5
212
1
5
6
The equation of the desired line is therefore
y 1 4 5 16 (x 1 3) or x 2 6y 2 21 5 0.
b. The equation 3x 2 2y 5 5 can be rewritten as
2y 5 3x 2 5 or y 5 32x 2 52, which has slope 32.
The equation of the desired line is therefore
y 1 5 5 32 (x 1 2) or 3x 2 2y 2 4 5 0.
c. The line perpendicular to y 5 34 x 2 6 will have
21
slope m 5 3 5 2 43. The equation of the desired line
4. a. This line has slope m 5
4
is therefore y 1 3 5 2 43 (x 2 4) or 4x 1 3y 2 7 5 0.
5. a. (x 2 3y)(2x 1 y) 5 2x2 1 xy 2 6xy 2 3y2
5 2x2 2 5xy 2 3y2
2
b. (x 2 2)(x 2 3x 1 4)
5 x3 2 3x2 1 4x 2 2x2 1 6x 2 8
5 x3 2 5x2 1 10x 2 8
c. (6x 2 3)(2x 1 7) 5 12x2 1 42x 2 6x 2 21
5 12x2 1 36x 2 21
d. 2(x 1 y) 2 5(3x 2 8y) 5 2x 1 2y 2 15x 1 40y
5 213x 1 42y
2
2
e. (2x 2 3y) 1 (5x 1 y)
5 4x2 2 12xy 1 9y2 1 25x2 1 10xy 1 y2
5 29x2 2 2xy 1 10y2
f. 3x(2x 2 y)2 2 x(5x 2 y)(5x 1 y)
5 3x(4x2 2 4xy 1 y2 ) 2 x(25x2 2 y2 )
5 12x3 2 12x2y 1 3xy2 2 25x3 1 xy2
5 213x3 2 12x2y 1 4xy2
2-1
3x(x 1 2)
5x3
15x4 (x 1 2)
3
5
x2
2x(x 1 2)
2x3 (x 1 2)
15
5 x423
2
15
5 x
2
x 2 0, 22
y
(y 2 5)2
3
b.
(y 1 2)(y 2 5)
4y3
y(y 2 5)(y 2 5)
5 3
4y (y 1 2)(y 2 5)
y25
5 2
4y (y 1 2)
y 2 22, 0, 5
4
9
4
2(h 1 k)
c.
4
5
3
h 1 k 2(h 1 k)
h1k
9
8(h 1 k)
5
9(h 1 k)
8
5
9
h 2 2k
(x 1 y)(x 2 y) (x 1 y)3
d.
4
5(x 2 y)
10
(x 1 y)(x 2 y)
10
5
3
5(x 2 y)
(x 1 y)3
10(x 1 y)(x 2 y)
5
5(x 2 y)(x 1 y)3
2
5
(x 1 y)2
x 2 2y, 1y
x27
5x
(x 2 7)(x 2 1)
(5x)(2x)
1
5
1
e.
2x
x21
2x(x 2 1)
2x(x 2 1)
x2 2 7x 2 x 1 7 1 10x2
5
2x(x 2 1)
11x2 2 8x 1 7
5
2x(x 2 1)
x 2 0, 1
x12
x11
2
f.
x22
x13
(x 1 2)(x 2 2)
(x 1 1)(x 1 3)
2
5
(x 2 2)(x 1 3)
(x 1 3)(x 2 2)
x2 1 x 1 3x 1 3 2 x2 1 4
5
(x 1 3)(x 2 2)
4x 1 7
5
(x 1 3)(x 2 2)
x 2 23, 2
7. a. 4k2 2 9 5 (2k 1 3)(2k 2 3)
6. a.
2-2
b. x2 1 4x 2 32 5 x2 1 8x 2 4x 2 32
5 x(x 1 8) 2 4(x 1 8)
5 (x 2 4)(x 1 8)
c. 3a2 2 4a 2 7 5 3a2 2 7a 1 3a 2 7
5 a(3a 2 7) 1 1(3a 2 7)
5 (a 1 1)(3a 2 7)
d. x4 2 1 5 (x2 1 1)(x2 2 1)
5 (x2 1 1)(x 1 1)(x 2 1)
3
3
e. x 2 y 5 (x 2 y)(x2 1 xy 1 y2 )
f. r4 2 5r 2 1 4 5 r4 2 4r 2 2 r 2 1 4
5 r 2 (r 2 2 4) 2 1(r 2 2 4)
5 (r2 2 1)(r2 2 4)
5 (r 1 1)(r 2 1)(r 1 2)(r 2 2)
8. a. Letting f(a) 5 a3 2 b3, f(b) 5 b3 2 b3
50
So b is a root of f (a), and so by the factor theorem,
a 2 b is a factor of a3 2 b3. Polynomial long
division provides the other factor:
a2 1 ab 1 b2
a 2 bq a3 1 0a2 1 0a 2 b3
a3 2 a2b
a2b 1 0a 2 b3
a2b 2 ab2
ab2 2 b3
ab2 2 b3
0
So a3 2 b3 5 (a 2 b)(a2 1 ab 1 b2 ).
b. Using long division or recognizing a pattern from
the work in part a.:
a5 2 b5 5 (a 2 b)(a4 1 a3b 1 a2b2 1 ab3 1 b4 ).
c. Using long division or recognizing a pattern from
the work in part a.: a7 2 b7
5 (a 2 b)(a6 1 a5b 1 a4b2 1 a3b3
1 a2b4 1 ab5 1 b6 ).
d. Using the pattern from the previous parts:
an 2 bn 5 (a 2 b)(an21 1 an22b 1 an23b2 1 c
1 a2bn23 1 abn22 1 bn21 ).
9. a. f(2) 5 22(24 ) 1 3(22 ) 1 7 2 2(2)
5 232 1 12 1 7 2 4
5 217
b. f(21) 5 22(21)4 1 3(21)2 1 7 2 2(21)
5 22 1 3 1 7 1 2
5 10
1 4
1 2
1
1
c. f a b 5 22a b 1 3a b 1 7 2 2a b
2
2
2
2
3
1
52 1 1721
8
4
53
5
8
Chapter 2: Derivatives
1
d. f(20.25) 5 f a2 b
4
1 4
1 2
1
5 2a2 b 1 3a2 b 1 7 2 2a2 b
4
4
4
1
3
1
52
1
171
128
16
2
983
5
128
8 7.68
3
3 !2
5
10. a.
!2
( !2)( !2)
3 !2
5
2
4 2 !2
(4 2 !2)( !3)
5
b.
!3
( !3)(!3)
4!3 2 !6
5
3
2 1 3 !2
(2 1 3 !2)(3 1 4 !2)
5
c.
3 2 4 !2
(3 2 4 !2)(3 1 4 !2)
6 1 9!2 1 8!2 1 12(2)
5
32 2 A4 !2B 2
30 1 17 !2
5
9 2 16(2)
30 1 17!2
52
23
3!2 2 4!3
(3!2 2 4 !3)(3 !2 2 4 !3)
5
d.
3!2 1 4!3
(3 !2 1 4 !3)(3"2 2 4 !3)
(3!2)2 2 24!6 1 (4 !3)2
5
(3!2)2 2 (4!3)2
9(2) 2 24!6 1 16(3)
5
9(2) 2 16(3)
66 2 24 !6
52
30
11 2 4 !6
52
5
11. a. f(x) 5 3x2 2 2x
When a 5 2,
f(a 1 h) 2 f(a)
f(2 1 h) 2 f(2)
5
h
h
2
3(2 1 h) 2 2(2 1 h) 2 C3(2)2 2 2(2)D
5
h
3(4 1 4h 1 h2 ) 2 4 2 2h 2 8
5
h
12 1 12h 1 3h2 2 2h 2 12
5
h
Calculus and Vectors Solutions Manual
3h2 1 10h
h
5 3h 1 10
This expression can be used to determine the slope of
the secant line between (2, 8) and (2 1 h, f(2 1 h)).
b. For h 5 0.01: 3(0.01) 1 10 5 10.03
c. The value in part b. represents the slope of the
secant line through (2, 8) and (2.01, 8.1003).
5
2.1 The Derivative Function, pp. 73–75
1. A function is not differentiable at a point where its
graph has a cusp, a discontinuity, or a vertical tangent:
a. The graph has a cusp at x 5 22, so f is
differentiable on 5xPR Z x 2 226.
b. The graph is discontinuous at x 5 2, so f is
differentiable on 5xPR Z x 2 26.
c. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
d. The graph has a cusp at x 5 1, so f is
differentiable on 5xPR Z x 2 16.
e. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
f. The function does not exist for x , 2, but has
no cusps, discontinuities, or vertical tangents
elsewhere, so f is differentiable on 5xPR Z x . 26.
2. The derivative of a function represents the slope of
the tangent line at a given value of the independent
variable or the instantaneous rate of change of the
function at a given value of the independent variable.
3.
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
f(x) 5 5x 2 2
f(a 1 h) 5 5(a 1 h) 2 2
5 5a 1 5h 2 2
f(a 1 h) 2 f(a) 5 5a 1 5h 2 2 2 (5a 2 2)
5 5h
b.
f(x) 5 x2 1 3x 2 1
f(a 1 h) 5 (a 1 h)2 1 3(a 1 h) 2 1
5 a2 1 2ah 1 h2 1 3a
1 3h 2 1
f(a 1 h) 2 f(a) 5 a2 1 2ah 1 h2 1 3a 1 3h
2 1 2 (a2 1 3a 2 1)
5 2ah 1 h2 1 3h
4. a.
2-3
1
d. f(20.25) 5 f a2 b
4
1 4
1 2
1
5 2a2 b 1 3a2 b 1 7 2 2a2 b
4
4
4
1
3
1
52
1
171
128
16
2
983
5
128
8 7.68
3
3 !2
5
10. a.
!2
( !2)( !2)
3 !2
5
2
4 2 !2
(4 2 !2)( !3)
5
b.
!3
( !3)(!3)
4!3 2 !6
5
3
2 1 3 !2
(2 1 3 !2)(3 1 4 !2)
5
c.
3 2 4 !2
(3 2 4 !2)(3 1 4 !2)
6 1 9!2 1 8!2 1 12(2)
5
32 2 A4 !2B 2
30 1 17 !2
5
9 2 16(2)
30 1 17!2
52
23
3!2 2 4!3
(3!2 2 4 !3)(3 !2 2 4 !3)
5
d.
3!2 1 4!3
(3 !2 1 4 !3)(3"2 2 4 !3)
(3!2)2 2 24!6 1 (4 !3)2
5
(3!2)2 2 (4!3)2
9(2) 2 24!6 1 16(3)
5
9(2) 2 16(3)
66 2 24 !6
52
30
11 2 4 !6
52
5
11. a. f(x) 5 3x2 2 2x
When a 5 2,
f(a 1 h) 2 f(a)
f(2 1 h) 2 f(2)
5
h
h
2
3(2 1 h) 2 2(2 1 h) 2 C3(2)2 2 2(2)D
5
h
3(4 1 4h 1 h2 ) 2 4 2 2h 2 8
5
h
12 1 12h 1 3h2 2 2h 2 12
5
h
Calculus and Vectors Solutions Manual
3h2 1 10h
h
5 3h 1 10
This expression can be used to determine the slope of
the secant line between (2, 8) and (2 1 h, f(2 1 h)).
b. For h 5 0.01: 3(0.01) 1 10 5 10.03
c. The value in part b. represents the slope of the
secant line through (2, 8) and (2.01, 8.1003).
5
2.1 The Derivative Function, pp. 73–75
1. A function is not differentiable at a point where its
graph has a cusp, a discontinuity, or a vertical tangent:
a. The graph has a cusp at x 5 22, so f is
differentiable on 5xPR Z x 2 226.
b. The graph is discontinuous at x 5 2, so f is
differentiable on 5xPR Z x 2 26.
c. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
d. The graph has a cusp at x 5 1, so f is
differentiable on 5xPR Z x 2 16.
e. The graph has no cusps, discontinuities, or
vertical tangents, so f is differentiable on 5xPR6.
f. The function does not exist for x , 2, but has
no cusps, discontinuities, or vertical tangents
elsewhere, so f is differentiable on 5xPR Z x . 26.
2. The derivative of a function represents the slope of
the tangent line at a given value of the independent
variable or the instantaneous rate of change of the
function at a given value of the independent variable.
3.
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
4
2
–3 –2 –1 0
–2
–4
y
x
1 2 3
f(x) 5 5x 2 2
f(a 1 h) 5 5(a 1 h) 2 2
5 5a 1 5h 2 2
f(a 1 h) 2 f(a) 5 5a 1 5h 2 2 2 (5a 2 2)
5 5h
b.
f(x) 5 x2 1 3x 2 1
f(a 1 h) 5 (a 1 h)2 1 3(a 1 h) 2 1
5 a2 1 2ah 1 h2 1 3a
1 3h 2 1
f(a 1 h) 2 f(a) 5 a2 1 2ah 1 h2 1 3a 1 3h
2 1 2 (a2 1 3a 2 1)
5 2ah 1 h2 1 3h
4. a.
2-3
f(x) 5 x3 2 4x 1 1
f(a 1 h) 5 (a 1 h)3 2 4(a 1 h) 1 1
5 a3 1 3a2h 1 3ah2 1 h3
2 4a 2 4h 1 1
f(a 1 h) 2 f(a) 5 a3 1 3a2h 1 3ah2 1 h3 2 4a
2 4h 1 1 2 (a3 2 4a 1 1)
5 3a2h 1 3ah2 1 h3 2 4h
f(x) 5 x2 1 x 2 6
d.
f(a 1 h) 5 (a 1 h)2 1 (a 1 h) 2 6
5 a2 1 2ah 1 h2 1 a 1 h 2 6
f(a 1 h) 2 f(a) 5 a2 1 2ah 1 h2 1 a 1 h 2 6
2 (a2 1 a 2 6)
5 2ah 1 h2 1 h
e.
f(x) 5 27x 1 4
f(a 1 h) 5 27(a 1 h) 1 4
5 27a 2 7h 1 4
f(a 1 h) 2 f(a) 5 27a 2 7h 1 4 2 (27a 1 4)
5 27h
f(x) 5 4 2 2x 2 x2
f.
f(a 1 h) 5 4 2 2(a 1 h) 2 (a 1 h)2
5 4 2 2a 2 2h 2 a2 2 2ah 2 h2
f(a 1 h) 2 f(a) 5 4 2 2a 2 2h 2 a2 2 2ah
2 h2 2 4 1 2a 1 a2
5 22h 2 h2 2 2ah
f(1 1 h) 2 f(1)
5. a. f r (1) 5 lim
hS0
h
(1 1 h)2 2 12
5 lim
hS0
h
1 1 2h 1 h2 2 1
5 lim
hS0
h
2h 1 h2
5 lim
hS0
h
5 lim (2 1 h)
c.
hS0
52
f(3 1 h) 2 f(3)
b. f r(3) 5 lim
hS0
h
(3 1 h)2 1 3(3 1 h) 1 1
5 lim c
hS0
h
2
(3 1 3(3) 1 1)
d
2
h
9 1 6h 1 h2 1 9 1 3h 1 1 2 19
5 lim
hS0
h
9h 1 h2
5 lim
hS0
h
5 lim (9 1 h)
hS0
c.
f(0 1 h) 2 f(0)
hS0
h
!h 1 1 2 !0 1 1
5 lim
hS0
h
!h 1 1 2 1
5 lim
hS0
h
( ! h 1 1 2 1)( ! h 1 1 1 1)
5 lim
hS0
h( ! h 1 1 1 1)
f r (0) 5 lim
( "h 1 1) 2 2 1
hS0 h( !h 1 1 1 1)
5 lim
h1121
hS0 h( !h 1 1 1 1)
1
5 lim
hS0 ( !h 1 1 1 1)
1
5 lim
hS0 ( !1 1 1)
1
5
2
f(21 1 h) 2 f(21)
d. f r(21) 5 lim
hS0
h
5 lim
5 lim
hS0
5 lim
hS0
5
5
2 21
21 1 h
h
5
21 1 h 1 5
h
5
5(21 1 h)
1 21 1 h
21
1
h
5 lim
h
5 2 5 1 5h
5 lim
hS0 h(21 1 h)
5h
5 lim
hS0 h(21 1 h)
5
5 lim
hS0 (21 1 h)
5
5
21
5 25
f(x 1 h) 2 f(x)
6. a. f r (x) 5 lim
hS0
h
25(x 1 h) 2 8 2 (25x 2 8)
5 lim
hS0
h
25x 2 5h 2 8 1 5x 1 8
5 lim
hS0
h
hS0
59
2-4
Chapter 2: Derivatives
25h
hS0 h
5 lim 25
5 lim
hS0
5 25
f(x 1 h) 2 f(x)
b. f r(x) 5 lim
hS0
h
2(x 1 h)2 1 4(x 1 h)
5 lim c
hS0
h
A2x2 1 4xB
d
h
2x2 1 4xh 1 2h2 1 4x
5 lim c
hS0
h
4h 2 2x2 2 4x
1
d
h
4xh 1 2h2 1 4h
5 lim
hS0
h
5 lim (4x 1 2h 1 4)
2
hS0
5 4x 1 4
f(x 1 h) 2 f(x)
c. f r(x) 5 lim
hS0
h
6(x 1 h)3 2 7(x 1 h)
5 lim c
hS0
h
(6x3 2 7x)
2
d
h
6x3 1 18x2h 1 18xh2 1 6h3
5 lim c
hS0
h
27x 2 7h 2 6x3 1 7x
1
d
h
18x2h 1 18xh2 1 6h3 2 7h
5 lim
hS0
h
2
5 lim (18x 1 18xh 1 6h2 2 7)
hS0
5 18x2 2 7
f(x 1 h) 2 f(x)
d. f r(x) 5 lim
hS0
h
5 lim
hS0
5 lim
hS0
A !3x 1 3h 1 2 B 2 2 A !3x 1 2 B 2
hA !3x 1 3h 1 2 1 !3x 1 2 B
3x 1 3h 1 2 2 3x 2 2
hS0 hA !3x 1 3h 1 2 1 !3x 1 2B
3
5 lim
hS0 !3x 1 3h 1 2 1 !3x 1 2
3
5
2 !3x 1 2
7. a. Let y 5 f(x), then
f(x 1 h) 2 f(x)
dy
5 f r(x) 5 lim
dx
hS0
h
6 2 7(x 1 h) 2 (6 2 7x)
5 lim
hS0
h
6 2 7x 2 7h 2 6 1 7x
5 lim
hS0
h
27h
5 lim
hS0
h
5 lim 27
5 lim
hS0
5 27
b. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
5 lim
!3x 1 3h 1 2 2 !3x 1 2
hS0
h
A !3x 1 3h 1 2 2 !3x 1 2 B
5 lim c
hS0
h
3
A !3x 1 3h 1 2 1 !3x 1 2B
d
A !3x 1 3h 1 2 1 !3x 1 2B
Calculus and Vectors Solutions Manual
h
hS0
5 lim
hS0
(x 1 h 1 1)(x 2 1)
£ (x 1 h 2 1)(x 2 1)
h
(x 1 1)(x 1 h 2 1)
(x 2 1)(x 1 h 2 1) §
2
h
x2
5 lim £
2
5 lim
1 hx 1 x 2 x 2 h 2 1
(x 1 h 2 1)(x 2 1)
h
hS0
!3(x 1 h) 1 2 2 !3x 1 2
h
5 lim
x1h11
x11
2x21
x1h21
x2 1 hx 2 x 1 x 1 h 2 1
(x 1 h 2 1)(x 2 1)
§
h
22h
(x 1 h 2 1)(x 2 1)
h
22
5 lim
hS0 (x 1 h 2 1)(x 2 1)
2
52
(x 2 1)2
hS0
2-5
c. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
3(x 1 h)2 2 3x2
5 lim
hS0
h
3x2 1 6xh 1 3h2 2 3x2
5 lim
hS0
h
6xh 1 3h2
5 lim
hS0
h
5 lim 6x 1 3h
9. a.
5 6x
8. Let y 5 f(x), then the slope of the tangent at
each point x can be found by calculating f r(x)
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
2(x 1 h)2 2 4(x 1 h) 2 2x2 1 4x
5 lim
hS0
h
2
2x 1 4xh 1 2h2 2 4x 2 4h
5 lim c
hS0
h
2
22x 1 4x
1
d
h
4xh 1 h2 2 4h
5 lim
hS0
h
5 lim 4x 1 h 2 4
b. Let y 5 f(x), then the slope of the tangent at
each point x can be found by calculating f(x)
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
(x 1 h)3 2 x3
5 lim
hS0
h
x3 1 3x2h 1 3xh2 1 h3 2 x3
5 lim
hS0
h
3x2h 1 3xh2 1 h3
5 lim
hS0
h
5 lim 3x2 1 3xh 1 h2
hS0
hS0
5 4x 2 4
So the slope of the tangent at x 5 0 is
f r(0) 5 4(0) 2 4
5 24
At x 5 1, the slope of the tangent is
f r(1) 5 4(1) 2 4
50
At x 5 2, the slope of the tangent is
f r(2) 5 4(2) 2 4
54
y
4
3
2
1
x
0
–2 –1
1 2 3 4
–1
–2
2-6
12
10
8
6
4
2
–6 –4 –2 0
–2
–4
y
x
2 4 6
hS0
5 3x2
So the slope of the tangent at x 5 22 is
f r(22) 5 3(22)2
5 12
At x 5 21, the slope of the tangent is
f r(21) 5 3(21)2
53
At x 5 0, the slope of the tangent is
f r(0) 5 3(0)2
50
At x 5 1, the slope of the tangent is
f r(1) 5 3(1)2
53
At x 5 2, the slope of the tangent is
f r(2) 5 3(2)2
5 12
c.
y
12
10
8
6
4
2
x
0
–6 –4 –2
2 4 6
–2
Chapter 2: Derivatives
d. The graph of f(x) is a cubic. The graph of f r(x)
seems to be a parabola.
10. The velocity the particle at time t is given by sr(t)
s(t 1 h) 2 s(t)
sr(t) 5 lim
hS0
h
2 (t 1 h)2 1 8(t 1 h) 2 (2t2 1 8t)
5 lim
hS0
h
2t2 2 2th 2 h2 1 8t 1 8h 1 t2 2 8t
5 lim
hS0
h
22th 2 h2 1 8h
5 lim
hS0
h
5 lim 2 2t 2 h 1 8
hS0
5 22t 1 8
So the velocity at t 5 0 is
sr(0) 5 22(0) 1 8
5 8 m>s
At t 5 4, the velocity is
sr(4) 5 22(4) 1 8
5 0 m>s
At t 5 6, the velocity is
sr(6) 5 22(6) 1 8
5 24 m>s
f(x 1 h) 2 f(x)
11. f r(x) 5 lim
hS0
h
!x 1 h 1 1 2 !x 1 1
hS0
h
5 lim
5 lim c
hS0
A !x 1 h 1 1 2 !x 1 1 B
h
3
A !x 1 h 1 1 1 !x 1 1 B
d
A !x 1 h 1 1 1 !x 1 1 B
A !x 1 h 1 1 B 2 2 A !x 1 1 B 2
hS0 hA !x 1 h 1 1 1 !x 1 1 B
5 lim
x1h112x21
hS0 hA !x 1 h 1 1 1 !x 1 1B
5 lim
h
hS0 hA !x 1 h 1 1 1 !x 1 1B
5 lim
5 lim
hS0
5
1
A !x 1 h 1 1 1 !x 1 1 B
1
2 !x 1 1
The equation x 2 6y 1 4 5 0 can be rewritten as
y 5 16 x 1 23, so this line has slope 16. The value of x
where the tangent to f(x) has slope 16 will satisfy
f r(x) 5 16.
Calculus and Vectors Solutions Manual
1
1
5
2 !x 1 1
6
6 5 2 !x 1 1
32 5 A !x 1 1B 2
95x11
85x
f(8) 5 !8 1 1
5 !9
53
So the tangent passes through the point (8, 3), and its
equation is y 2 3 5 16 (x 2 8) or x 2 6y 1 10 5 0.
12. a. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
c2c
5 lim
hS0
h
0
5 lim
hS0 h
50
b. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
(x 1 h) 2 x
5 lim
hS0
h
h
5 lim
hS0 h
5 lim 1
hS0
51
c. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
m(x 1 h) 1 b 2 mx 2 b
5 lim
hS0
h
mx 1 mh 1 b 2 mx 2 b
5 lim
hS0
h
mh
5 lim
hS0 h
5 lim m
hS0
5m
d. Let y 5 f(x), then
dy
f(x 1 h) 2 f(x)
5 f r(x) 5 lim
dx
hS0
h
a(x 1 h)2 1 b(x 1 h) 1 c
5 lim c
hS0
h
(ax2 1 bx 1 c)
2
d
h
2-7
5 lim c
ax2 1 2axh 1 ah2 1 bx 1 bh
hS0
h
2ax2 2 bx 2 c
1
d
h
2axh 1 ah2 1 bh
5 lim
hS0
h
5 lim (2ax 1 ah 1 b)
hS0
5 2ax 1 b
13. The slope of the function at a point x is given by
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
(x 1 h)3 2 x3
5 lim
hS0
h
3
x 1 3x2h 1 3xh2 1 h3 2 x3
5 lim
hS0
h
2
2
3x h 1 3xh 1 h3
5 lim
hS0
h
2
5 lim 3x 1 3xh 1 h2
hS0
2
5 3x
Since 3x2 is nonnegative for all x, the original
function never has a negative slope.
14. h(t) 5 18t 2 4.9t2
h(t 1 k) 2 h(t)
a. hr(t) 5 lim
kS0
k
18(t 1 k) 2 4.9(t 1 k)2
5 lim
kS0
k
A18t 2 4.9t2 B
2
k
18t 1 18k 2 4.9t 2 2 9.8tk 2 4.9k2
5 lim
kS0
k
18t 1 4.9t2
2
k
18k 2 9.8tk 2 4.9k2
5 lim
kS0
k
5 lim (18 2 9.8t 2 4.9k)
kS0
5 18 2 9.8t 2 4.9(0)
5 18 2 9.8t
Then hr(2) 5 18 2 9.8(2) 5 21.6 m>s.
b. hr(2) measures the rate of change in the height
of the ball with respect to time when t 5 2.
15. a. This graph has positive slope for x , 0, zero
slope at x 5 0, and negative slope for x . 0, which
corresponds to graph e.
b. This graph has positive slope for x , 0, zero
slope at x 5 0, and positive slope for x . 0, which
corresponds to graph f.
2-8
c. This graph has negative slope for x , 22,
positive slope for 22 , x , 0, negative slope for
0 , x , 2, positive slope for x . 2, and zero slope
at x 5 22, x 5 0, and x 5 2, which corresponds to
graph d.
16. This function is defined piecewise as f(x) 5 2x2
for x , 0, and f(x) 5 x2 for x $ 0. The derivative
will exist if the left-side and right-side derivatives are
the same at x 5 0:
f(0 1 h) 2 f(0)
2 (0 1 h)2 2 A20 2 B
lim2
5 lim2
hS0
h
hS0
h
2h2
5 lim2
hS0
h
5 lim2 (2h)
hS0
50
lim1
hS0
f(0 1 h) 2 f(0)
(0 1 h)2 2 A02 B
5 lim1
h
hS0
h
h2
5 lim1
hS0 h
5 lim1 (h)
hS0
50
Since the limits are equal for both sides, the derivative
exists and f r(0) 5 0.
17. Since f r(a) 5 6 and f(a) 5 0,
f(a 1 h) 2 f(a)
6 5 lim
hS0
h
f(a 1 h) 2 0
6 5 lim
hS0
h
f(a 1 h)
3 5 lim
hS0
2h
18.
y
6
4
2
x
–1
–2
1 2 3 4 5
f(x) is continuous.
f(3) 5 2
But f r(3) 5 `.
(Vertical tangent)
19. y 5 x2 2 4x 2 5 has a tangent parallel to
2x 2 y 5 1.
Let f(x) 5 x2 2 4x 2 5. First, calculate
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
Chapter 2: Derivatives
5 lim c
(x 1 h)2 2 4(x 1 h) 2 5
hS0
h
2
Ax 2 4x 2 5B
2
d
h
x2 1 2xh 1 h2 2 4x 2 4h 2 5
5 lim c
hS0
h
2x2 1 4x 1 5
1
d
h
2xh 1 h2 2 4h
5 lim
hS0
h
5 lim (2x 1 h 2 4)
hS0
5 2x 1 0 2 4
5 2x 2 4
Thus, 2x 2 4 is the slope of the tangent to the curve
at x. We want the tangent parallel to 2x 2 y 5 1.
Rearranging, y 5 2x 2 1.
If the tangent is parallel to this line,
2x 2 4 5 2
x53
When x 5 3, y 5 (3)2 2 4(3) 2 5 5 28.
The point is (3, 28).
20. f(x) 5 x2
The slope of the tangent at any point Ax, x2 B is
f(x 1 h) 2 f(x)
f r 5 lim
hS0
h
(x 1 h)2 2 x2
5 lim
hS0
h
(x 1 h 1 x)(x 1 h 2 x)
5 lim
hS0
h
h(2x 1 h)
5 lim
hS0
h
5 lim (2x 1 h)
hS0
5 2x 1 0
5 2x
Let (a, a2 ) be a point of tangency. The equation of
the tangent is
y 2 a2 5 (2a)(x 2 a)
y 5 (2a)x 2 a2
Suppose the tangent passes through (1, 23).
Substitute x 5 1 and y 5 23 into the equation of
the tangent:
23 5 (2a)(1) 2 a2
2
a 2 2a 2 3 5 0
(a 2 3)(a 1 1) 5 0
a 5 21, 3
So the two tangents are y 5 22x 2 1 or
2x 1 y 1 1 5 0 and y 5 6x 2 9 or 6x 2 y 2 9 5 0.
Calculus and Vectors Solutions Manual
2.2 The Derivatives of Polynomial
Functions, pp. 82–84
1. Answers may vary. For example:
d
constant function rule:
(5) 5 0
dx
d 3
Ax B 5 3x2
power rule:
dx
d
A4x3 B 5 12x2
constant multiple rule:
dx
d 2
Ax 1 xB 5 2x 1 1
sum rule:
dx
d 3
Ax 2 x2 1 3xB 5 3x2 2 2x 1 3
difference rule:
dx
d
d
2. a. f r(x) 5 (4x) 2 (7)
dx
dx
d
d
5 4 (x) 2
(7)
dx
dx
5 4Ax0 B 2 0
54
d
d
b. f r(x) 5 Ax3 B 2 Ax2 B
dx
dx
5 3x2 2 2x
d
d
d
c. f r(x) 5 A2x2 B 1 (5x) 1 (8)
dx
dx
dx
d
d
d
5 2 Ax2 B 1 5 (x) 1
(8)
dx
dx
dx
5 2 (2x) 1 5 1 0
5 22x 1 5
d 3
x)
d. f r(x) 5 ("
dx
d 1
5 ( x3 )
dx
1 1
5 ( x(3 21))
3
1 2
5 (x23)
3
1
5 3 2
3"x
d x 4
e. f r(x) 5 aa b b
dx 2
1 4d 4
5a b
Ax B
2 dx
1
5 A4x3 B
16
x3
5
4
2-9
5 lim c
(x 1 h)2 2 4(x 1 h) 2 5
hS0
h
2
Ax 2 4x 2 5B
2
d
h
x2 1 2xh 1 h2 2 4x 2 4h 2 5
5 lim c
hS0
h
2x2 1 4x 1 5
1
d
h
2xh 1 h2 2 4h
5 lim
hS0
h
5 lim (2x 1 h 2 4)
hS0
5 2x 1 0 2 4
5 2x 2 4
Thus, 2x 2 4 is the slope of the tangent to the curve
at x. We want the tangent parallel to 2x 2 y 5 1.
Rearranging, y 5 2x 2 1.
If the tangent is parallel to this line,
2x 2 4 5 2
x53
When x 5 3, y 5 (3)2 2 4(3) 2 5 5 28.
The point is (3, 28).
20. f(x) 5 x2
The slope of the tangent at any point Ax, x2 B is
f(x 1 h) 2 f(x)
f r 5 lim
hS0
h
(x 1 h)2 2 x2
5 lim
hS0
h
(x 1 h 1 x)(x 1 h 2 x)
5 lim
hS0
h
h(2x 1 h)
5 lim
hS0
h
5 lim (2x 1 h)
hS0
5 2x 1 0
5 2x
Let (a, a2 ) be a point of tangency. The equation of
the tangent is
y 2 a2 5 (2a)(x 2 a)
y 5 (2a)x 2 a2
Suppose the tangent passes through (1, 23).
Substitute x 5 1 and y 5 23 into the equation of
the tangent:
23 5 (2a)(1) 2 a2
2
a 2 2a 2 3 5 0
(a 2 3)(a 1 1) 5 0
a 5 21, 3
So the two tangents are y 5 22x 2 1 or
2x 1 y 1 1 5 0 and y 5 6x 2 9 or 6x 2 y 2 9 5 0.
Calculus and Vectors Solutions Manual
2.2 The Derivatives of Polynomial
Functions, pp. 82–84
1. Answers may vary. For example:
d
constant function rule:
(5) 5 0
dx
d 3
Ax B 5 3x2
power rule:
dx
d
A4x3 B 5 12x2
constant multiple rule:
dx
d 2
Ax 1 xB 5 2x 1 1
sum rule:
dx
d 3
Ax 2 x2 1 3xB 5 3x2 2 2x 1 3
difference rule:
dx
d
d
2. a. f r(x) 5 (4x) 2 (7)
dx
dx
d
d
5 4 (x) 2
(7)
dx
dx
5 4Ax0 B 2 0
54
d
d
b. f r(x) 5 Ax3 B 2 Ax2 B
dx
dx
5 3x2 2 2x
d
d
d
c. f r(x) 5 A2x2 B 1 (5x) 1 (8)
dx
dx
dx
d
d
d
5 2 Ax2 B 1 5 (x) 1
(8)
dx
dx
dx
5 2 (2x) 1 5 1 0
5 22x 1 5
d 3
x)
d. f r(x) 5 ("
dx
d 1
5 ( x3 )
dx
1 1
5 ( x(3 21))
3
1 2
5 (x23)
3
1
5 3 2
3"x
d x 4
e. f r(x) 5 aa b b
dx 2
1 4d 4
5a b
Ax B
2 dx
1
5 A4x3 B
16
x3
5
4
2-9
d 23
(x )
dx
5 (23)( x2321)
5 23x24
d
3. a. hr(x) 5 ((2x 1 3)(x 1 4))
dx
d
5
A2x2 1 8x 1 3x 1 12B
dx
d
d
d
5
A2x2 B 1
(11x) 1
(12)
dx
dx
dx
d
d
d
5 2 Ax2 B 1 11 (x) 1
(12)
dx
dx
dx
5 2(2x) 1 11(1) 1 0
5 4x 1 11
d
b. f r(x) 5 A2x3 1 5x2 2 4x 2 3.75B
dx
d
d
d
5
(2x3 ) 1
A5x2 B 2
(4x)
dx
dx
dx
d
2
(3.75)
dx
d
d
d
5 2 Ax3 B 1 5 Ax2 B 2 4 (x)
dx
dx
dx
d
2
(3.75)
dx
5 2A3x2 B 1 5(2x) 2 4(1) 2 0
5 6x2 1 10x 2 4
ds
d
c.
5 At 2 At 2 2 2tB B
dt
dt
d
5 At 4 2 2t 3 B
dt
d
d
5 At 4 B 2 A2t 3 B
dt
dt
d 4
d
5 At B 2 2 At 3 B
dt
dt
5 4t 3 2 2A3t 2 B
5 4t 3 2 6t 2
dy
d 1 5 1 3 1 2
d.
5
a x 1 x 2 x 1 1b
dx
dx 5
3
2
d 1 5
d 1 3
d 1 2
d
5
a xb1
a xb2
a xb1
(1)
dx 5
dx 3
dx 2
dx
1 d
1 d
1 d
5 a b Ax5 B 1 a b Ax3 B 2 a b Ax2 B
5 dx
3 dx
2 dx
d
1
(1)
dx
1
1
1
5 A5x4 B 1 A3x2 B 2 (2x) 1 0
5
3
2
5 x4 1 x2 2 x
f. f r(x) 5
2-10
d
A5Ax2 B 4 B
dx
d
5 5 Ax234 B
dx
d
5 5 Ax8 B
dx
5 5A8x7 B
5 40x7
d t5 2 3t2
f. sr(t) 5 a
b
dt
2t
1 d
5 a b At 4 2 3tB
2 dt
1 d
d
5 a b a At4 B 2 (3t)b
2 dt
dt
1 d
d
5 a b a At4 B 2 3 (t)b
2 dt
dt
1
5 a b A4t 3 2 3(1)B
2
3
5 2t 3 2
2
dy
d
5
4. a.
5
A3x3 B
dx
dx
d 5
5 3 ( x3)
dx
5
5
5 a b3( xA3 21))
3
2
5 5x3
dy
6
d
1
b.
5
a4x22 2 b
x
dx
dx
d
d
1
5 4 ( x22) 2 6 Ax21 B
dx
dx
21
1
5 4a
b ( x22 21) 2 6(21)Ax2121 B
2
3
5 22x22 1 6x22
dy
d 6
2
5
a 3 1 2 2 3b
c.
dx
dx x
x
d
d
d
5 6 Ax23 B 1 2 Ax22 B 2
(3)
dx
dx
dx
5 6(23)Ax2321 B 1 2(22)Ax2221 B 2 0
e. gr(x) 5
5 218x24 2 4x23
218
4
5 4 2 3
x
x
Chapter 2: Derivatives
d.
dy
d
5
9x22 1 3"x
dx
dx
d
d 1
5 9 Ax22 B 1 3 ( x2)
dx
dx
(
)
(
1
1
5 9(22)Ax2221 B 1 3a b ( x2 21)
2
3
1
5 218x23 1 x22
2
dy
d
e.
5
"x 1 6"x3 1 "2
dx
dx
d 1
d 3
d
5 ( x2) 1 6 ( x2) 1
"2
dx
dx
dx
1 1
3
3
5 ( x2 21) 1 6a b ( x2 21) 1 0
2
2
1
1
1
5 ( x22) 1 9x2
2
(
5 2t 2 6(1) 1 0
5 2t 2 6
d 3
6. a. f r(x) 5
x 2 "x
dx
d 3
d 1
5
Ax B 2 ( x2)
dx
dx
1 1
5 3x2 2 ( x2 21)
2
1 1
5 3x2 2 x22
2
1
1
so f r(a) 5 f r(4) 5 3(4)2 2 (4)22
2
1 1
5 3(16) 2
2 "4
1 1
5 48 2 a b a b
2 2
5 47.75
d
2
7 2 6"x 1 5x3
b. f r(x) 5
dx
d
d 1
d 2
5
(7) 2 6 ( x2) 1 5 ( x3)
dx
dx
dx
1
2
1
2
5 0 2 6a b ( x2 21) 1 5a b ( x3 21)
2
3
10
1
1
5 23x22 1 a b ( x23)
3
10
1
1
so f r(a) 5 f r(64) 5 23( 6422) 1 a b ( 6423)
3
1
10 1
5 23a b 1 a b
8
3 4
11
5
24
dy
d
5
A3x4 B
7. a.
dx
dx
d
5 3 Ax4 B
dx
5 3A4x3 B
5 12x3
The slope at (1, 3) is found by substituting x 5 1 into
)
( )
dy
d 1 1 "x
5
a
b
x
dx
dx
1
d 1
d x2
5
a b1
a b
dx x
dx x
d 21
d
1
5
(x ) 1 ( x22)
dx
dx
21 212 21
5 (21)x2121 1
(x )
2
1 3
5 2x22 2 x22
2
ds
d
5. a.
5 A22t 2 1 7tB
dt
dt
d
d
5 (22)a At2 B b 1 7a (t)b
dt
dt
5 (22)(2t) 1 7(1)
5 24t 1 7
ds
d
1
b.
5 a18 1 5t 2 t 3 b
dt
dt
3
d
d
1 d
5 (18) 1 5 (t) 2 a b At 3 B
dt
dt
3 dt
1
5 0 1 5(1) 2 a b A3t 2 B
3
5 5 2 t2
d
ds
c.
5 A(t 2 3)2B
dt
dt
d
5 At 2 2 6t 1 9B
dt
d
d
d
5 At 2 B 2 (6) (t) 1 (9)
dt
dt
dt
f.
Calculus and Vectors Solutions Manual
(
)
)
dy
the equation for dx. So the slope 5 12(1)3
5 12
dy
d 1
b
5
a
b.
dx
dx x25
d 5
5
Ax B
dx
5 5x4
2-11
The slope at (21, 21) is found by substituting x 5 21
The slope at x 5 4 is found by substituting x 5 4
dy
into the equation for dx. So the
into the equation for dx. So the slope is (4) 2 5 12.
16
c. y 5 2
x
dy
d 16
5
a b
dx
dx x2
d
5 16 Ax22 B
dx
5 16(22)x2221
5 232x23
The slope at x 5 22 is found by substituting
dy
slope 5 5(21)4
55
dy
d 2
c.
5
a b
dx
dx x
d
5 2 Ax21 B
dx
5 2(21)x2121
5 22x22
The slope at (22 , 21) is found by substituting
dy
x 5 22 into the equation for dx. So the
slope 5 22(22)22
1
52
2
dy
d
d.
5 ("16x3)
dx
dx
d 3
5 "16 ( x2)
dx
3 3
5 4a bx2 21
2
1
2
5 6x
The slope at (4, 32) is found by substituting x 5 4
into the equation for
1
dy
. So the
dx
slope 5 6(4)2
5 12
8. a. y 5 2x3 1 3x
dy
d
5
A2x3 1 3xB
dx
dx
d
d
5 2 Ax3 B 1 3 (x)
dx
dx
5 2A3x2 B 1 3(1)
5 6x2 1 3
The slope at x 5 1 is found by substituting x 5 1
dy
into the equation for dx. So the slope is
6(1)2 1 3 5 9.
b. y 5 2"x 1 5
dy
d
5 (2"x 1 5)
dx
dx
d 1
d
5 2 ( x2 ) 1
(5)
dx
dx
1
1
5 2a b ( x2 21) 1 0
2
21
5 x2
2-12
21
x 5 22 into the equation for
23
232(22)
5
(232)
(22)3
dy
.
dx
So the slope is
5 4.
d. y 5 x23 (x21 1 1)
5 x24 1 x23
dy
d 24
5
(x 1 x23 )
dx
dx
5 24x25 2 3x24
3
4
52 52 4
x
x
The slope at x 5 1 is found by substituting
dy
x 5 1 into the equation for dx. So the slope is
2 145 2 134 5 27.
dy
d
1
5
a2x 2 b
x
dx
dx
d
d 21
5 2 (x) 2
Ax B
dx
dx
5 2(1) 2 (21)x2121
5 2 1 x22
The slope at x 5 0.5 is found by substituting
9. a.
dy
x 5 0.5 into the equation for dx.
So the slope is 2 1 (0.5)22 5 6.
The equation of the tangent line is therefore
y 1 1 5 6(x 2 0.5) or 6x 2 y 2 4 5 0.
dy
d 3
4
b.
5
a 2 3b
dx
dx x2
x
d
d
5 3 Ax22 B 2 4 Ax23 B
dx
dx
5 3(22)x2221 2 4(23)x2321
5 12x24 2 6x23
The slope at x 5 21 is found by substituting
dy
x 5 21 into the equation for dx. So the slope is
12(21)24 2 6(21)23 5 18.
Chapter 2: Derivatives
The equation of the tangent line is therefore
y 2 7 5 18(x 1 1) or 18x 2 y 1 25 5 0.
d
dy
c.
5 ("3x3)
dx
dx
d 3
5 "3 ( x2)
dx
3 32 21
5 "3a bx
2
1
3"3x2
5
2
The slope at x 5 3 is found by substituting x 5 3
dy
into the equation for dx.
1
3"3(3)2
9
So the slope is
5 .
2
2
The equation of the tangent line is therefore
y 2 9 5 92 (x 2 3) or 9x 2 2y 2 9 5 0.
dy
d 1 2 1
d.
5
a ax 1 bb
x
dx
dx x
d
1
5
ax 1 2 b
dx
x
d
d 22
5
(x) 1
Ax B
dx
dx
5 1 1 (22)x2221
5 1 2 2x23
The slope at x 5 1 is found by substituting
dy
into the equation for dx.
So the slope is 1 2 2(1)23 5 21.
The equation of the tangent line is therefore
y 2 2 5 2 (x 2 1) or x 1 y 2 3 5 0.
dy
d
e.
5 (("x 2 2)(3"x 1 8))
dx
dx
d
5 (3("x)2 1 8"x 2 6"x 2 16)
dx
d
5 (3x 1 2"x 2 16)
dx
d
d 1
d
5
(3x) 1 2 ( x2) 2
(16)
dx
dx
dx
1 1
5 3(1) 1 2a bx2 21 2 0
2
1
5 3 1 x22
The slope at x 5 4 is found by substituting x 5 4
The equation of the tangent line is therefore
y 5 3.5(x 2 4) or 7x 2 2y 2 28 5 0.
dy
d "x 2 2
f.
5
a 3
b
dx
dx
"x
1
d x2 2 2
5
a
b
1
dx
x3
d 1 1
1
5 ( x2 2 3 2 2x23)
dx
d 1
d
1
5 ( x6) 2 2 Ax23 B
dx
dx
1 1
1
1
5 ( x6 21) 2 2a2 bx23 21 2 0
6
3
2 243
1 256
5 (x ) 1 x
6
3
The slope at x 5 1 is found by substituting x 5 1
dy
into the equation for dx.
5
4
So the slope is 16 (1)26 1 23 (1)23 5 56.
The equation of the tangent line is therefore
y 1 1 5 56 (x 2 1) or 5x 2 6y 2 11 5 0.
10. A normal to the graph of a function at a point is
a line that is perpendicular to the tangent at the
given point.
3
4
y 5 2 2 3 at P(21, 7)
x
x
Slope of the tangent is 18, therefore, the slope of
the normal is 2 181 .
1
Equation is y 2 7 5 2 (x 1 1).
18
x 1 18y 2 125 5 0
3
1
11. y 5 3 5 3x23
"x
Parallel to x 1 16y 1 3 2 0
Slope of the line is 2 161 .
dy
4
5 2x23
dx
1
4
x23 5
16
1
1
4 5
3
16
x
4
x3 5 16
3
x 5 (16)4 5 8
dy
into the equation for dx.
1
So the slope is 3 1 (4)22 5 3.5.
Calculus and Vectors Solutions Manual
2-13
1
5 x21 : y 5 x3
x
dy
1 dy
52 2:
5 3x2
dx
x dx
1
Now, 2 2 5 3x2
x
1
x4 5 2 .
3
No real solution. They never have the same slope.
dy
5 2x
13. y 5 x2,
dx
The slope of the tangent at A(2, 4) is 4 and at
B Q 2 18 , 641 R is 2 14.
Since the product of the slopes is 21, the tangents
at A(2, 4) and B Q 2 18 , 641 R will be perpendicular.
12. y 5
4 y
3
2
1
–3 –2 –1 0
–1
x
1 2 3
14. y 5 2x2 1 3x 1 4
dy
5 22x 1 3
dx
dy
55
For
dx
5 5 22x 1 3
x 5 21.
The point is (21, 0).
6
5
4
3
2
1
–2 –1 0
–1
–2
y
x
1
2 3 4
15. y 5 x3 1 2
dy
5 3x2, slope is 12
dx
x2 5 4
x 5 2 or x 5 22
Points are (2, 10) and (22, 26).
2-14
16. y 5 15x5 2 10x, slope is 6
dy
5 x4 2 10 5 6
dx
x4 5 16
x2 5 4 or x2 5 24
x 5 62 non-real
Tangents with slope 6 are at the points Q 2, 2 685 R
and Q2 2, 685 R .
17. y 5 2x2 1 3
a. Equation of tangent from A(2, 3):
If x 5 a, y 5 2x2 1 3.
Let the point of tangency be PAa, 2a2 1 3B.
dy
dy
Now, dx 5 4x and when x 5 a, dx 5 4a.
The slope of the tangent is the slope of AP.
2a2
5 4a.
a22
2a2 5 4a2 2 8a
2
2a 2 8a 5 0
2a(a 2 4) 5 0
a 5 0 or a 5 4.
Point (2, 3):
Slope is 0.
Slope is 16.
Equation of tangent is
Equation of tangent is
y 2 3 5 0.
y 2 3 5 16(x 2 2) or
16x 2 y 2 29 5 0.
b. From the point B(2, 27):
2a2 1 10
Slope of BP:
5 4a
a22
2a2 1 10 5 4a2 2 8a
2
2a 2 8a 2 10 5 0
a2 2 4a 2 5 5 0
(a 2 5)(a 1 1) 5 0
a55
a 5 21
Slope is 4a 5 20.
Slope is 4a 5 24.
Equation is
Equation is
y 1 7 5 20(x 2 2)
y 1 7 5 24(x 2 2)
or 20x 2 y 2 47 5 0.
or 4x 1 y 2 1 5 0.
a
18. ax 2 4y 1 21 5 0 is tangent to y 5 x2 at x 5 22.
Therefore, the point of tangency is a22, 4 b,
This point lies on the line, therefore,
a
a(22) 2 4a b 1 21 5 0
4
23a 1 21 5 0
a 5 7.
a
Chapter 2: Derivatives
19. a. When h 5 200,
d 5 3.53"200
8 49.9
Passengers can see about 49.9 km.
1
b. d 5 3.53!h 5 3.53h2
1 1
dr 5 3.53a h 2 2 b
2
3.53
5
2!h
When h 5 200,
3.53
dr 5
2 !200
8 0.12
The rate of change is about 0.12 km>m.
20. d(t) 5 4.9t2
a. d(2) 5 4.9(2)2 5 19.6 m
d(5) 5 4.9(5)2 5 122.5 m
The average rate of change of distance with respect
to time from 2 s to 5 s is
Dd
122.5 2 19.6
5
Dt
522
5 34.3 m>s
b. dr(t) 5 9.8t
Thus, dr(4) 5 9.8(4) 5 39.2 m>s.
c. When the object hits the ground, d 5 150.
Set d(t) 5 150:
4.9t2 5 150
1500
t2 5
49
10
t 5 6 "15
7
10
Since t $ 0, t 5 "15
7
Then,
10
10
dra "15b 5 9.8a "15b
7
7
8 54.2 m>s
21. v(t) 5 sr(t) 5 2t 2 t2
0.5 5 2t 2 t2
t2 2 2t 1 0.5 5 0
2t2 2 4t 1 1 5 0
4 6 "8
t5
4
t 8 1.71, 0.29
The train has a velocity of 0.5 km>min at about
0.29 min and 1.71 min.
Calculus and Vectors Solutions Manual
22. v(t) 5 Rr(t) 5 210t
v(2) 5 220
The velocity of the bolt at t 5 2 is 220 m>s.
23.
y
3 (0, 3)
2
1
–3 –2 –1 0
–1
–2
–3
1
x
2 3
Let the coordinates of the points of tangency be
AAa,23a2 B.
dy
5 26x, slope of the tangent at A is 26a
dx
23a2 2 3
5 26a
a
23a2 2 3 5 26a2
3a2 5 3
a 5 1 or a 5 21
Coordinates of the points at which the tangents
touch the curve are (1, 23) and (21, 23).
24. y 5 x3 2 6x2 1 8x, tangent at A(3, 23)
dy
5 3x2 2 12x 1 8
dx
When x 5 3,
dy
5 27 2 36 1 8 5 21
dx
The slope of the tangent at A(3, 23) is 21.
Equation will be
y 1 3 5 21(x 2 3)
y 5 2x.
2x 5 x3 2 6x2 1 8x
x3 2 6x2 1 9x 5 0
xAx2 2 6x 1 9B 5 0
x(x 2 3)2 5 0
x 5 0 or x 5 3
Coordinates are B(0, 0).
Slope of PA:
3
2
1
–1 0
–11
–2
–3
y
x
1
2 3 4
2-15
25. a. i. f(x) 5 2x 2 5x2
f r(x) 5 2 2 10x
Set f r(x) 5 0:
2 2 10x 5 0
10x 5 2
1
x5
5
Then,
1
1
1 2
f a b 5 2a b 2 5a b
5
5
5
2
1
5 2
5
5
1
5
5
Thus the point is Q 15, 15 R .
ii. f(x) 5 4x2 1 2x 2 3
f r(x) 5 8x 1 2
Set f r(x) 5 0:
8x 1 2 5 0
8x 5 22
1
x52
4
Then,
1
1 2
1
f a2 b 5 4a2 b 1 2a2 b 2 3
4
4
4
1
2
12
5 2 2
4
4
4
13
52
4
Thus the point is Q 2 14, 2 134 R .
iii. f(x) 5 x3 2 8x2 1 5x 1 3
f r(x) 5 3x2 2 16x 1 5
Set f r(x) 5 0:
3x2 2 16x 1 5 5 0
2
3x 2 15x 2 x 1 5 5 0
3x(x 2 5) 2 (x 2 5) 5 0
(3x 2 1)(x 2 5) 5 0
1
x5 ,5
3
1 3
1 2
1
1
f a b 5 a b 2 8a b 1 5a b 1 3
3
3
3
3
1
24
45
81
5
2
1
1
27
27
27
27
103
5
27
f(5) 5 (5)3 2 8(5)2 1 5(5) 1 3
5 25 2 200 1 25 1 3
5 247
Thus the two points are Q 13, 103
27 R and (5, 247).
b. At these points, the slopes of the tangents are
zero, meaning that the rate of change of the value
of the function with respect to the domain is zero.
These points are also local maximum or minimum
points.
26. "x 1 "y 5 1
P(a, b) is on the curve, therefore a $ 0, b $ 0.
!y 5 1 2 !x
y 5 1 2 2 !x 1 x
1
dy
1
5 2 ? 2x22 1 1
dx
2
21 1 !a
1
At x 5 a, slope is 2
115
.
!a
!a
But !a 1 !b 5 1
2 !b 5 !a 2 1.
"b
b
Therefore, slope is 2
52
.
Åa
"a
27. f(x) 5 xn, f r(x) 5 nxn21
Slope of the tangent at x 5 1 is f r(1) 5 n,
The equation of the tangent at (1, 1) is:
y 2 1 5 n(x 2 1)
nx 2 y 2 n 1 1 5 0
Let y 5 0, nx 5 n 2 1
n21
1
x5
512 .
n
n
1
1
The x-intercept is 1 2 ; as n S `, S 0, and
n
n
the x-intercept approaches 1. As n S `, the slope
of the tangent at (1, 1) increases without bound, and
the tangent approaches a vertical line having equation
x 2 1 5 0.
28. a.
y
9
8
7
6
5
4
3
2
1
–2 –1 0
2-16
x
1
2 3 4
Chapter 2: Derivatives
f(x) 5 e
x2, if x , 3
x 1 6, if x $ 3
f r(3) does not exist.
b.
7
6
5
4
3
2
1
2x, if x , 3
1, if x $ 3
y
x
–3 –2 –1 0
–1
f(x) 5 e
f '(x) 5 e
1 2 3
3x2 2 6, if x , 2"2 or x . "2
6 2 3x2, if 2"2 , x , "2
f r(x) 5 e
6x, if x , 2"2 or x . "2
26x, if 2"2 # x # "2
f r "2 and f r 2"2 do not exist.
( )
(
c.
3
2
1
–3 –2 –1 0
–1
)
y
c. h(x) 5 (3x 1 2)(2x 2 7)
hr(x) 5 (3x 1 2)(2) 1 (3)(2x 2 7)
5 12x 2 17
d. h(x) 5 A5x7 1 1B Ax2 2 2xB
hr(x) 5 A5x7 1 1B (2x 2 2) 1 A35x6 BAx2 2 2xB
5 45x8 2 80x7 1 2x 2 2
e. s(t) 5 At2 1 1B A3 2 2t2 B
sr(t) 5 At2 1 1B (24t) 1 (2t)A3 2 2t2 B
5 28t3 1 2t
x23
f. f(x) 5
x13
f(x) 5 (x 2 3)(x 1 3)21
f r(x) 5 (x 2 3)(21)(x 1 3)22 1 (1)(x 1 3)21
5 (x 1 3)22 (2x 1 3 1 x 1 3)
6
5
(x 1 3)2
2. a. y 5 (5x 1 1)3 (x 2 4)
dy
5 (5x 1 1)3 (1) 1 3(5x 1 1)2 (5)(x 2 4)
dx
5 (5x 1 1)3 1 15(5x 1 1)2 (x 2 4)
b. y 5 A3x2 1 4BA3 1 x3 B 5
dy
5 A3x2 1 4B (5)A3 1 x3 B 4 A3x2 B
dx
1 (6x)A3 1 x3 B 5
5 15x2 (3x2 1 4)(3 1 x3 )4 1 6x(3 1 x3 )5
x
1 2 3
x 2 1, if x $ 1
1 2 x, if 0 # x , 1
f(x) 5 μ
x 1 1, if 21 , x , 0
2x 2 1, if x # 21
since
since
since
since
1, if x . 1
21, if 0 , x , 1
f'(x) 5 μ
1, if 21 , x , 0
21, if x , 21
f r(0), f r(21), and f r(1) do not exist.
2.3 The Product Rule, pp. 90–91
1. a. h(x) 5 x(x 2 4)
hr(x) 5 x(1) 1 (1)(x 2 4)
5 2x 2 4
b. h(x) 5 x2 (2x 2 1)
hr(x) 5 x2 (2) 1 (2x)(2x 2 1)
5 6x2 2 2x
Calculus and Vectors Solutions Manual
Zx 2 1 Z 5 x 2 1
Zx 2 1 Z 5 1 2 x
Z2x 2 1 Z 5 x 1 1
Z2x21 Z 5 2x 2 1
y 5 A1 2 x2 B 4 (2x 1 6)3
dy
5 4A1 2 x2 B 3 (22x)(2x 1 6)3
dx
1 A1 2 x2 B 4 3(2x 1 6)2 (2)
5 28xA1 2 x2 B 3 (2x 1 6)3
1 6A1 2 x2 B 4 (2x 1 6)2
d. y 5 Ax2 2 9B 4 (2x 2 1)3
dy
5 Ax2 2 9B 4 (3)(2x 2 1)2 (2)
dx
1 4Ax2 2 9B 3 (2x)(2x 2 1)3
5 6(x2 2 9)4 (2x 2 1)2
1 8x(x2 2 9)3 (2x 2 1)3
c.
2-17
f(x) 5 e
x2, if x , 3
x 1 6, if x $ 3
f r(3) does not exist.
b.
7
6
5
4
3
2
1
2x, if x , 3
1, if x $ 3
y
x
–3 –2 –1 0
–1
f(x) 5 e
f '(x) 5 e
1 2 3
3x2 2 6, if x , 2"2 or x . "2
6 2 3x2, if 2"2 , x , "2
f r(x) 5 e
6x, if x , 2"2 or x . "2
26x, if 2"2 # x # "2
f r "2 and f r 2"2 do not exist.
( )
(
c.
3
2
1
–3 –2 –1 0
–1
)
y
c. h(x) 5 (3x 1 2)(2x 2 7)
hr(x) 5 (3x 1 2)(2) 1 (3)(2x 2 7)
5 12x 2 17
d. h(x) 5 A5x7 1 1B Ax2 2 2xB
hr(x) 5 A5x7 1 1B (2x 2 2) 1 A35x6 BAx2 2 2xB
5 45x8 2 80x7 1 2x 2 2
e. s(t) 5 At2 1 1B A3 2 2t2 B
sr(t) 5 At2 1 1B (24t) 1 (2t)A3 2 2t2 B
5 28t3 1 2t
x23
f. f(x) 5
x13
f(x) 5 (x 2 3)(x 1 3)21
f r(x) 5 (x 2 3)(21)(x 1 3)22 1 (1)(x 1 3)21
5 (x 1 3)22 (2x 1 3 1 x 1 3)
6
5
(x 1 3)2
2. a. y 5 (5x 1 1)3 (x 2 4)
dy
5 (5x 1 1)3 (1) 1 3(5x 1 1)2 (5)(x 2 4)
dx
5 (5x 1 1)3 1 15(5x 1 1)2 (x 2 4)
b. y 5 A3x2 1 4BA3 1 x3 B 5
dy
5 A3x2 1 4B (5)A3 1 x3 B 4 A3x2 B
dx
1 (6x)A3 1 x3 B 5
5 15x2 (3x2 1 4)(3 1 x3 )4 1 6x(3 1 x3 )5
x
1 2 3
x 2 1, if x $ 1
1 2 x, if 0 # x , 1
f(x) 5 μ
x 1 1, if 21 , x , 0
2x 2 1, if x # 21
since
since
since
since
1, if x . 1
21, if 0 , x , 1
f'(x) 5 μ
1, if 21 , x , 0
21, if x , 21
f r(0), f r(21), and f r(1) do not exist.
2.3 The Product Rule, pp. 90–91
1. a. h(x) 5 x(x 2 4)
hr(x) 5 x(1) 1 (1)(x 2 4)
5 2x 2 4
b. h(x) 5 x2 (2x 2 1)
hr(x) 5 x2 (2) 1 (2x)(2x 2 1)
5 6x2 2 2x
Calculus and Vectors Solutions Manual
Zx 2 1 Z 5 x 2 1
Zx 2 1 Z 5 1 2 x
Z2x 2 1 Z 5 x 1 1
Z2x21 Z 5 2x 2 1
y 5 A1 2 x2 B 4 (2x 1 6)3
dy
5 4A1 2 x2 B 3 (22x)(2x 1 6)3
dx
1 A1 2 x2 B 4 3(2x 1 6)2 (2)
5 28xA1 2 x2 B 3 (2x 1 6)3
1 6A1 2 x2 B 4 (2x 1 6)2
d. y 5 Ax2 2 9B 4 (2x 2 1)3
dy
5 Ax2 2 9B 4 (3)(2x 2 1)2 (2)
dx
1 4Ax2 2 9B 3 (2x)(2x 2 1)3
5 6(x2 2 9)4 (2x 2 1)2
1 8x(x2 2 9)3 (2x 2 1)3
c.
2-17
3. It is not appropriate or necessary to use the product
rule when one of the factors is a constant or when it
would be easier to first determine the product of the
factors and then use other rules to determine the
derivative. For example, it would not be best to
use the product rule for f(x) 5 3Ax2 1 1B or
g(x) 5 (x 1 1) (x 2 1).
4. F(x) 5 3b(x)43c(x)4
F r(x) 5 3b(x)43cr(x)4 1 3br(x)43c(x)4
5. a. y 5 (2 1 7x)(x 2 3)
dy
5 (2 1 7x)(1) 1 7(x 2 3)
dx
At x 5 2,
dy
5 (2 1 14) 1 7(21)
dx
5 16 2 7
59
b.
y 5 (1 2 2x)(1 1 2x)
dy
5 (1 2 2x)(2) 1 (22)(1 1 2x)
dx
1
At x 5 ,
2
dy
5 (0)(2) 2 2(2)
dx
5 24
c.
y 5 A3 2 2x 2 x2 B Ax2 1 x 2 2B
dy
5 A3 2 2x 2 x2 B A2x 1 1B
dx
1 (22 2 2x)Ax2 1 x 2 2B
At x 5 22,
dy
5 (3 1 4 2 4)(24 1 1)
dx
1 (22 1 4)(4 2 2 2 2)
5 (3)(23) 1 (2)(0)
5 29
d. y 5 x3 (3x 1 7)2
dy
5 3x2 (3x 1 7)2 1 x36(3x 1 7)
dx
At x 5 22,
dy
5 12(1)2 1 (28)(6)(1)
dx
5 12 2 48
5 236
e.
y 5 (2x 1 1)5 (3x 1 2)4, x 5 21
dy
5 5(2x 1 1)4 (2)(3x 1 2)4
dx
1 (2x 1 1)54(3x 1 2)3 (3)
At x 5 21,
2-18
dy
5 5(21)4 (2)(21)4
dx
1 (21)5 (4)(21)3 (3)
5 10 1 12
5 22
f.
y 5 x(5x 2 2)(5x 1 2)
5 xA25x2 2 4B
dy
5 x(50x) 1 (25x2 2 4)(1)
dx
At x 5 3,
dy
5 3(150) 1 (25 ? 9 2 4)
dx
5 450 1 221
5 671
6. Tangent to y 5 Ax3 2 5x 1 2B A3x2 2 2xB
at (1, 22)
dy
5 A3x2 2 5B A3x2 2 2xB
dx
1 Ax3 2 5x 1 2B (6x 2 2)
when x 5 1,
dy
5 (22)(1) 1 (22)(4)
dx
5 22 1 28
5 210
Slope of the tangent at (1, 22) is 210.
The equation is y 1 2 5 210(x 2 1);
10x 1 y 2 8 5 0.
7. a. y 5 2(x 2 29)(x 1 1)
dy
5 2(x 2 29)(1) 1 2(1)(x 1 1)
dx
2x 2 58 1 2x 1 2 5 0
4x 2 56 5 0
4x 5 56
x 5 14
Point of horizontal tangency is (14, 2450).
b. y 5 Ax2 1 2x 1 1B Ax2 1 2x 1 1B
5 Ax2 1 2x 1 1B 2
dy
5 2Ax2 1 2x 1 1B (2x 1 2)
dx
Ax2 1 2x 1 1B (2x 1 2) 5 0
2(x 1 1)(x 1 1)(x 1 1) 5 0
x 5 21
Point of horizontal tangency is (21, 0).
8. a. y 5 (x 1 1)3 (x 1 4)(x 2 3)2
dy
5 3(x 1 1)2 (x 1 4)(x 2 3)2
dx
1 (x 1 1)3 (1)(x 2 3)2
1 (x 1 1)3 (x 1 4)32(x 2 3)4
Chapter 2: Derivatives
b. y 5 x2 A3x2 1 4B 2 A3 2 x3 B 4
dy
5 2xA3x2 1 4B 2 A3 2 x3 B 4
dx
1 x2 32A3x2 1 4B (6x)4 A3 2 x3 B 4
1 x2 A3x2 1 4B 2 34A3 2 x3 B 3 A23x2 B4
t 2
9. V(t) 5 75a1 2 b , 0 # t # 24
24
75 L 3 60% 5 45 L
t 2
45
5 a1 2 b
Set
75
24
t
3
512
6
24
Å5
3
t5a6
2 1b (224)
Å5
t 8 42.590 (inadmissable) or t 8 5.4097
t 2
V(t) 5 75a1 2 b
24
t
t
V(t) 5 75a1 2 b a1 2 b
24
24
t
1
Vr(t) 5 75 c a1 2 b a2 b
24
24
1
t
1 a2 b a1 2 b d
24
24
t
1
5 (75)(2)a1 2 b a2 b
24
24
Vr(5.4097) 5 24.84 L>h
10. Determine the point of tangency, and then find the
negative reciprocal of the slope of the tangent. Use
this information to find the equation of the normal.
h(x) 5 2x(x 1 1)3 Ax2 1 2x 1 1B 2
hr(x) 5 2(x 1 1)3 Ax2 1 2x 1 1B 2
1 (2x)(3)(x 1 1)2 Ax2 1 2x 1 1B 2
1 2x(x 1 1)3 2Ax2 1 2x 1 1B (2x 1 2)
hr(22) 5 2(21)3 (1)2
1 2(22)(3)(21)2 (1)2
1 2(22)(21)3 (2)(1)(22)
5 22 2 12 2 16
5 230
11.
a. f(x) 5 g1 (x)g2 (x)g3 (x) c gn21 (x)gn (x)
f r(x) 5 g1r(x)g2 (x)g3 (x) c gn21 (x)gn (x)
1 g1 (x)g2r(x)g3 (x) c gn21 (x)gn (x)
1 g1 (x)g2 (x)g3r(x) c gn21 (x)gn (x)
1 c 1 g1 (x)g2 (x)g3 (x) cgn21 (x)gnr(x)
Calculus and Vectors Solutions Manual
f(x) 5 (1 1 x)(1 1 2x)(1 1 3x) c
(1 1 nx)
f r(x) 5 1(1 1 2x)(1 1 3x) c(1 1 nx)
1 (1 1 x)(2)(1 1 3x) c(1 1 nx)
1 (1 1 x)(1 1 2x)(3) c(1 1 nx)
1 c 1 (1 1 x)(1 1 2x)(1 1 3x)
c (n)
f r(0) 5 1(1)(1)(1) c (1)
1 1(2)(1)(1) c (1)
1 1(1)(3)(1) c (1)
1 c 1 (1)(1)(1) c (n)
5112131c1n
n(n 1 1)
f r(0) 5
2
12. f(x) 5 ax2 1 bx 1 c
f r(x) 5 2ax 1 b
(1)
Horizontal tangent at (21, 28)
f r(x) 5 0 at x 5 21
22a 1 b 5 0
Since (2, 19) lies on the curve,
4a 1 2b 1 c 5 19
(2)
Since (21, 28) lies on the curve,
a 2 b 1 c 5 28
(3)
4a 1 2b 1 c 5 19
23a 2 3b 5 227
a1b59
22a 1 b 5 0
3a 5 9
a 5 3, b 5 6
3 2 6 1 c 5 28
c 5 25
The equation is y 5 3x2 1 6x 2 5.
13.
y
3
2
1
x
0
–3 –2 –1
1 2 3
–1
b.
a. x 5 1 or x 5 21
b. f r(x) 5 2x, x , 21 or x . 1
f r(x) 5 22x, 21 , x , 1
6
4
2
–3 –2 –1 0
–2
–4
–6
y
x
1 2 3
2-19
c. f r(22) 5 2(22) 5 24
f r(0) 5 22(0) 5 0
f r(3) 5 2(3) 5 6
16
14. y 5 2 2 1
x
dy
32
52 3
dx
x
Slope of the line is 4.
32
2 3 54
x
4x3 5 232
x3 5 28
x 5 22
16
y5
21
4
53
Point is at (22, 3).
Find intersection of line and curve:
4x 2 y 1 11 5 0
y 5 4x 1 11
Substitute,
16
4x 1 11 5 2 2 1
x
4x3 1 11x2 5 16 2 x2 or 4x3 1 12x2 2 16 5 0.
Let x 5 22
RS 5 4(22)3 1 12(22)2 2 16
50
Since x 5 22 satisfies the equation, therefore it is
a solution.
When x 5 22, y 5 4(22) 1 11 5 3.
Intersection point is (22, 3). Therefore, the line is
tangent to the curve.
Mid-Chapter Review, pp. 92–93
1. a.
6
4
2
y
x
–2 0
–2
–4
–6
2 4 6
˛
x
Slope of Tangent
f 9(x)
0
25
1
23
2
21
3
1
4
3
5
5
y
c.
6
4
2
–2 0
–2
–4
–6
x
2 4 6
d. f(x) is quadratic; f r(x) is linear.
(6(x 1 h) 1 15) 2 (6x 1 15)
2. a. f r(x) 5 lim
hS0
h
6h
5 lim
hS0 h
5 lim 6
˛
hS0
56
A2(x 1 h)2 2 4B 2 A2x2 2 4B
b. f r(x) 5 lim
hS0
h
(x 1 h) 2 2 x2
5 lim2
hS0
h
((x 1 h) 2 x)((x 1 h) 1 x)
5 lim2
hS0
h
h(2x 1 h)
5 lim2
hS0
h
5 lim2(2x 1 h)
hS0
5 4x
((x 1 h)2 2 5(x 1 h)) 2 (x2 2 5x)
b. f r(x) 5 lim
hS0
h
x2 1 2hx 1 h2 2 5x 2 5h 2 x2 1 5x
5 lim
hS0
h
h2 1 2hx 2 5h
5 lim
hS0
h
˛
˛
˛
2-20
h(h 1 2x 2 5)
hS0
h
5 2x 2 5
Use the derivative function to calculate the slopes of
the tangents.
5 lim
c. f r(x) 5 lim
5
5
2x15
(x 1 h) 1 5
h
5(x 1 5) 2 5((x 1 h) 1 5)
5 lim
hS0
((x 1 h) 1 5)(x 1 5)h
25h
5 lim
h S 0 ((x 1 h) 1 5) (x 1 5)h
hS0
Chapter 2: Derivatives
c. f r(22) 5 2(22) 5 24
f r(0) 5 22(0) 5 0
f r(3) 5 2(3) 5 6
16
14. y 5 2 2 1
x
dy
32
52 3
dx
x
Slope of the line is 4.
32
2 3 54
x
4x3 5 232
x3 5 28
x 5 22
16
y5
21
4
53
Point is at (22, 3).
Find intersection of line and curve:
4x 2 y 1 11 5 0
y 5 4x 1 11
Substitute,
16
4x 1 11 5 2 2 1
x
4x3 1 11x2 5 16 2 x2 or 4x3 1 12x2 2 16 5 0.
Let x 5 22
RS 5 4(22)3 1 12(22)2 2 16
50
Since x 5 22 satisfies the equation, therefore it is
a solution.
When x 5 22, y 5 4(22) 1 11 5 3.
Intersection point is (22, 3). Therefore, the line is
tangent to the curve.
Mid-Chapter Review, pp. 92–93
1. a.
6
4
2
y
x
–2 0
–2
–4
–6
2 4 6
˛
x
Slope of Tangent
f 9(x)
0
25
1
23
2
21
3
1
4
3
5
5
y
c.
6
4
2
–2 0
–2
–4
–6
x
2 4 6
d. f(x) is quadratic; f r(x) is linear.
(6(x 1 h) 1 15) 2 (6x 1 15)
2. a. f r(x) 5 lim
hS0
h
6h
5 lim
hS0 h
5 lim 6
˛
hS0
56
A2(x 1 h)2 2 4B 2 A2x2 2 4B
b. f r(x) 5 lim
hS0
h
(x 1 h) 2 2 x2
5 lim2
hS0
h
((x 1 h) 2 x)((x 1 h) 1 x)
5 lim2
hS0
h
h(2x 1 h)
5 lim2
hS0
h
5 lim2(2x 1 h)
hS0
5 4x
((x 1 h)2 2 5(x 1 h)) 2 (x2 2 5x)
b. f r(x) 5 lim
hS0
h
x2 1 2hx 1 h2 2 5x 2 5h 2 x2 1 5x
5 lim
hS0
h
h2 1 2hx 2 5h
5 lim
hS0
h
˛
˛
˛
2-20
h(h 1 2x 2 5)
hS0
h
5 2x 2 5
Use the derivative function to calculate the slopes of
the tangents.
5 lim
c. f r(x) 5 lim
5
5
2x15
(x 1 h) 1 5
h
5(x 1 5) 2 5((x 1 h) 1 5)
5 lim
hS0
((x 1 h) 1 5)(x 1 5)h
25h
5 lim
h S 0 ((x 1 h) 1 5) (x 1 5)h
hS0
Chapter 2: Derivatives
25
h S 0 ((x 1 h) 1 5) (x 1 5)
25
5
(x 1 5) 2
!(x 1 h) 2 2 2 !x 2 2
d. f r(x) 5 lim
hS0
h
5 lim
5 lim c
hS0
!(x 1 h) 2 2 2 !x 2 2
h
3
!(x 1 h) 2 2 1 !x 2 2
d
!(x 1 h) 2 2 1 !x 2 2
((x 1 h) 2 2) 2 (x 2 2)
1 h) 2 2 1 !x 2 2B
hS0 hA !(x
5 lim
h
hS0 hA !(x 1 h) 2 2 1 !x 2 2B
5 lim
1
hS0 !(x 1 h) 2 2 1 !x 2 2
1
5
2 !x 2 2
3. a. yr 5 2x 2 4
When x 5 1,
yr 5 2(1) 2 4
5 22.
When x 5 1,
y 5 (1)2 2 4(1) 1 3
5 0.
Equation of the tangent line:
y 2 0 5 22(x 2 1), or y 5 22x 1 2
b.
y
6
4
2
x
0
–4 –2
2 4 6
–2
–4
–6
5 lim
dy
5 24x3
dx
dy
1
b.
5 5x22
dx
5
5
!x
c. gr(x) 5 26x24
6
52 4
x
4. a.
Calculus and Vectors Solutions Manual
dy
5 5 2 6x23
dx
6
552 3
x
dy
5 2(11t 1 1)(11)
e.
dt
5 242t 1 22
1
f. y 5 1 2
x
5 1 2 x21
dy
5 x22
dx
1
5 2
x
5. f r(x) 5 8x3
8x3 5 1
1
x3 5
8
1
x5
2
1
1 4
f a b 5 2a b
2
2
1
5
8
Equation of the tangent line:
3
1
1
y 2 5 1ax 2 b, or y 5 x 2
8
2
8
6. a. f r(x) 5 8x 2 7
b. f r(x) 5 26x2 1 8x 1 5
c. f(x) 5 5x22 2 3x23
f r(x) 5 210x23 1 9x24
10
9
52 3 1 4
x
x
1
1
d. f(x) 5 x2 1 x3
1 1
1 2
f r(x) 5 x22 1 x23
2
3
1
1
5 12 1 23
2x
3x
1
e. f(x) 5 7x22 2 3x2
3 1
f r(x) 5 214x23 2 x22
2
14
3
5 2 3 2 12
x
2x
f. f r(x) 5 4x22 1 5
4
5 215
x
d.
2-21
7. a. yr 5 26x 1 6
When x 5 1,
yr 5 26(1) 1 6
5 0.
When x 5 1,
y 5 23A12 B 1 6(1) 1 4
5 7.
Equation of the tangent line:
y 2 7 5 0(x 2 1), or
y57
1
b. y 5 3 2 2x2
1
yr 5 2x22
21
5
!x
When x 5 9,
21
yr 5
!9
1
52 .
3
When x 5 9,
y 5 3 2 2 !9
5 23.
Equation of the tangent line:
1
1
y 2 (23) 5 2 (x 2 9), or y 5 2 x
3
3
c. f r(x) 5 28x3 1 12x2 2 4x 2 8
f r(3) 5 28(3)3 1 12(3)2 2 4(3) 2 8
5 2216 1 108 2 12 2 8
5 2218
f(3) 5 22(3)4 1 4(3)3 2 2(3)2 2 8(3) 1 9
5 2162 1 108 2 18 2 24 1 9
5 287
Equation of the tangent line:
y 2 (287) 5 2128(x 2 3), or
y 5 2128x 1 297
d
8. a. f r(x) 5 A4x2 2 9xB A3x2 1 5B
dx
d
1 A4x2 2 9xB A3x2 1 5B
dx
5 (8x 2 9)A3x2 1 5B 1 A4x2 2 9xB (6x)
5 24x3 2 27x2 1 40x 2 45
1 24x3 2 54x2
5 48x3 2 81x2 1 40x 2 45
d
b. f r(t) 5 A23t2 2 7t 1 8B (4t 2 1)
dt
d
1 A23t2 2 7t 1 8B (4t 2 1)
dt
5 (26t 2 7)(4t 2 1)
1 A23t2 2 7t 1 8B (4)
2-22
5 224t2 2 28t 1 6t 1 7 2 12t2 2 28t 1 32
5 236t2 2 50t 1 39
dy
d
c.
5
A3x2 1 4x 2 6B (2x2 2 9)
dx
dx
d
1 A3x2 1 4x 2 6B A2x2 2 9B
dx
5 (6x 1 4)A2x2 2 9B 1 A3x2 1 4x 2 6B (4x)
5 12x3 2 54x 1 8x2 2 36 1 12x3
1 16x2 2 24x
5 24x3 1 24x2 2 78x 2 36
dy
d
d.
5
A3 2 2x3 B 2 A3 2 2x3 B
dx
dx
d
1 A3 2 2x3 B 2 A3 2 2x3 B
dx
d
5 c A3 2 2x3 B A3 2 2x3 B
dx
d
1 A3 2 2x3 B A3 2 2x3 B d A3 2 2x3 B
dx
1 A3 2 2x3 B 2 A26x2 B
5 S 2A26x2 B A3 2 2x3 B T A3 2 2x3 B
1 A3 2 2x3 B 2 A26x2 B
5 3A3 2 2x3 B 2 A26x2 B
5 A3 2 2x3 B 2 A218x2 B
5 A9 2 12x3 1 4x6 BA218x2 B
5 2162x2 1 216x5 2 72x8
d
9. yr 5 A5x2 1 9x 2 2B A2x2 1 2x 1 3B
dx
d
1 A5x2 1 9x 2 2B A2x2 1 2x 1 3B
dx
5 (10x 1 9)A2x2 1 2x 1 3B
1 A5x2 1 9x 2 2B (2 2 2x)
yr(1) 5 (10(1) 1 9)(2 (1)2 1 2(1) 1 3)
1 (5(1)2 1 9(1) 2 2)(2 2 2(1))
5 (19)(4)
5 76
Equation of the tangent line:
y 2 48 5 76(x 2 1), or 76x 2 y 2 28 5 0
dy
d
10.
5 2 (x 2 1)(5 2 x)
dx
dx
d
1 2(x 2 1) (5 2 x)
dx
5 2(5 2 x) 2 2(x 2 1)
5 12 2 4x
dy
The tangent line is horizontal when dx 5 0.
12 2 4x 5 0
12 5 4x
x53
Chapter 2: Derivatives
When x 5 3,
y 5 2((3) 2 1)(5 2 (3))
5 8.
Point where tangent line is horizontal: (3, 8)
dy
(5(x 1 h)2 2 8(x 1 h) 1 4)
11.
5 lim c
dx hS0
h
A5x2 2 8x 1 4B
2
d
h
5(x 1 h)2 2 5x2 2 8h
5 lim
hS0
h
5((x 1 h) 2 x)((x 1 h) 1 x) 2 8h
5 lim
hS0
h
5h(2x 1 h) 2 8h
5 lim
hS0
h
5 lim (5(2x 1 h) 2 8)
t1h
5 lim
3
5 lim
30 2
b
90
1 2
5 500a b
3
500
5
L
9
b. V(0) 5 500(1 2 0)2 5 500 L
500
V(60) 5
L
9
The average rate of change of volume with respect
to time from 0 min to 60 min is
500
DV
2 500
5 9
Dt
60 2 0
28
(500)
5 9
60
200
52
L>min
27
c. Calculate Vr(t):
V(t 1 h) 2 V(t)
Vr(t) 5 lim
hS0
h
5 lim
hS0
500a1 2 90 b 2 500a21 1 90 b
t
h
Calculus and Vectors Solutions Manual
h
500a2 b a2 2
h
90
2t 1 h
b
90
h
500
2t 1 h
5 lim 2
a2 2
b
hS0
90
90
250
2t
5
a2 2 b
9
90
2900 1 10t
5
81
hS0
2900 1 10(30)
81
200
52
L>min
27
4
13. V(r) 5 pr3
3
4
4
V(15) 5 p(15)3
a. V(10) 5 p(10)3
3
3
4
4
5 p(3375)
5 p(1000)
3
3
4000
p
5
5 4500p
3
Then, the average rate of change of volume with
respect to radius is
DV
4500p 2 4000
3 p
5
Dr
15 2 10
500p Q 9 2 83 R
5
5
19
5 100pa b
3
1900
p cm3>cm
5
3
b. First calculate Vr(r):
V(r 1 h) 2 V(r)
Vr(r) 5 lim
hS0
h
4
p3(r
1
h)3 2 r34
5 lim 3
hS0
h
4
3
p
r
1
3r2h 1 3rh2 1 h3 2 r3 R
Q
3
5 lim
hS0
h
4
2
2
3
3 p Q 3r h 1 3rh 1 h R
5 lim
hS0
h
Vr(30) 5
5 500a
2
t1h
t
a1 2 90 1 1 2 90 b
Then,
t 2
12. V(t) 5 500a1 2 b . 0 # t # 90
90
a. After 1 h, t 5 60, and the volume is
2
V(60) 5 500 Q 1 2 60
90 R
t1h
t
h
hS0
hS0
5 10x 2 8
500a1 2 90 2 1 1 90 b
2
2-23
4
5 lim pA3r2 1 3rh 1 h2 B
hS0 3
4
5 pA3r2 1 3r(0) 1 (0)2 B
3
5 4pr2
Then, Vr(8) 5 4p(8)2
5 4p(64)
5 256p cm3>cm
14. This statement is always true. A cubic polynomial
function will have the form f(x) 5 ax3 1 bx2 1
cx 1 d, a 2 0. So the derivative of this cubic is
f r(x) 5 3ax2 1 2bx 1 c, and since 3a 2 0, this
derivative is a quadratic polynomial function. For
example, if f(x) 5 x3 1 x2 1 1,
we get
f r(x) 5 3x2 1 2x,
and if
f(x) 5 2x3 1 3x2 1 6x 1 2,
we get
f r(x) 5 6x2 1 6x 1 6
x2a13b
15. y 5 a2b , a, bPI
x
Simplifying,
y 5 x2a13b2 (a2b) 5 xa14b
Then,
yr 5 (a 1 4b)a14b21
16. a. f(x) 5 26x3 1 4x 2 5x2 1 10
f r(x) 5 218x2 1 4 2 10x
Then, f r(x) 5 218(3)2 1 4 2 10(3)
5 2188
b. f r(3) is the slope of the tangent line to f(x) at
x 5 3 and the rate of change in the value of f(x)
with respect to x at x 5 3.
17. a. P(t) 5 100 1 120t 1 10t2 1 2t3
P(t) 5 100 1 120t 1 10t2 1 2t3
P(0) 5 100 1 120(0) 1 10(0)2 1 2(0)3
5 100 bacteria
b. At 5 h, the population is
P(5) 5 100 1 120(5) 1 10(5)2 1 2(5)3
5 1200 bacteria
c. Pr(t) 5 120 1 20t 1 6t2
At 5 h, the colony is growing at
Pr(5) 5 120 1 20(5) 1 6(5)2
5 370 bacteria> h
100
18. C(t) 5
,t.2
t
Simplifying, C(t) 5 100t21.
100
Then, Cr(t) 5 2100t22 5 2 2 .
t
2-24
Cr(5)
Cr(50)
Cr(100)
100
100
100
52
52
52
2
2
(5)
(50)
(100)2
1
100
100
52
52
52
25
2500
100
5 24
5 20.04
5 20.01
These are the rates of change of the percentage with
respect to time at 5, 50, and 100 min. The percentage
of carbon dioxide that is released per unit time from
the pop is decreasing. The pop is getting flat.
2.4 The Quotient Rule, pp. 97–98
1. For x, a, b real numbers,
xaxb 5 xa1b
For example,
x9x26 5 x3
Also,
Axa B b 5 xab
For example,
Ax2 B 3 5 x6
Also,
xa
5 xa2b, x 2 0
xb
For example,
x5
5 x2
x3
2.
Function
x2 1 3x
,
x
x20
f(x) 5
Rewrite
Differentiate
and Simplify,
If Necessary
f(x) 5 x 1 3
f r(x) 5 1
5
g(x) 5
3x3
,x20
x
1
,
10x5
x20
h(x) 5
y5
8x3 1 6x
,
2x
h(x) 5
1 25
x
10
hr(x) 5
21 26
x
2
y 5 4x2 1 3
dy
5 8x
dx
s5t13
ds
51
dt
x20
t2 2 9
s5
,t23
t23
1
gr(x) 5 2x23
2
g(x) 5 3x3
Chapter 2: Derivatives
4
5 lim pA3r2 1 3rh 1 h2 B
hS0 3
4
5 pA3r2 1 3r(0) 1 (0)2 B
3
5 4pr2
Then, Vr(8) 5 4p(8)2
5 4p(64)
5 256p cm3>cm
14. This statement is always true. A cubic polynomial
function will have the form f(x) 5 ax3 1 bx2 1
cx 1 d, a 2 0. So the derivative of this cubic is
f r(x) 5 3ax2 1 2bx 1 c, and since 3a 2 0, this
derivative is a quadratic polynomial function. For
example, if f(x) 5 x3 1 x2 1 1,
we get
f r(x) 5 3x2 1 2x,
and if
f(x) 5 2x3 1 3x2 1 6x 1 2,
we get
f r(x) 5 6x2 1 6x 1 6
x2a13b
15. y 5 a2b , a, bPI
x
Simplifying,
y 5 x2a13b2 (a2b) 5 xa14b
Then,
yr 5 (a 1 4b)a14b21
16. a. f(x) 5 26x3 1 4x 2 5x2 1 10
f r(x) 5 218x2 1 4 2 10x
Then, f r(x) 5 218(3)2 1 4 2 10(3)
5 2188
b. f r(3) is the slope of the tangent line to f(x) at
x 5 3 and the rate of change in the value of f(x)
with respect to x at x 5 3.
17. a. P(t) 5 100 1 120t 1 10t2 1 2t3
P(t) 5 100 1 120t 1 10t2 1 2t3
P(0) 5 100 1 120(0) 1 10(0)2 1 2(0)3
5 100 bacteria
b. At 5 h, the population is
P(5) 5 100 1 120(5) 1 10(5)2 1 2(5)3
5 1200 bacteria
c. Pr(t) 5 120 1 20t 1 6t2
At 5 h, the colony is growing at
Pr(5) 5 120 1 20(5) 1 6(5)2
5 370 bacteria> h
100
18. C(t) 5
,t.2
t
Simplifying, C(t) 5 100t21.
100
Then, Cr(t) 5 2100t22 5 2 2 .
t
2-24
Cr(5)
Cr(50)
Cr(100)
100
100
100
52
52
52
2
2
(5)
(50)
(100)2
1
100
100
52
52
52
25
2500
100
5 24
5 20.04
5 20.01
These are the rates of change of the percentage with
respect to time at 5, 50, and 100 min. The percentage
of carbon dioxide that is released per unit time from
the pop is decreasing. The pop is getting flat.
2.4 The Quotient Rule, pp. 97–98
1. For x, a, b real numbers,
xaxb 5 xa1b
For example,
x9x26 5 x3
Also,
Axa B b 5 xab
For example,
Ax2 B 3 5 x6
Also,
xa
5 xa2b, x 2 0
xb
For example,
x5
5 x2
x3
2.
Function
x2 1 3x
,
x
x20
f(x) 5
Rewrite
Differentiate
and Simplify,
If Necessary
f(x) 5 x 1 3
f r(x) 5 1
5
g(x) 5
3x3
,x20
x
1
,
10x5
x20
h(x) 5
y5
8x3 1 6x
,
2x
h(x) 5
1 25
x
10
hr(x) 5
21 26
x
2
y 5 4x2 1 3
dy
5 8x
dx
s5t13
ds
51
dt
x20
t2 2 9
s5
,t23
t23
1
gr(x) 5 2x23
2
g(x) 5 3x3
Chapter 2: Derivatives
3. In the previous problem, all of these rational
examples could be differentiated via the power rule
after a minor algebraic simplification.
A second approach would be to rewrite a rational
example
f(x)
h(x) 5
g(x)
using the exponent rules as
h(x) 5 f(x)(g(x))21,
and then apply the product rule for differentiation
(together with the power of a function rule to find
hr(x).
A third (and perhaps easiest) approach would be to
just apply the quotient rule to find hr(x).
(x 1 1)(1) 2 x(1)
4. a. hr(x) 5
Ax 1 1B 2
1
5
Ax 1 1B 2
(t 1 5)(2) 2 (2t 2 3)(1)
b. hr(t) 5
At 1 5B 2
13
5
At 1 5B 2
A2x2 2 1B A3x2 B 2 x3 A4xB
c. hr(x) 5
A2x2 2 1B 2
4
2x 2 3x2
5
A2x2 2 1B 2
Ax2 1 3B (0) 2 1(2x)
d. hr(x) 5
Ax2 1 3B 2
22x
5 2
Ax 1 3B 2
x(3x 1 5)
3x2 1 5x
e. y 5
5
2
A1 2 x B
1 2 x2
dy
(6x 1 5)A1 2 x2 B 2 A3x2 1 5xB (22x)
5
dx
A1 2 x2 B 2
6x 1 5 2 6x3 2 5x2 1 6x3 1 10x2
5
A1 2 x2 B 2
5x2 1 6x 1 5
5
A1 2 x2 B 2
dy
Ax2 1 3B A2x 2 1B 2 Ax2 2 x 1 1B (2x)
f.
5
dx
Ax2 1 3B 2
2x3 1 6x 2 x2 2 3 2 2x3 1 2x2 2 2x
5
Ax2 1 3B 2
x2 1 4x 2 3
5
Ax2 1 3B 2
Calculus and Vectors Solutions Manual
3x 1 2
, x 5 23
x15
dy
(x 1 5)(3) 2 (3x 1 2)(1)
5
dx
(x 1 5)2
At x 5 23:
dy
(2)(3) 2 (27)(1)
5
dx
A2B 2
13
5
4
x3
b. y 5 2
,x51
x 19
dy
Ax2 1 9B A3x2 B 2 Ax3 B (2x)
5
dx
Ax2 1 9B 2
At x 5 1:
dy
(10)(3) 2 (1)(2)
5
dx
A10B 2
28
5
100
7
5
25
x2 2 25
c. y 5 2
,x52
x 1 25
dy
2xAx2 1 25B 2 Ax2 2 25B (2x)
5
dx
Ax2 1 25B 2
At x 5 2:
dy
4(29) 2 (221)(4)
5
dx
A29B 2
116 1 84
5
292
200
5
841
(x 1 1)(x 1 2)
,x54
d. y 5
(x 2 1)(x 2 2)
x2 1 3x 1 2
5 2
x 2 3x 1 2
dy
(2x 1 3)Ax2 2 3x 1 2B
5
dx
Ax 2 1B 2 Ax 2 2B 2
Ax2 1 3x 1 2B (2x 2 3)
2
Ax 2 1B 2 Ax 2 2B 2
At x 5 4:
dy
(11)(6) 2 (30)(5)
5
dx
(9)(4)
84
52
36
7
52
3
5. a. y 5
2-25
x3
x2 2 6
dy
3x2 Ax2 2 6B 2 x3 A2xB
5
dx
Ax2 2 6B 2
At (3, 9):
3(9)(3) 2 (27)(6)
dy
5
dx
A3B 2
5 9 2 18
5 29
The slope of the tangent to the curve at (3, 9) is 29.
3x
7. y 5
x24
dy
3(x 2 4) 2 3x
12
5
52
2
dx
Ax 2 4B
Ax 2 4B 2
12
Slope of the tangent is 2 25.
6. y 5
Therefore,
12
2 5
(x 2 4)
dy
12
25
x 2 4 5 5 or x 2 4 5 25
x 5 9 or x 5 21
Points are Q 9, 275 R and Q 21, 35 R .
5x 1 2
8. f(x) 5
x12
(x 1 2)(5) 2 (5x 1 2)(1)
f r(x) 5
Ax 1 2B 2
8
f r(x) 5
Ax 1 2B 2
Since Ax 1 2B 2 is positive or zero for all xPR,
8
.0
(x 1 2)2
for x 2 22. Therefore, tangents to
5x 1 2
the graph of f(x) 5 x 1 2 do not have a negative
slope.
2x2
9. a. y 5
,x24
x24
dy
(x 2 4)(4x) 2 A2x2 B (1)
5
dx
Ax 2 4B 2
2
4x 2 16x 2 2x2
5
(x 2 4)2
2
2x 2 16x
5
(x 2 4)2
2x(x 2 8)
5
(x 2 4)2
dy
Curve has horizontal tangents when dx 5 0, or
when x 5 0 or 8. At x 5 0:
0
y5
24
50
At x 5 8:
2-26
2(8)2
4
5 32
So the curve has horizontal tangents at the points
(0, 0) and (8, 32).
x2 2 1
b. y 5 2
x 1x22
(x 2 1)(x 1 1)
5
(x 1 2)(x 2 1)
x11
5
,x21
x12
dy
(x 1 2) 2 (x 1 1)
5
dx
Ax 1 2B 2
1
5
Ax 1 2B 2
y5
Curve has horizontal tangents when dx 5 0.
No value of x will produce a slope of 0, so there
are no horizontal tangents.
4t
10. p(t) 5 1000a1 1 2
b
t 1 50
4At 2 1 50B 2 4t(2t)
pr(t) 5 1000a
b
At 2 1 50B 2
1000A200 2 4t 2 B
5
At 2 1 50B 2
1000(196)
5 75.36
pr(1) 5
A51B 2
1000(184)
5 63.10
pr(2) 5
A54B 2
Population is growing at a rate of 75.4 bacteria per
hour at t 5 1 and at 63.1 bacteria per hour at t 5 2.
x2 2 1
11. y 5
3x
1
1
5 x 2 x21
3
3
dy
1
1 22
5 1 x
dx
3
3
1
1
5 1 2
3
3x
At x 5 2:
A2B 2 2 1
y5
3(2)
1
5
2
and
dy
1
1
5 1
dx
3
3A2B 2
Chapter 2: Derivatives
1
1
1
3
12
5
5
12
So the equation of the tangent to the curve at x 5 2 is:
1
5
y 2 5 (x 2 2), or 5x 2 12y 2 4 5 0.
2
12
10(6 2 t)
, 0 # t # 6, t 5 0,
12. a. s(t) 5
t13
s(0) 5 20
The boat is initially 20 m from the dock.
(t 1 3)(21) 2 (6 2 t)(1)
b. v(t) 5 sr(t) 5 10 c
d
At 1 3B 2
290
v(t) 5
At 1 3B 2
At t 5 0, v(0) 5 210, the boat is moving towards
the dock at a speed of 10 m> s. When s(t) 5 0, the
boat will be at the dock.
10(6 2 t)
5 0, t 5 6.
t13
290
10
v(6) 5 2 5 2
9
9
The speed of the boat when it bumps into the dock
is 109 m> s.
13. a. i. t 5 0
1 1 2(0)
r(0) 5
110
5 1 cm
1 1 2t
ii.
5 1.5
11t
1 1 2t 5 1.5(1 1 t)
1 1 2t 5 1.5 1 1.5t
0.5t 5 0.5
t 5 1s
(1 1 t)(2) 2 (1 1 2t)(1)
iii. rr(t) 5
A1 1 tB 2
2 1 2t 2 1 2 2t
5
A1 1 tB 2
1
5
A1 1 tB 2
1
rr(1.5) 5
A1 1 1B 2
1
5
4
5 0.25 cm> s
b. No, the radius will never reach 2 cm, because
y 5 2 is a horizontal asymptote of the graph of the
function. Therefore, the radius approaches but never
equals 2 cm.
5
Calculus and Vectors Solutions Manual
ax 1 b
(x 2 1)(x 2 4)
(x 2 1)(x 2 4)(a)
f r(x) 5
Ax 2 1B 2 Ax 2 4B 2
d
(ax 1 b) 3(x 2 1)(x 2 4)4
dx
2
Ax 2 1B 2 Ax 2 4B 2
(x 2 1)(x 2 4)(a)
5
Ax 2 1B 2 Ax 2 4B 2
(ax 1 b)3(x 2 1) 1 (x 2 4)4
2
Ax 2 1B 2 Ax 2 4B 2
2
(x 2 5x 1 4)(a) 2 (ax 1 b)(2x 2 5)
5
(x 2 1)2 (x 2 4)2
2
2ax 2 2bx 1 4a 1 5b
5
(x 2 1)2 (x 2 4)2
Since the point (2, 21) is on the graph (as it’s on
the tangent line) we know that
21 5 f(2)
2a 1 b
5
(1)(22)
2 5 2a 1 b
b 5 2 2 2a
Also, since the tangent line is horizontal at (2, 21),
we know that
0 5 f r(2)
2aA2B 2 2 2b(2) 1 4a 1 5b
5
A1B 2 A22B 2
b50
0 5 2 2 2a
a51
So we get
x
f(x) 5
(x 2 1)(x 2 4)
Since the tangent line is horizontal at the point
(2, 21), the equation of this tangent line is
y 2 (21) 5 0(x 2 2), or y 5 21
Here are the graphs of both f(x) and this horizontal
tangent line:
x
f (x) =
(x – 1) (x –4)
y
8
6
4
2
x
0
–2 –1
1 2 3 4 5
–2
y =–1
–4
–6
14. f(x) 5
2-27
A2t 2 1 7B (5) 2 (5t)(4t)
A2t 2 1 7B 2
10t 2 1 35 2 20t 2
5
A2t 2 1 7B 2
210t 2 1 35
5
A2t 2 1 7B 2
Set cr(t) 5 0 and solve for t.
210t 2 1 35
50
(2t 2 1 7)2
210t 2 1 35 5 0
10t 2 5 35
t 2 5 3.5
15. cr(t) 5
t 5 6"3.5
t 8 61.87
To two decimal places, t 5 21.87 or t 5 1.87,
because sr(t) 5 0 for these values. Reject the
negative root in this case because time is positive
(t $ 0). Therefore, the concentration reaches its
maximum value at t 5 1.87 hours.
16. When the object changes direction, its velocity
changes sign.
At 2 1 8B (1) 2 t(2t)
sr(t) 5
At 2 1 8B 2
2
t 1 8 2 2t 2
5
(t 2 1 8)2
2t 2 1 8
5 2
(t 1 8)2
solve for t when sr(t) 5 0.
2t 2 1 8
50
At 2 1 8B 2
2t 2 1 8 5 0
t2 5 8
t 5 6"8
t 8 62.83
To two decimal places, t 5 2.83 or t 5 22.83,
because sr(t) 5 0 for these values. Reject the
negative root because time is positive (t $ 0).
The object changes direction when t 5 2.83 s.
d
ax 1 b
17. f(x) 5
,x22
c
cx 1 d
(cx 1 d)(a) 2 (ax 1 b)(c)
f r(x) 5
Acx 1 dB 2
ad 2 bc
f r(x) 5
Acx 1 dB 2
For the tangents to the graph of y 5 f(x) to have
positive slopes, f r(x) . 0. (cx 1 d)2 is positive for
all xPR. ad 2 bc . 0 will ensure each tangent has
a positive slope.
2-28
2.5 The Derivatives of Composite
Functions, pp. 105–106
1. f(x) 5 !x, g(x) 5 x2 2 1
a. f(g(1)) 5 f(1 2 1)
5 f(0)
50
b. g(f(1)) 5 g(1)
50
c. g(f(0)) 5 g(0)
5021
5 21
d. f(g(24)) 5 f(16 2 1)
5 f(15)
5 !15
e. f(g(x)) 5 f Ax2 2 1B
5 "x2 2 1
f. g(f(x)) 5 gA !xB
5 A !xB 2 2 1
5x21
2. a. f(x) 5 x2, g(x) 5 !x
(f + g)(x) 5 f(g(x))
5 f A !xB
5 A !xB 2
5x
Domain 5 5x $ 06
(g + f)(x) 5 g(f(x))
5 gAx2 B
5 "x2
5 Zx Z
Domain 5 5xPR6
The composite functions are not equal for negative
x-values (as (f + g) is not defined for these x), but
are equal for non-negative x-values.
1
b.
f(x) 5 , g(x) 5 x2 1 1
x
(f + g)(x) 5 f(g(x))
5 f Ax2 1 1B
1
5 2
x 11
Domain 5 5xPR6
(g + f)(x) 5 g(f(x))
1
5 ga b
x
1 2
5a b 11
x
Chapter 2: Derivatives
A2t 2 1 7B (5) 2 (5t)(4t)
A2t 2 1 7B 2
10t 2 1 35 2 20t 2
5
A2t 2 1 7B 2
210t 2 1 35
5
A2t 2 1 7B 2
Set cr(t) 5 0 and solve for t.
210t 2 1 35
50
(2t 2 1 7)2
210t 2 1 35 5 0
10t 2 5 35
t 2 5 3.5
15. cr(t) 5
t 5 6"3.5
t 8 61.87
To two decimal places, t 5 21.87 or t 5 1.87,
because sr(t) 5 0 for these values. Reject the
negative root in this case because time is positive
(t $ 0). Therefore, the concentration reaches its
maximum value at t 5 1.87 hours.
16. When the object changes direction, its velocity
changes sign.
At 2 1 8B (1) 2 t(2t)
sr(t) 5
At 2 1 8B 2
2
t 1 8 2 2t 2
5
(t 2 1 8)2
2t 2 1 8
5 2
(t 1 8)2
solve for t when sr(t) 5 0.
2t 2 1 8
50
At 2 1 8B 2
2t 2 1 8 5 0
t2 5 8
t 5 6"8
t 8 62.83
To two decimal places, t 5 2.83 or t 5 22.83,
because sr(t) 5 0 for these values. Reject the
negative root because time is positive (t $ 0).
The object changes direction when t 5 2.83 s.
d
ax 1 b
17. f(x) 5
,x22
c
cx 1 d
(cx 1 d)(a) 2 (ax 1 b)(c)
f r(x) 5
Acx 1 dB 2
ad 2 bc
f r(x) 5
Acx 1 dB 2
For the tangents to the graph of y 5 f(x) to have
positive slopes, f r(x) . 0. (cx 1 d)2 is positive for
all xPR. ad 2 bc . 0 will ensure each tangent has
a positive slope.
2-28
2.5 The Derivatives of Composite
Functions, pp. 105–106
1. f(x) 5 !x, g(x) 5 x2 2 1
a. f(g(1)) 5 f(1 2 1)
5 f(0)
50
b. g(f(1)) 5 g(1)
50
c. g(f(0)) 5 g(0)
5021
5 21
d. f(g(24)) 5 f(16 2 1)
5 f(15)
5 !15
e. f(g(x)) 5 f Ax2 2 1B
5 "x2 2 1
f. g(f(x)) 5 gA !xB
5 A !xB 2 2 1
5x21
2. a. f(x) 5 x2, g(x) 5 !x
(f + g)(x) 5 f(g(x))
5 f A !xB
5 A !xB 2
5x
Domain 5 5x $ 06
(g + f)(x) 5 g(f(x))
5 gAx2 B
5 "x2
5 Zx Z
Domain 5 5xPR6
The composite functions are not equal for negative
x-values (as (f + g) is not defined for these x), but
are equal for non-negative x-values.
1
b.
f(x) 5 , g(x) 5 x2 1 1
x
(f + g)(x) 5 f(g(x))
5 f Ax2 1 1B
1
5 2
x 11
Domain 5 5xPR6
(g + f)(x) 5 g(f(x))
1
5 ga b
x
1 2
5a b 11
x
Chapter 2: Derivatives
1
11
x2
Domain 5 5x 2 06
The composite functions are not equal here. For
instance, (f + g)(1) 5 12 and (g + f )(1) 5 2.
1
c.
f(x) 5 , g(x) 5 !x 1 2
x
( f + g)(x) 5 f(g(x))
5
5 f( !x 1 2)
1
5
!x 1 2
Domain 5 5x . 226
(g + f)(x) 5 g(f(x))
1
5 ga b
x
1
12
Åx
The domain is all x such that
5
1
12$0
x
and x 2 0, or equivalently
Domain 5 5x # 2 12 or x . 06
The composite functions are not equal here. For
instance, ( f + g)(2) 5 12 and (g + f )(2) 5 # 52.
3. If f(x) and g(x) are two differentiable functions
of x, and
h(x) 5 (f + g)(x)
5 f(g(x))
is the composition of these two functions, then
hr(x) 5 f r(g(x)) ? gr(x)
This is known as the “chain rule” for differentiation of
composite functions. For example, if f(x) 5 x10 and
g(x) 5 x2 1 3x 1 5, then h(x) 5 Ax2 1 3x 1 5B 10,
and so
hr(x) 5 f r(g(x)) ? gr(x)
5 10Ax2 1 3x 1 5B 9 (2x 1 3)
2
As another example, if f(x) 5 x3 and
2
g(x) 5 x2 1 1, then h(x) 5 Ax2 1 1B 3,
and so
2
1
hr(x) 5 Ax2 1 1B 23 (2x)
3
4. a. f(x) 5 (2x 1 3)4
f r(x) 5 4A2x 1 3B 3 (2)
5 8A2x 1 3B 3
b. g(x) 5 Ax2 2 4B 3
gr(x) 5 3Ax2 2 4B 2 (2x)
5 6xAx2 2 4B 2
Calculus and Vectors Solutions Manual
c. h(x) 5 A2x2 1 3x 2 5B 4
hr(x) 5 4A2x2 1 3x 2 5B 3 (4x 1 3)
d. f(x) 5 Ap2 2 x2 B 3
f r(x) 5 3Ap2 2 x2 B 2 (22x)
5 26xAp2 2 x2 B 2
e. y 5 "x2 2 3
1
5 Ax2 2 3B 2
1
1
yr 5 Ax2 2 3B 2 (2x)
2
x
5
2
"x 2 3
1
f. f(x) 5 2
Ax 2 16B 5
5 Ax2 2 16B 25
f r(x) 5 25Ax2 2 16B 26 (2x)
210x
5 2
Ax 2 16B 6
2
5. a. y 5 2 3
x
5 22x23
dy
5 (22)(23)x24
dx
6
5 4
x
1
b. y 5
x11
5 (x 1 1)21
dy
5 (21)(x 1 1)22 (1)
dx
21
5
Ax 1 1B 2
1
c. y 5 2
x 24
5 Ax2 2 4B 21
dy
5 (21)Ax2 2 4B 22 (2x)
dx
22x
5 2
Ax 2 4B 2
3
5 3A9 2 x2 B 21
d. y 5
9 2 x2
dy
6x
5
dx
A9 2 x2 B 2
2-29
1
5x 1 x
5 A5x2 1 xB 21
e. y 5
2
dy
5 (21)A5x2 1 xB 22 (10x 1 1)
dx
10x 1 1
52 2
A5x 1 xB 2
1
f. y 5 2
Ax 1 x 1 1B 4
5 Ax2 1 x 1 1B 24
dy
5 (24)Ax2 1 x 1 1B 25 (2x 1 1)
dx
8x 1 4
52 2
Ax 1 x 1 1B 5
h5g+f
6.
5 g(f(x))
h(21) 5 g(f(21))
5 g(1)
5 24
h(x) 5 g(f(x))
hr(x) 5 gr(f(x))f r(x)
hr(21) 5 gr(f(21))f r(21)
5 gr(1)(25)
5 (27)(25)
5 35
1
7. f(x) 5 (x 2 3)2, g(x) 5 , h(x) 5 f(g(x)),
x
1
f r(x) 5 2(x 2 3), gr(x) 5 2 2
x
hr(x) 5 f r(g(x))gr(x)
1
1
5 f ra b a2 2 b
x
x
1
1
5 2a 2 3b a2 2 b
x
x
2 1
5 2 2 a 2 3b
x x
8. a. f(x) 5 Ax 1 4B 3 Ax 2 3B 6
d
3Ax 1 4B 34 ? Ax 2 3B 6
f r(x) 5
dx
d
1 Ax 1 4B 3 3Ax 2 3B 64
dx
5 3Ax 1 4B 2 Ax 2 3B 6
1 Ax 1 4B 3 (6)Ax 2 3B 5
5 Ax 1 4B 2 Ax 2 3B 5
3 33(x 2 3) 1 6(x 1 4)4
5 Ax 1 4B 2 Ax 2 3B 5 (9x 1 15)
2-30
b. y 5 Ax2 1 3B 3 Ax3 1 3B 2
dy
d 2
5
3Ax 1 3B 34 ? Ax3 1 3B 2
dx
dx
d 3
1 Ax2 1 3B 3 ?
3Ax 1 3B 24
dx
5 3Ax2 1 3B 2 (2x)Ax3 1 3B 2
1 Ax2 1 3B 3 (2)Ax3 1 3BA3x2 B
5 6xAx2 1 3B 2 Ax3 1 3B 3Ax3 1 3B 1 xAx2 1 3B4
5 6xAx2 1 3B 2 Ax3 1 3B A2x3 1 3x 1 3B
3x2 1 2x
c. y 5 2
x 11
dy
A6x 1 2B Ax2 1 1B 2 A3x2 1 2xBA2xB
5
dx
Ax2 1 1B 2
3
2
6x 1 2x 1 6x 1 2 2 6x3 2 4x2
5
Ax2 1 1B 2
2
22x 1 6x 1 2
5
Ax2 1 1B 2
d. h(x) 5 x3 A3x 2 5B 2
d 3
d
hr(x) 5
3x 4 ? A3x 2 5B 2 1 x3 3A3x 2 5B 24
dx
dx
5 3x2 A3x 2 5B 2 1 x3 (2)(3x 2 5)(3)
5 3x2 (3x 2 5)3(3x 2 5) 1 2x4
5 3x2 (3x 2 5)(5x 2 5)
5 15x2 (3x 2 5)(x 2 1)
4
e. y 5 x A1 2 4x2 B 3
dy
d 4
d
5
3x 4A1 2 4x2 B 3 1 x4 ?
3A1 2 4x2 B 34
dx
dx
dx
5 4x3 A1 2 4x2 B 3 1 x4 (3)A1 2 4x2 B 2 A28xB
5 4x3 A1 2 4x2 B 2 3A1 2 4x2 B 2 6x24
5 4x3 A1 2 4x2 B 2 A1 2 10x2 B
x2 2 3 4
b
f. y 5 a 2
x 13
dy
x2 2 3 3 d x2 2 3
5 4a 2
b
c
d
dx
x 1 2 dx x2 1 3
x2 2 3 3 Ax2 1 3B (2x) 2 Ax2 2 3B (2x)
5 4a 2
b ?
x 13
Ax2 1 3B 2
x2 2 3 3
12x
5 4a 2
b ? 2
x 13
Ax 1 3B 2
48xAx2 2 3B 3
5
Ax2 1 3B 5
1
2
9. a. s(t) 5 t3 (4t 2 5)3
1
1
5 t3 3(4t 2 5)24 3
1
5 3t(4t 2 5)24 3
1
5 3t A16t2 2 40t 1 25B4 3
1
5 A16t3 2 40t2 1 25tB 3, t 5 8
Chapter 2: Derivatives
1
2
sr(t) 5 A16t3 2 40t2 1 25tB 23
3
3 A48t2 2 80t 1 25B
A48t2 2 80t 1 25B
5
2
3A16t3 2 40t2 1 25tB 3
Rate of change at t 5 8:
(48(8)2 2 80(8) 1 25)
sr(8) 5
2
3(16(8)3 2 40(8)2 1 25(8))3
2457
5
972
91
5
36
1
t2p 3
b. s(t) 5 a
b , t 5 2p
t 2 6p
2
1 t 2 p 23 d t 2 p
sr(t) 5 a
b ? c
d
3 t 2 6p
dt t 2 6p
2
1 t 2 6p 3 (t 2 6p) 2 (t 2 p)
5 a
b ?
3 t2p
(t 2 6p)2
2
1 t 2 6p 3
25p
5 a
b ?
3 t2p
(t 2 6p)2
Rate of change at t 5 2p:
1
25p
2
sr(2p) 5 (24)3 ?
3
16p2
3
5"2
52
24p
y 5 2x6
10. y 5 A1 1 x3 B 2
dy
dy
5 2A1 1 x3 B (3x2 )
5 12x5
dx
dx
For the same slope,
6x2 A1 1 x3 B 5 12x5
6x2 1 6x5 5 12x5
6x2 2 6x5 5 0
6x2 Ax3 2 1B 5 0
x 5 0 or x 5 1.
Curves have the same slope at x 5 0 and x 5 1.
11. y 5 A3x 2 x2 B 22
dy
5 22A3x 2 x2 B 23 (3 2 2x)
dx
At x 5 2,
dy
5 2236 2 44 23 (3 2 4)
dx
5 2(2)23
1
5
4
The slope of the tangent line at x 5 2 is 41.
Calculus and Vectors Solutions Manual
y 5 Ax3 2 7B 5 at x 5 2
dy
5 5Ax3 2 7B 4 A3x2 B
dx
When x 5 2,
dy
5 5(1)4 (12)
dx
5 60
Slope of the tangent is 60.
Equation of the tangent at (2, 1) is
y 2 1 5 60(x 2 2)
60x 2 y 2 119 5 0.
13. a. y 5 3u2 2 5u 1 2
u 5 x2 2 1, x 5 2
u53
du
dy
5 6u 2 5,
5 2x
du
dx
dy
dy
du
5
3
dx
du
dx
5 (6u 2 5)(2x)
5 (18 2 5)(4)
5 13(4)
5 52
1
b. y 5 2u3 1 3u2, u 5 x 1 x2, x 5 1
dy
dy du
5
?
dx
du dx
1
5 (6u2 1 6u)a1 1
b
2!x
At x 5 1:
1
u 5 1 1 12
52
dy
1
b
5 (6(2)2 1 6(2))a1 1
dx
2 !1
3
5 36 3
2
5 54
c. y 5 uAu2 1 3B 3, u 5 (x 1 3)2, x 5 22
du
dy
5 Au2 1 3B 3 1 6u2 Au2 1 3B 2,
5 2(x 1 3)
du
dx
dy
dy du
5
5 373 1 6(4)2432(1)4
dx
du dx
5 439 3 2
5 878
d. y 5 u3 2 5Au3 2 7uB 2,
12.
u 5 "x
1
5 x2, x 5 4
dy
dy du
5
?
dx
du dx
2-31
1 1
5 333u2 2 10Au3 2 7uB A3u2 2 7B4 ? a x2 b
2
1
5 33u2 2 10Au3 2 7uB A3u2 2 7B4 ?
2"x
At x 5 4:
u 5 "4
52
dy
1
5 33(2)2 2 10( (2)3 2 7(2))(3(2)2 2 7)4
dx
2(2)
5 78
14. h(x) 5 f(g(x)), therefore
hr(x) 5 f r(g(x)) 3 gr(x)
f(u) 5 u2 2 1, g(2) 5 3, gr(2) 5 21
Now, hr(2) 5 f r(g(2)) 3 gr(2)
5 f r(3) 3 gr(2).
Since f(u) 5 u2 2 1, f r(u) 5 2u, and f r(3) 5 6,
hr(2) 5 6(21)
5 26.
t 2
15. V(t) 5 50 000a1 2 b
30
1
t
Vr(t) 5 50 000 c2a1 2 b a2 b d
30
30
10
1
Vr(10) 5 50 000 c2a1 2 b a2 b d
30
30
2
1
5 50 000 c2a b a2 b d
3
30
8 2222
At t 5 10 minutes, the water is flowing out of the
tank at a rate of 2222 L> min.
16. The velocity function is the derivative of the
position function.
1
s(t) 5 At 3 1 t 2 B 2
1
1
v(t) 5 sr(t) 5 At 3 1 t 2 B 22 A3t 2 1 2tB
2
3t 2 1 2t
5
2"t 3 1 t 2
3(3)2 1 2(3)
v(3) 5
2"33 1 32
27 1 6
5
2"36
33
5
12
5 2.75
The particle is moving at 2.75 m/s.
17. a. h(x) 5 p(x)q(x)r(x)
hr(x) 5 pr(x)q(x)r(x) 1 p(x)qr(x)r(x)
1 p(x)q(x)rr(x)
2-32
b. h(x) 5 x(2x 1 7)4 (x 2 1)2
Using the result from part a.,
hr(x) 5 (1)(2x 1 7)4 (x 2 1)2
1 x34(2x 1 7)3 (2)4 (x 2 1)2
1 x(2x 1 7)4 32(x 2 1)4
hr(23) 5 1(16) 1 (23)34(1)(2)4 (16)
1 (23)(1)32(24)4
5 16 2 384 1 24
5 2344
18. y 5 Ax2 1 x 2 2B 3 1 3
dy
5 3Ax2 1 x 2 2B 2 (2x 1 1)
dx
At the point (1, 3), x 5 1 and the slope of the
tangent will be 3(1 1 1 2 2)2 (2 1 1) 5 0.
Equation of the tangent at (1, 3) is y 2 3 5 0.
Solving this equation with the function, we have
Ax2 1 x 2 2B 3 1 3 5 3
(x 1 2)3 (x 2 1)3 5 0
x 5 22 or x 5 1
Since 22 and 1 are both triple roots, the line with
equation y 2 3 5 0 will be a tangent at both x 5 1
and x 5 22. Therefore, y 2 3 5 0 is also a tangent
at (22, 3).
x2 (1 2 x)3
19. y 5
(1 1 x)3
12x 3
5 x2 c a
bd
11x
dy
12x 3
12x 2
5 2xa
b 1 3x2 a
b
dx
11x
11x
2 (1 1 x) 2 (1 2 x)(1)
3 c
d
(1 1 x)2
12x 3
12x 2
22
5 2xa
b 1 3x2 a
b c
d
11x
11x
(1 1 x)2
12x 2 12x
3x
5 2xa
b c
2
d
11x
11x
(1 1 x)2
1 2 x 2 1 2 x2 2 3x
5 2xa
b c
d
11x
(1 1 x)2
2xAx2 1 3x 2 1B (1 2 x)2
52
(1 1 x)4
Review Exercise, pp. 110–113
1. To find the derivative f r(x), the limit
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
must be computed, provided it exists. If this limit
does not exist, then the derivative of f (x) does not
Chapter 2: Derivatives
1 1
5 333u2 2 10Au3 2 7uB A3u2 2 7B4 ? a x2 b
2
1
5 33u2 2 10Au3 2 7uB A3u2 2 7B4 ?
2"x
At x 5 4:
u 5 "4
52
dy
1
5 33(2)2 2 10( (2)3 2 7(2))(3(2)2 2 7)4
dx
2(2)
5 78
14. h(x) 5 f(g(x)), therefore
hr(x) 5 f r(g(x)) 3 gr(x)
f(u) 5 u2 2 1, g(2) 5 3, gr(2) 5 21
Now, hr(2) 5 f r(g(2)) 3 gr(2)
5 f r(3) 3 gr(2).
Since f(u) 5 u2 2 1, f r(u) 5 2u, and f r(3) 5 6,
hr(2) 5 6(21)
5 26.
t 2
15. V(t) 5 50 000a1 2 b
30
1
t
Vr(t) 5 50 000 c2a1 2 b a2 b d
30
30
10
1
Vr(10) 5 50 000 c2a1 2 b a2 b d
30
30
2
1
5 50 000 c2a b a2 b d
3
30
8 2222
At t 5 10 minutes, the water is flowing out of the
tank at a rate of 2222 L> min.
16. The velocity function is the derivative of the
position function.
1
s(t) 5 At 3 1 t 2 B 2
1
1
v(t) 5 sr(t) 5 At 3 1 t 2 B 22 A3t 2 1 2tB
2
3t 2 1 2t
5
2"t 3 1 t 2
3(3)2 1 2(3)
v(3) 5
2"33 1 32
27 1 6
5
2"36
33
5
12
5 2.75
The particle is moving at 2.75 m/s.
17. a. h(x) 5 p(x)q(x)r(x)
hr(x) 5 pr(x)q(x)r(x) 1 p(x)qr(x)r(x)
1 p(x)q(x)rr(x)
2-32
b. h(x) 5 x(2x 1 7)4 (x 2 1)2
Using the result from part a.,
hr(x) 5 (1)(2x 1 7)4 (x 2 1)2
1 x34(2x 1 7)3 (2)4 (x 2 1)2
1 x(2x 1 7)4 32(x 2 1)4
hr(23) 5 1(16) 1 (23)34(1)(2)4 (16)
1 (23)(1)32(24)4
5 16 2 384 1 24
5 2344
18. y 5 Ax2 1 x 2 2B 3 1 3
dy
5 3Ax2 1 x 2 2B 2 (2x 1 1)
dx
At the point (1, 3), x 5 1 and the slope of the
tangent will be 3(1 1 1 2 2)2 (2 1 1) 5 0.
Equation of the tangent at (1, 3) is y 2 3 5 0.
Solving this equation with the function, we have
Ax2 1 x 2 2B 3 1 3 5 3
(x 1 2)3 (x 2 1)3 5 0
x 5 22 or x 5 1
Since 22 and 1 are both triple roots, the line with
equation y 2 3 5 0 will be a tangent at both x 5 1
and x 5 22. Therefore, y 2 3 5 0 is also a tangent
at (22, 3).
x2 (1 2 x)3
19. y 5
(1 1 x)3
12x 3
5 x2 c a
bd
11x
dy
12x 3
12x 2
5 2xa
b 1 3x2 a
b
dx
11x
11x
2 (1 1 x) 2 (1 2 x)(1)
3 c
d
(1 1 x)2
12x 3
12x 2
22
5 2xa
b 1 3x2 a
b c
d
11x
11x
(1 1 x)2
12x 2 12x
3x
5 2xa
b c
2
d
11x
11x
(1 1 x)2
1 2 x 2 1 2 x2 2 3x
5 2xa
b c
d
11x
(1 1 x)2
2xAx2 1 3x 2 1B (1 2 x)2
52
(1 1 x)4
Review Exercise, pp. 110–113
1. To find the derivative f r(x), the limit
f(x 1 h) 2 f(x)
f r(x) 5 lim
hS0
h
must be computed, provided it exists. If this limit
does not exist, then the derivative of f (x) does not
Chapter 2: Derivatives
exist at this particular value of x. As an alternative
to this limit, we could also find f r(x) from the
definition by computing the equivalent limit
f(z) 2 f(x)
.
z2x
zSx
f r(x) 5 lim
These two limits are seen to be equivalent by
substituting z 5 x 1 h.
2. a. y 5 2x2 2 5x
dy
(2(x 1 h)2 2 5(x 1 h)) 2 A2x2 2 5xB
5 lim
dx hS0
h
2A(x 1 h)2 2 x2 B 2 5h
5 lim
hS0
h
2((x 1 h) 2 x)((x 1 h) 1 x) 2 5h
5 lim
hS0
h
2h(2x 1 h) 2 5h
5 lim
hS0
h
5 lim (2(2x 1 h) 2 5)
hS0
5 4x 2 5
b. y 5 !x 2 6
!(x 1 h) 2 6 2 !x 2 6
dy
5 lim
dx hS0
h
!(x 1 h) 2 6 2 !x 2 6
5 lim c
hS0
h
!(x 1 h) 2 6 1 !x 2 6
d
3
!(x 1 h) 2 6 1 !x 2 6
( (x 1 h) 2 6) 2 (x 2 6)
5 lim
hS0 h( !(x 1 h) 2 6 1 !x 2 6)
1
5 lim
hS0 !(x 1 h) 2 6 1 !x 2 6
1
5
2 !x 2 6
x
c. y 5
42x
x1h
x
2
dy
4 2 (x 1 h)
42x
5 lim
dx hS0
h
(x 1 h)(4 2 x) 2 x(4 2 (x 1 h))
(4 2 (x 1 h))(4 2 x)
5 lim
hS0
h
4h
5 lim
hS0 h(4 2 (x 1 h))(4 2 x)
4
5 lim
hS0 (4 2 (x 1 h))(4 2 x)
4
5
(4 2 x)2
Calculus and Vectors Solutions Manual
3. a. y 5 x2 2 5x 1 4
dy
5 2x 2 5
dx
3
b. f(x) 5 x4
3 1
f r(x) 5 x24
4
3
5 14
4x
7
c. y 5 4
3x
7
5 x24
3
dy
228 25
5
x
dx
3
28
52 5
3x
1
d. y 5 2
x 15
5 Ax2 1 5B 21
dy
5 (21)Ax2 1 5B 22 ? (2x)
dx
2x
52 2
Ax 1 5B 2
3
e. y 5
A3 2 x2 B 2
5 3A3 2 x2 B 22
dy
5 (26)A3 2 x2 B 23 ? (22x)
dx
12x
5
A3 2 x2 B 3
f. y 5 "7x2 1 4x 1 1
1
5 A7x2 1 4x 1 1B 2
dy
1
1
5 A7x2 1 4x 1 1B 22 A14x 1 4B
dx
2
7x 1 2
5
"7x2 1 4x 1 1
2x3 2 1
4. a. f(x) 5
x2
1
5 2x 2 2
x
5 2x 2 x22
f r(x) 5 2 1 2x23
2
521 3
x
2-33
b. g(x) 5 !xAx3 2 xB
1
5 x2 Ax3 2 xB
7
3
5 x2 2 x2
7 5
3 1
gr(x) 5 x2 2 x2
2
2
!x 2
5
A7x 2 3B
2
x
c. y 5
3x 2 5
dy
(3x 2 5)(1) 2 (x)(3)
5
dx
(3x 2 5)2
5
52
(3x 2 5)2
1
d. y 5 (x 2 1)2 (x 1 1)
1
1
1
yr 5 (x 2 1)2 1 (x 1 1)a b (x 2 1)22
2
x11
5 !x 2 1 1
2!x 2 1
2x 2 2 1 x 1 1
5
2!x 2 1
3x 2 1
5
2!x 2 1
2
e. f(x) 5 A !x 1 2B 23
1
2
5 ( x2 1 2) 23
1
22 12
5 1
f r(x) 5
x 1 2) 23 # x2 2
(
3
2
1
52
5
3 !xA !x 1 2B 3
x2 1 5x 1 4
f. y 5
x14
(x 1 4)(x 1 1)
5
x14
5 x 1 1, x 2 24
dy
51
dx
5. a. y 5 x4 (2x 2 5)6
yr 5 x4 36(2x 2 5)5 (2)4 1 4x3 (2x 2 5)6
5 4x3 (2x 2 5)5 33x 1 (2x 2 5)4
5 4x3 (2x 2 5)5 (5x 2 5)
5 20x3 (2x 2 5)5 (x 2 1)
b. y 5 x"x2 1 1
1
1
yr 5 x c Ax2 1 1B 22 (2x)d 1 (1)"x2 1 1
2
x2
5
1 "x2 1 1
"x2 1 1
2-34
(2x 2 5)4
(x 1 1)3
(x 1 1)34(2x 2 5)3 (2)
yr 5
(x 1 1)6
3(2x 2 5)4 (x 1 1)2
2
(x 1 1)6
(x 1 1)2 (2x 2 5)3 38x 1 8 2 6x 1 154
5
(x 1 1)6
3
(2x 2 5) (2x 1 23)
yr 5
(x 1 1)4
10x 2 1 6
b 5 (10x 2 1)6 (3x 1 5)26
d. y 5 a
3x 1 5
yr 5 (10x 2 1)6 326(3x 1 5)27 (3)4
1 6(10x 2 1)5 (10)(3x 1 5)26
5 (10x 2 1)5 (3x 1 5)27 3x 2 18(10x 2 1)4
1 60(3x 1 5)
5 (10x 2 1)5 (3x 1 5)27
3 (2180x 1 18 1 180x 1 300)
318(10x 2 1)5
5
(3x 1 5)7
e. y 5 (x 2 2)3 Ax2 1 9B 4
yr 5 (x 2 2)3 C4Ax2 1 9B 3 (2x)D
1 3(x 2 2)2 (1)Ax2 1 9B 4
5 (x 2 2)2 Ax2 1 9B 3 C8x(x 2 2) 1 3Ax2 1 9B D
5 (x 2 2)2 Ax2 1 9B 3 A11x2 2 16x 1 27B
f. y 5 A1 2 x2 B 3 (6 1 2x)23
1 2 x2 3
5a
b
6 1 2x
1 2 x2 2
b
yr 5 3a
6 1 2x
c. y 5
3 c
(6 1 2x)(22x) 2 A1 2 x2 B (2)
d
(6 1 2x)2
3A1 2 x2 B 2 A212x 2 4x2 2 2 1 2x2 B
(6 1 2x)4
2 2
3A1 2 x B A2x2 1 12x 1 2B
52
(6 1 2x)4
3A1 2 x2 B 2 Ax2 1 6x 1 1B
52
8(3 2 x)4
2
6. a. g(x) 5 f(x )
gr(x) 5 f(x2 ) 3 2x
b. h(x) 5 2xf(x)
hr(x) 5 2xf r(x) 1 2f(x)
18
7. a. y 5 5u2 1 3u 2 1, u 5 2
x 15
x52
u52
5
Chapter 2: Derivatives
dy
5 10u 1 3
du
du
36x
52 2
dx
Ax 1 5B 2
When x 5 2,
du
72
8
52 52
dx
81
9
When u 5 2,
dy
5 20 1 3
du
5 23
dy
8
5 23a2 b
dx
9
184
52
9
!x 1 x
u14
,u5
,
b. y 5
u24
10
x54
3
u5
5
dy
(u 2 4) 2 (u 1 4)
5
du
(u 2 4)2
du
1 1 1
5 a x22 1 1b
dx
10 2
When x 5 4,
du
8
1 5
52
5 a b
(u 2 4)2 dx
10 4
1
5
8
3
When u 5 ,
5
dy
8
52
du
3
20 2
a 2 b
5
5
8(25)
52
(217)2
When x 5 4,
dy
8(25)
1
5
3
dx
172
8
25
5
289
c. y 5 f("x2 1 9), f r(5) 5 22, x 5 4
dy
1
1
5 f r("x2 1 9) 3 Ax2 1 9B 22 (2x)
dx
2
dy
1 1
5 f r(5) ? ? ? 8
dx
2 5
Calculus and Vectors Solutions Manual
5 22 ?
4
5
8
5
2
8. f(x) 5 A9 2 x2 B 3
2
1
f r(x) 5 A9 2 x2 B 23 (22x)
3
24x
5
1
3(9 2 x2 )3
2
f r(1) 5 2
3
The slope of the tangent line at (1, 4) is 2 23.
9. y 5 2x3 1 6x2
yr 5 23x2 1 12x
23x2 1 12x 5 215
23x2 1 12x 5 212
2
x2 2 4x 2 5 5 0
x 2 4x 2 4 5 0
52
4 6 !16 1 16
2
4 6 4!2
5
2
x 5 2 6 2 !2
x5
(x 2 5)(x 1 1) 5 0
x 5 5, x 5 21
10. a. i. y 5 Ax2 2 4B 5
yr 5 5Ax2 2 4B 4 (2x)
Horizontal tangent,
10xAx2 2 4B 4 5 0
x 5 0, x 5 62
ii. y 5 Ax3 2 xB 2
yr 5 2Ax3 2 xB A3x2 2 1B
Horizontal tangent,
2x(x2 2 1)(3x2 2 1) 5 0
!3
x 5 0, x 5 61, x 5 6
.
3
2-35
b. i.
ii.
11. a. y 5 Ax2 1 5x 1 2B 4 at (0, 16)
yr 5 4Ax2 1 5x 1 2B 3 (2x 1 5)
At x 5 0,
yr 5 4(2)3 (5)
5 160
Equation of the tangent at (0, 16) is
y 2 16 5 160(x 2 0)
y 5 160x 1 16
or 160x 2 y 1 16 5 0
b. y 5 A3x22 2 2x3 B 5 at (1, 1)
yr 5 5A3x22 2 2x3 B 4 A26x23 2 6x2 B
At x 5 1,
yr 5 5(1)4 (26 2 6)
5 260
Equation of the tangent at (1, 1) is
y 2 1 5 260(x 2 1)
60x 1 y 2 61 5 0.
12. y 5 3x2 2 7x 1 5
dy
5 6x 2 7
dx
2-36
Slope of x 1 5y 2 10 5 0 is 2 15.
Since perpendicular, 6x 2 7 5 5
x52
y 5 3(4) 2 14 1 5
5 3.
Equation of the tangent at (2, 3) is
y 2 3 5 5(x 2 2)
5x 2 y 2 7 5 0.
13. y 5 8x 1 b is tangent to y 5 2x2
dy
5 4x
dx
Slope of the tangent is 8, therefore 4x 5 8, x 5 2.
Point of tangency is (2, 8).
Therefore, 8 5 16 1 b, b 5 28.
Or 8x 1 b 5 2x2
2x2 2 8x 2 b 5 0
8 6 !64 1 8b
x5
.
2(2)
For tangents, the roots are equal, therefore
64 1 8b 5 0, b 5 28.
Point of tangency is (2, 8), b 5 28.
14. a.
b.
The equation of the tangent is y 5 0.
The equation of the tangent is y 5 6.36.
The equation of the tangent is y 5 26.36.
Chapter 2: Derivatives
c.
Ax2 2 6B A3x2 B 2 x3 (2x)
Ax2 2 6B 2
x4 2 18x2
5 2
Ax 2 6B 2
f r(x) 5
x4 2 18x2
50
Ax2 2 6B 2
x2 Ax2 2 18B 5 0
x2 5 0 or x2 2 18 5 0
x50
x 5 63 !2
The coordinates of the points where the slope is 0
9 !2
are (0, 0), Q 3 !2, 9 !2
2 R , and Q 23!2, 2 2 R .
d. Substitute into the expression for f r(x) from
part b.
16 2 72
f r(2) 5
(22)2
256
5
4
5 214
5
2
15. a. f(x) 5 2x3 2 5x3
5 2
2 1
f r(x) 5 2 3 x3 2 5 3 x3
3
3
10 23
10
5 x 2 13
3
3x
2
f(x) 5 0 6 x3 32x 2 54 5 0
5
x 5 0 or x 5
2
y 5 f(x) crosses the x-axis at x 5 52, and
10 x 2 1
f r(x) 5 a 13 b
3
x
5
10
3
1
f ra b 5
3 3 5 13
2
3
2
Q2 R
553
3
!
2
2
1
5 53 3 23
3
!5
1
5 (25 3 2)3
3
5!
50
b. To find a, let f(x) 5 0.
10 23
10
x 2 13 5 0
3
3x
Calculus and Vectors Solutions Manual
30x 5 30
x51
Therefore a 5 1.
16. M 5 0.1t2 2 0.001t3
a. When t 5 10,
M 5 0.1(100) 2 0.001(1000)
59
When t 5 15,
M 5 0.1(225) 2 0.001(3375)
5 19.125
One cannot memorize partial words, so 19 words
are memorized after 15 minutes.
b. Mr 5 0.2t 2 0.003t2
When t 5 10,
Mr 5 0.2(10) 2 0.003(100)
5 1.7
The number of words memorized is increasing by
1.7 words> min.
When t 5 15,
Mr 5 0.2(15) 2 0.003(225)
5 2.325
The number of words memorized is increasing by
2.325 words> min.
30
17. a. N(t) 5 20 2
"9 1 t2
30t
3
A9 1 t2 B 2
b. No, according to this model, the cashier never
stops improving. Since t . 0, the derivative is always
positive, meaning that the rate of change in the
cashier’s productivity is always increasing. However,
these increases must be small, since, according to the
model, the cashier’s productivity can never exceed 20.
1
18. C(x) 5 x3 1 40x 1 700
3
a. Cr(x) 5 x2 1 40
b. Cr(x) 5 76
x2 1 40 5 76
x2 5 36
x56
Production level is 6 gloves> week.
x2
2
19. R(x) 5 750x 2 2 x3
6
3
a. Marginal Revenue
x
Rr(x) 5 750 2 2 2x2
3
Nr(t) 5
2-37
10
2 2(100)
3
5 $546.67
20
20. D(p) 5
,p.1
!p 2 1
1
3
Dr(p) 5 20a2 b (p 2 1)22
2
10
52
3
(p 2 1)2
10
10
Dr(5) 5
52
3
8
"4
5
52
4
Slope of demand curve at (5, 10) is 2 54.
21. B(x) 5 20.2x2 1 500, 0 # x # 40
a. B(0) 5 20.2(0)2 1 500 5 500
B(30) 5 20.2(30)2 1 500 5 320
b. Br(x) 5 20.4x
Br(0) 5 20.4(0) 5 0
Br(30) 5 20.4(30) 5 212
c. B(0) 5 blood sugar level with no insulin
B(30) 5 blood sugar level with 30 mg of insulin
Br(0) 5 rate of change in blood sugar level
with no insulin
Br(30) 5 rate of change in blood sugar level
with 30 mg of insulin
d. Br(50) 5 20.4(50) 5 220
B(50) 5 20.2(50)2 1 500 5 0
Br(50) 5 220 means that the patient’s blood sugar
level is decreasing at 20 units per mg of insulin 1 h
after 50 mg of insulin is injected.
B(50) 5 0 means that the patient’s blood sugar level
is zero 1 h after 50 mg of insulin is injected. These
values are not logical because a person’s blood sugar
level can never reach zero and continue to decrease.
3x
22. a. f(x) 5
1 2 x2
3x
5
(1 2 x)(1 1 x)
f(x) is not differentiable at x 5 1 because it is not
defined there (vertical asymptote at x 5 1).
x21
b. g(x) 5 2
x 1 5x 2 6
x21
5
(x 1 6)(x 2 1)
1
5
for x 2 1
(x 1 6)
g(x) is not differentiable at x 5 1 because it is not
defined there (hole at x 5 1).
b. Rr(10) 5 750 2
2-38
3
c. h(x) 5 "
(x 2 2)2
The graph has a cusp at (2, 0) but it is differentiable
at x 5 1.
d. m(x) 5 Z3x 2 3 Z 2 1.
The graph has a corner at x 5 1, so m(x) is not
differentiable at x 5 1.
3
23. a. f(x) 5 2
4x 2 x
3
5
x(4x 2 1)
f(x) is not defined at x 5 0 and x 5 0.25. The
graph has vertical asymptotes at x 5 0 and
x 5 0.25. Therefore, f(x) is not differentiable at
x 5 0 and x 5 0.25.
x2 2 x 2 6
b. f(x) 5
x2 2 9
(x 2 3)(x 1 2)
5
(x 2 3)(x 1 3)
(x 1 2)
5
for x 2 3
(x 1 3)
f(x) is not defined at x 5 3 and x 5 23. At
x 5 23, the graph as a vertical symptote and at
x 5 3 it has a hole. Therefore, f(x) is not
differentiable at x 5 3 and x 5 23.
c. f(x) 5 "x2 2 7x 1 6
5 !(x 2 6)(x 2 1)
f(x) is not defined for 1 , x , 6. Therefore,
f(x) is not differentiable for 1 , x , 6.
(t 1 1)(25) 2 (25t)(t)
24. pr(t) 5
(t 1 1)2
25t 1 25 2 25t
5
(t 1 1)2
25
5
(t 1 1)2
25. Answers may vary. For example,
f(x) 5 2x 1 3
1
y5
2x 1 3
(2x 1 3)(0) 2 (1)(2)
yr 5
(2x 1 3)2
2
52
(2x 1 3)2
f(x) 5 5x 1 10
1
y5
5x 1 10
(5x 1 10)(0) 2 (1)(5)
yr 5
(5x 1 10)2
Chapter 2: Derivatives
52
5
(5x 1 10)2
1
Rule: If f(x) 5 ax 1 b and y 5 f (x), then
yr 5
2a
(ax 1 b)2
1
1
1
yr 5 lim c
2
d
hS0 h a(x 1 h) 1 b
ax 1 b
1 ax 1 b 2 3a(x 1 h)b4
d
5 lim c
hS0 h 3a(x 1 h) 1 b4 (ax 1 h)
1 ax 1 b 2 ax 2 ah 2 b
5 lim c
d
hS0 h 3a(x 1 h) 1 b4 (ax 1 b)
1
2ah
5 lim c
d
hS0 h 3a(x 1 h) 1 b4 (ax 1 b)
2a
5 lim
hS0 3a(x 1 h) 1 b4 (ax 1 b)
2a
5
(ax 1 b)2
26. a. Let y 5 f(x)
(2x 2 3)2 1 5
y5
2x 2 3
Let u 5 2x 2 3.
u2 1 5
Then y 5
.
u
y 5 u 1 5u21
dy
b. f r(x) 5
dx
dy
dy
du
5
3
dx
du
dx
5 (1 2 5u22 )(2)
5 2(1 2 5(2x 2 3)22 )
27. g(x) 5 !2x 2 3 1 5(2x 2 3)
a. Let y 5 g(x).
y 5 !2x 2 3 1 5(2x 2 3)
Let u 5 2x 2 3.
Then y 5 !u 1 5u.
dy
dy
du
b. gr(x) 5
5
3
dx
du
dx
1 212
5 a u 1 5b (2)
2
212
5 u 1 10
1
5 (2x 2 3)22 1 10
Calculus and Vectors Solutions Manual
28. a. f(x) 5 (2x 2 5)3 (3x2 1 4)5
f r(x) 5 (2x 2 5)3 (5)A3x2 1 4B 4 (6x)
1 A3x2 1 4B 5 (3)(2x 2 5)2 (2)
5 30x(2x 2 5)3 A3x2 1 4B 4
1 6(3x2 1 4)5 (2x 2 5)2
5 6(2x 2 5)2 A3x2 1 4B 4
3 C5x(2x 2 5) 1 A3x2 1 4B D
5 6(2x 2 5)2 A3x2 1 4B 4
3 A10x2 2 25x 1 3x2 1 4B
5 6(2x 2 5)2 (3x2 1 4)4
3 (13x2 2 25x 1 4)
3
b. g(x) 5 (8x )(4x2 1 2x 2 3)5
gr(x) 5 (8x3 )(5)(4x2 1 2x 2 3)4 (8x 1 2)
1 (4x2 1 2x 2 3)5 (24x2 )
5 40x3 (4x2 1 2x 2 3)4 (8x 1 2)
1 24x2 (4x2 1 2x 2 3)5
5 8x2 (4x2 1 2x 2 3)4 35x(8x 1 2)
1 3(4x2 1 2x 2 3)4
5 8x2 (4x2 1 2x 2 3)4
(40x2 1 10x 1 12x2 1 6x 2 9)
5 8x2 (4x2 1 2x 2 3)4 (52x2 1 16x 2 9)
c. y 5 (5 1 x)2 (4 2 7x3 )6
yr 5 (5 1 x)2 (6)(4 2 7x3 )5 (221x2 )
1 (4 2 7x3 )6 (2)(5 1 x)
5 2126x2 (5 1 x)2 (4 2 7x3 )5
1 2(5 1 x)(4 2 7x3 )6
5 2(5 1 x)(4 2 7x3 )5 3263x2 (5 1 x)
1 4 2 7x34
5 2(5 1 x)(4 2 7x3 )5 (4 2 315x2 2 70x3 )
6x 2 1
d. h(x) 5
(3x 1 5)4
(3x 1 5)4 (6) 2 (6x 2 1)(4)(3x 1 5)3 (3)
hr(x) 5
((3x 1 5)4 )2
3
6(3x 1 5) 3(3x 1 5) 2 2(6x 2 1)4
5
(3x 1 5)8
6(29x 1 7)
5
(3x 1 5)5
(2x2 2 5)3
e. y 5
(x 1 8)2
dy
(x 1 8)2 (3)(2x2 2 5)2 (4x)
5
dx
A(x 1 8)2 B 2
(2x2 2 5)3 (2)(x 1 8)
2
A(x 1 8)2 B 2
2(x 1 8)(2x2 2 5)2 36x(x 1 8) 2 (2x2 2 5)4
(x 1 8)4
2(2x2 2 5)2 (4x2 1 48x 1 5)
5
(x 1 8)3
5
2-39
f. f(x) 5
23x4
"4x 2 8
23x4
5
1
(4x 2 8)2
1
(4x 2 8)2 (212x3 )
f r(x) 5
1
A(4x 2 8)2 B 2
1
1
(23x4 )a b (4x 2 8) 2 2 (4)
2
2
1
A(4x 2 8)2 B 2
26x3 (4x 2 8)22 32(4x 2 8) 2 x4
5
4x 2 8
26x3 (7x 2 16)
5
3
(4x 2 8)2
23x3 (7x 2 16)
5
3
(4x 2 8)2
2x 1 5 4
g. g(x) 5 a
b
6 2 x2
1
2x 1 5 3
b
6 2 x2
(6 2 x2 )(2) 2 (2x 1 5)(22x)
b
3a
(6 2 x2 )2
2x 1 5 3 2(6 1 x2 1 5x)
5 4a
b a
b
6 2 x2
(6 2 x2 )2
2x 1 5 3 (x 1 2)(x 1 3)
5 8a
b a
b
6 2 x2
(6 2 x2 )2
3
1
h. y 5 c
2 3d
(4x 1 x )
In (1),
4a 2 8a 5 16
24a 5 16
a 5 24
Using (1),
b 5 28(24) 5 32
a 5 24, b 5 32, c 5 0, f(x) 5 24x2 1 32x
30. a. A(t) 5 2t3 1 5t 1 750
Ar(t) 5 23t2 1 5
b. Ar(5) 5 23(25) 1 5
5 270
At 5 h, the number of ants living in the colony is
decreasing by 7000 ants> h.
c. A(0) 5 750, so there were 750 (100) or
75 000 ants living in the colony before it was
treated with insecticide.
d. Determine t so that A(t) 5 0. 2t3 1 5t 1 750
cannot easily be factored, so find the zeros by using
a graphing calculator.
gr(x) 5 4a
5 (4x 1 x2 )29
dy
5 29(4x 1 x2 )210 (4 1 2x)
dx
29. f(x) 5 ax2 1 bx 1 c,
It is given that (0, 0) and (8, 0) are on the curve,
and f r(2) 5 16.
Calculate f r(x) 5 2ax 1 b.
Then,
16 5 2a(2) 1 b
(1)
4a 1 b 5 16
Since (0, 0) is on the curve,
0 5 a(0)2 1 b(0) 1 c
c50
Since (8, 0) is on the curve,
0 5 a(8)2 1 b(8) 1 c
0 5 64a 1 8b 1 0
(2)
8a 1 b 5 0
Solve (1) and (2):
From (2), b 5 28a
(1)
2-40
All of the ants have been killed after about 9.27 h.
Chapter 2 Test, p. 114
1. You need to use the chain rule when the derivative
for a given function cannot be found using the sum,
difference, product, or quotient rules or when writing
the function in a form that would allow the use of
these rules is tedious. The chain rule is used when
a given function is a composition of two or more
functions.
2. f is the blue graph (it's a cubic). f' is the red graph
(it is quadratic). The derivative of a polynomial
function has degree one less than the derivative of
the function. Since the red graph is a quadratic
(degree 2) and the blue graph is cubic (degree 3),
the blue graph is f and the red graph is f r.
f(x 1 h) 2 f(x)
3. f(x) 5 lim
hS0
h
x 1 h 2 (x 1 h)2 2 (x 2 x2 )
5 lim
hS0
h
x 1 h 2 (x2 1 2hx 1 h2 ) 2 x 1 x2
5 lim
hS0
h
h 2 2hx 2 h2
5 lim
hS0
h
Chapter 2: Derivatives
f. f(x) 5
23x4
"4x 2 8
23x4
5
1
(4x 2 8)2
1
(4x 2 8)2 (212x3 )
f r(x) 5
1
A(4x 2 8)2 B 2
1
1
(23x4 )a b (4x 2 8) 2 2 (4)
2
2
1
A(4x 2 8)2 B 2
26x3 (4x 2 8)22 32(4x 2 8) 2 x4
5
4x 2 8
26x3 (7x 2 16)
5
3
(4x 2 8)2
23x3 (7x 2 16)
5
3
(4x 2 8)2
2x 1 5 4
g. g(x) 5 a
b
6 2 x2
1
2x 1 5 3
b
6 2 x2
(6 2 x2 )(2) 2 (2x 1 5)(22x)
b
3a
(6 2 x2 )2
2x 1 5 3 2(6 1 x2 1 5x)
5 4a
b a
b
6 2 x2
(6 2 x2 )2
2x 1 5 3 (x 1 2)(x 1 3)
5 8a
b a
b
6 2 x2
(6 2 x2 )2
3
1
h. y 5 c
2 3d
(4x 1 x )
In (1),
4a 2 8a 5 16
24a 5 16
a 5 24
Using (1),
b 5 28(24) 5 32
a 5 24, b 5 32, c 5 0, f(x) 5 24x2 1 32x
30. a. A(t) 5 2t3 1 5t 1 750
Ar(t) 5 23t2 1 5
b. Ar(5) 5 23(25) 1 5
5 270
At 5 h, the number of ants living in the colony is
decreasing by 7000 ants> h.
c. A(0) 5 750, so there were 750 (100) or
75 000 ants living in the colony before it was
treated with insecticide.
d. Determine t so that A(t) 5 0. 2t3 1 5t 1 750
cannot easily be factored, so find the zeros by using
a graphing calculator.
gr(x) 5 4a
5 (4x 1 x2 )29
dy
5 29(4x 1 x2 )210 (4 1 2x)
dx
29. f(x) 5 ax2 1 bx 1 c,
It is given that (0, 0) and (8, 0) are on the curve,
and f r(2) 5 16.
Calculate f r(x) 5 2ax 1 b.
Then,
16 5 2a(2) 1 b
(1)
4a 1 b 5 16
Since (0, 0) is on the curve,
0 5 a(0)2 1 b(0) 1 c
c50
Since (8, 0) is on the curve,
0 5 a(8)2 1 b(8) 1 c
0 5 64a 1 8b 1 0
(2)
8a 1 b 5 0
Solve (1) and (2):
From (2), b 5 28a
(1)
2-40
All of the ants have been killed after about 9.27 h.
Chapter 2 Test, p. 114
1. You need to use the chain rule when the derivative
for a given function cannot be found using the sum,
difference, product, or quotient rules or when writing
the function in a form that would allow the use of
these rules is tedious. The chain rule is used when
a given function is a composition of two or more
functions.
2. f is the blue graph (it's a cubic). f' is the red graph
(it is quadratic). The derivative of a polynomial
function has degree one less than the derivative of
the function. Since the red graph is a quadratic
(degree 2) and the blue graph is cubic (degree 3),
the blue graph is f and the red graph is f r.
f(x 1 h) 2 f(x)
3. f(x) 5 lim
hS0
h
x 1 h 2 (x 1 h)2 2 (x 2 x2 )
5 lim
hS0
h
x 1 h 2 (x2 1 2hx 1 h2 ) 2 x 1 x2
5 lim
hS0
h
h 2 2hx 2 h2
5 lim
hS0
h
Chapter 2: Derivatives
h(1 2 2x 2 h)
hS0
h
5 lim (1 2 2x 2 h)
5 lim
hS0
5 1 2 2x
d
Therefore,
(x 2 x2 ) 5 1 2 2x.
dx
1
4. a. y 5 x3 2 3x25 1 4p
3
dy
5 x2 1 15x26
dx
b. y 5 6(2x 2 9)5
dy
5 30(2x 2 9)4 (2)
dx
5 60(2x 2 9)4
2
x
3
1
1 6"
x
c. y 5
"x
"3
1
1
1
5 2x22 1
x 1 6x3
"3
dy
1
3
2
5 2x22 1
1 2x23
dx
"3
x2 1 6 5
b
d. y 5 a
3x 1 4
dy
x2 1 6 4 2x(3x 1 4) 2 (x2 1 6)3
5 5a
b
dx
3x 1 4
(3x 1 4)2
5(x2 1 6)4 (3x2 1 8x 2 18)
5
(3x 1 4)6
3
6x2 2 7
e. y 5 x2 "
dy
1
1
2
5 2x(6x2 2 7)3 1 x2 (6x2 2 7)23 (12x)
dx
3
2
5 2x(6x2 2 7)23 ( (6x2 2 7) 1 2x2 )
2
5 2x(6x2 2 7)23 (8x2 2 7)
4x5 2 5x4 1 6x 2 2
f. y 5
x4
5 4x 2 5 1 6x23 2 2x24
dy
5 4 2 18x24 1 8x25
dx
4x5 2 18x 1 8
5
x5
5. y 5 (x2 1 3x 2 2)(7 2 3x)
dy
5 (2x 1 3)(7 2 3x) 1 (x2 1 3x 2 2)(23)
dx
At (1, 8),
dy
5 (5)(4) 1 (2)(23)
dx
5 14.
The slope of the tangent to
y 5 (x2 1 3x 2 2)(7 2 3x) at (1, 8) is 14.
Calculus and Vectors Solutions Manual
6. y 5 3u2 1 2u
dy
5 6u 1 2
du
u 5 "x2 1 5
du
1
1
5 (x2 1 5)22 2x
dy
2
dy
x
5 (6u 1 2)a
b
2
dx
"x 1 5
At x 5 22, u 5 3.
dy
2
5 (20)a2 b
dx
3
40
52
3
7. y 5 (3x22 2 2x3 )5
dy
5 5(3x22 2 2x3 )4 (26x23 2 6x2 )
dx
At (1, 1),
dy
5 5(1)4 (26 2 6)
dx
5 260.
Equation of tangent line at (1, 1) is y 2 1 5 60(x 2 1)
y 2 1 5 260x 1 60
60x 1 y 2 61 5 0.
1
8. P(t) 5 (t 4 1 3)3
1 3
1
Pr(t) 5 3(t 4 1 3)2 a t24 b
4
1
1
3
Pr(16) 5 3(16 4 1 3)2 a 3 1624 b
4
1
1
5 3(2 1 3)2 a 3 b
4
8
75
5
32
The amount of pollution is increasing at a rate of
75
32 ppm>year.
9. y 5 x4
dy
5 4x3
dx
1
2 5 4x3
16
Normal line has a slope of 16. Therefore,
dy
1
52 .
dx
16
1
x3 5 2
64
2-41
1
4
1
y5
256
Therefore, y 5 x4 has a normal line with a slope of
1
16 at Q 2 14, 256
R.
x52
10. y 5 x3 2 x2 2 x 1 1
dy
5 3x2 2 2x 2 1
dx
dy
For a horizontal tangent line, dx 5 0.
3x2 2 2x 2 1 5 0
(3x 1 1)(x 2 1) 5 0
1
or
x52
x51
3
1
1
1
y52 2 1 11
y51212111
27
9
3
50
21 2 3 1 9 1 27
5
27
32
5
27
The required points are Q 2 13, 32
27 ), (1, 0 R .
2-42
11. y 5 x2 1 ax 1 b
dy
5 2x 1 a
dx
y 5 x3
dy
5 3x2
dx
Since the parabola and cubic function are tangent at
(1, 1), then 2x 1 a 5 3x2.
At (1, 1) 2(1) 1 a 5 3(1)2
a 5 1.
Since (1, 1) is on the graph of
y 5 x2 1 x 1 b, 1 5 12 1 1 1 b
b 5 21.
The required values are 1 and 21 for a and b,
respectively.
Chapter 2: Derivatives