132
Chapter 2 Differentiation
= Iax3, x <_ 2
117. f(x) ~x2 +b, x > 2
fmust be continuous at x = 2 to be differentiable at x = 2.
limf(x) = limax3 = 8a
8a = 4 + b
.~-+2-~J
x->Z8a-4=b
lim f(x)= lim(xZ+b)=4+b
x-->2+
x-+2+
}
I3axxx <> 22
f’(x) = [2x, ,
Forfto be differentiable at x = 2, the left derivative must equal the right derivative.
3a(2)2 = 2(2)
12a = 4
1
b = 8a - 4 = _4
3
120. Let f(x) = cos x.
118. f(x)=ICOS
lax+b,
X, Xx<>
00
f(0) = b = cos(0) = 1 ~ b = 1
f’(x)
f’(x) = lim f(x + kx)- f(x)
Av-~O
= lim
f-sinx, x < 0
x>o
la,
~
cos x cos Ax - sin x sin Ax - cos x
Ax
=~-,0tim cos x(cOsAxAX-1)_
So, a = 0.
Answer: a = 0, b = 1
~-,01im sin
= 0 - sin x(l’l = -sinx
119. f(x) : Isin x is differentiable for all x mr, n an
integer.
f2(x) = sin xI is differentia(91e for all x , 0.
You can verify this by graphing fl and f2 and observing
the locations of the sharp turns.
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives
I. g(x) : (x2 +3)(x2 -4x)
3. h(t) = x/7(1-t2) = t’/2(1-t2)
g’(x) : (x= + 3)(2x - 4)+ (x2 - 4x)(gx)
= 2x3 - 4x2 + 6x - 12 + 2x3 - 8x2
= 4x3 - 12x2 + 6x - 12
= 2(2x3- 6x~ + 3x- 6)
2. f(x) : (6x+5)(x3 -2)
f’(x) : (6x + 5)(3x2) + (x3 - 2)(6)
= 18X3 +15X~ +6X3 --12
if(t) = tl/2(-21)+ (I- 12)~t-’/2
1 !tY2
= -2t3/2 + ~2t1/2 -- 2
1
= _5_5_1312+
~
2
2tg2
_ 1-512 : 1-St_____~~
211/2
2N/7
= 24x3 + 15x9- - 12
© 2010 Brooks/Cole, Cengage Learning
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 133
~x _ x1/2
9. h(x) - x3 +1 x3 +1
~’(x) =
(x3 + 1)½x-l/~-- x’/2(3x2)
(x3 ÷ 0~
x3 + 1 - 6x3
2Xi12(X3 + 1)2
1 - 5x3
2~x(x~ + 1)2
5, f(x) : x3 COSX
S’(x) = x3(-sin x)+ cos x(3x:)
= 3x2cosx-x3sinx
= x2(3 cos x - x sin x)
s
10. h(s) = ~_1
h’(~) =
(x/7-l)(l)- s(½s-112]
6. g(x) = ~x sin x
=
2
_ -,~- - 2
= -,/x cos x +1 ~ sin x
>,/ x
11. g(x) - sinx
2
x
x
7. f(x) - x2 +1
f’(x) = (x2 + 1)(1)- x(2x)
g’(x) = x2(c°s x) - sin x(2x) = x cos x - 2 sin x
1 -- X2
(; ÷ 0
12. r(t)t2 +4
8. g(t) - 5t-3
(5t- 3)(2t)- (t2 + 4)(5)
(5t- 3)2
2
lOt - 6t - 5t2 - 20 5t2 - 6t - 20
(st- 3)2
f’(t) =
cos t
t3
,3(-sin 1) - cos 1(312) , sin, + 3 cost
=
(~t- 3)2
13. f(x) = (x3 + 4x)(3x2 + 2x- 5)
(x3 "t- 4x)(6x + 2)+ (3x2 + 2x - 5)(3x2 + 4)
S’(x) =
6x4 + 24x2 + 2x3 + 8x + 9x4 + 6x3 - 15x2 + 12x2 + 8x - 20
15x4 + 8x3 + 21x2 + 16x - 20
s’(o) : -20
14. f(x) = (x2- 2x + 1)(x~- 1)
S’(x) = (x2 - 2x + 1)(3x2) + (x3 -1)(2x - 2)
3x2(x- 1)2 + 2(x -1)2(x2 + x + 1)
(x- 1)2(5x2 + 2x + 2)
f’(1) = 0
© 2010 Brooks/Cole, Cengage Learning
134 Chapter 2 Differentiation
x2 - 4
15. f(x) = x-3
X’(~) :
17.
f(x) = x cos x
f’(x) : (x)(-sin x) + (cos x)(1) = cos x - x sin x
(x - 3)(2x) -(x2 - 4)(1)
(~-
3)~
8"
2 4~2)
2x2 -6x-x~- +4
(~- 3)2
18.
r(x~ - sin x
x
x2 - 6x + 4
f’(x) = (x)(cos x)
-(sin x)(1)
x2
f’O) - ~- 6 + 4 _
x cos x - sin x
0- 3)~
16.
x2
f,(~_) = (z/6)(~r~/2)- (1/2)
z2/36
x-5
3~’-~z - 18
s’(~) -- (x- ~)0)- (~ + ~10)
(~- 5/2
~2
x-5-x-5
(x- ~)~
-10
(xf’(4) - -10 - -10
(4- 5)2
Function
2
19. y= x + 3x
7
20. y-
5x2 - 3
4
6
21. y- 7x2
22. y23. y-
10
3x3
4 x3/~
Rewrite
y
= X2 +--X
7
D!fferentiate
7
1’ 5x~ 3
y=
4
4
y=
6
7
23
77
y’ = --X + --
y, - 2x+3
7
10
y’ = --x
4
y, = 12
7
10
y = __x-3
3
y’ =
y = 4x1/2, x > 0
y’ = 2x-ll2
30,xm
3
5x
y, -
12
7x3
y’ -
10
x4
x
24. y-
5x2 - 8
11
5
y = __x2
11
11
y’: ~l(2X)
y
lOx
11
t ~--- __
© 2010 Brooks/Cole, Cengage Learning
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 135
4 - 3x - x2
25. f(x) = x2 -1
3X -- 1
29. f(x) : ~ : 3x’/2-
f’(x) : (x2 - 1)(-3 - 2x) -(4 - 3x - x2)(2x)
-3x2 + 3 - 2x3 + 2x
- 8x
f’(x) = 3x-1/2 + 1--x412
2 2
-- 3x+l
+ 6x2 + 2x3
2X3/2
Alternate solution:
3x - 1 3x - 1
3x2 - 6x + 3
f(x) : ~ = xl/2
3(;-- ÷ 0
=
3(x - 1)2
(x - 1)2(x + 1)2
S’(x) :
3
x*l
(x -t-
26. f(x) : x2-1
X’/2(B) - (Bx -1)(-~)(X-1/2)
X
~
x-~12(3x + 1)
1)2,
X
--
X3 + 5X +
x_l,/2
2X3/2
3
ft(x) = (x2 - 1)(3X2 ÷ 5)- (X3 + 5X + 3)(2X)
3x+l
30. f(x): 3~x(x/7 + 3): xll3(x’/2 + 3)
f’(x) : xlia(~x-li2)+(X112 -1- 3)(31"~-X-2/3)
3X4 + 5X2 - 3X2 - 5 - 2X4 - l Ox2 - 6X
= 5X-1/6 + X-2/3
6
x4 - 8x2 - 6x - 5
5 1
= ~Xl/6 + X27
Alternate solution:
s(x) :
27. f(x) x(1 x-~) 4x
x+3
f’(x) = 1-(x + 3)4- 4x(1)
(x+3)2
(x2 + 6x+9)-12
: X5/6 +
3xl/3
f’(x) : 5--x-116 + x-213
6
5 1
= ~x~/6 + x2/----y
(X + 3)2
X2 + 6x - 3
(X + 3)2
+
31. h(s) : (s3 -2)2 : s6-4s3 +4
h’(s) : 6ss -12s2 : 6s2(s3 - 2)
32.
h(x) : (x2 -1)2 = x4 - 2x2 q- 1
if(x) = 4x3 - 4x = 4x(x2
- 1)
© 2010 Brooks/Cole, Cengage Learning
136
Chapter 2 Differentiation
33. f(x) - 2-(l/x) _ 2x-1_ 2x-1
x(x- 3) x~-3x
x- 3
2x2 - 6x - 4x2 + 8x - 3
f’(x) = (x~ - 3x)2 -(2x - 1)(2x - 3) =
(x~-- 3x)2
-2x~ +2x-3
2x~ - 2x+3
(.~-3./
(x2 - 3x)2
34. g(x)
= xa
\]
x+
x~(x- 3)~
= 2x-~
x+l
~
g’(x) : 2-(x+l)2x-x=(1) = 2(x2 +2x+l)-x~- -2xx + 2x + 2
(X + 1)2
(~ + ~)~
(x + 1)=
35. f(x)= (2x3 + 5x)(x- 3)(x + 2)
f’(x) : (6x2 + 5)(x - 3)(x + 2)+ (2x’ + ’x)(1)(x + 2)+ (2x’ + 5x)(x - 3)(1)
=
(6x2 + 5)(x2 - x - 6)+ (2x3 -1- 5x)(x + 2) q-(2x3 q- 5x)(x - 3)
= lOx4 - 8x~ - 21X2 - lOx- 30
Note: You could simpli@ first:
s x) : + - _
36. f(x) = (x3 - x)(x~ + 2)(x2 + x- 1)
(3x2 -1)(x2 + 2)(x2 + x -1)+ (x’ - x)(2x)(x~ + x -1)+ (xa - x)(x2 + 2)(2x + 1)
s’(x) =
(3x4
+ 5X2 -
2)(x2 + x -1)+ (2x4- 2xg)(x2 + x -1)+ (x5 + x’- 2x)(2x + 1)
(3X6 + 5X4 - 2X2 q- 3X5 q- 5X3 - 2x - 3x4 - 5x2 + 2)
+ (2X6- 2X4 +, 2Xs- 2X6 - 2X4 + 2X2)
(2X6 + 2X4 -4x2 + x~ + x3 - 2x)
7x6 +6xs +4x3-9x2-4x+2
+
37. f(x)- X2 C2X2 L c2
- c >x) _ +
39. f(t) = t~ sint
f’(t) = t~ cos t + 2t sin t = t(t cos t + 2 sin t)
40. s(o) = (o + l)oos o
4XC2
f’(O) = (0 + 1)(-sin O) + (cos 0)(1)
(x=_c=)
= cosO-(O+l) sinO
C2 _ X2
38. f(x) = c2 + x2
41. f(t) = cos,
t
f’(t) = -t sin t - cos t= t sin t + cos t
f’(x) = @2 + x2)(_2x)_ (172 _ X2)(2X)
t2
t2
4xc2
© 2010 Brooks/Cole, Cengage Learning
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 137
50. y = xsinx+cosx
y’ = xcosx+sinx-sinx = xcosx
42. f(x) - sinx
x3
x3 cos x - sin x(3x2]
\ !
x cos x - 3 sin x
X4
43. f(x) = -x + tan x
f’(x) : -1 + sec2 x : tan2 x
51. f(x) = x2 tan x
f’(x) = x2 sec2x + 2x tan x
= x(x sec2x + 2 tan x)
52. f(x) = sin x cos x
44. y = x + cot x
y’ = 1-csc2x =-cot2x
45. g(t) ="~-g,/t +6csct = t1/4 + 6csct
g’(t),: = 1 t_3/4 _ 6 csc t cot t
4
1
6 csc t cot t
4t3/4
f’(x) = sin x(-sin x) + cos x(cos x)
= cos 2x
53. y = 2x sin x + X2 GOSX
y’ = 2xcosx+2sinx+xz(-sinx)+2xcosx
= 4xcosx+(2-x2) sinx
54. h(O) = 50sec 0 + 0 tan 0
46. h(x) =-1 _12secx
x
= x-I -12secx
h’(x) = -x-~ - 12 see x tan x
-1
12 sec x tan x
2
h’(O) = 50 sec 0 tan 0 + 5 sec 0 + 0 sec2 0 + tan 0
55. g(x)= ~ (2x-5)
x+l)
X
3(1-sinx) _ 3-3sinx
2 cos x
2 cos x
y, = (-3 cos x)(2 cos x) - (3 - 3 sin x)(-2 sin x)
(2 cos x)2
-6 cos2 x + 6 sinx - 6 sin2 x
,
47. y-
g’(x) = (x@2)(2) + (2x - 5)[(x + 2)(~)~2~ + 1)(1),]
2x2 + 8x - 1
(x+2)2
(Form of answer may vary.)
4 COS2 X
= ~(-l + tanx secx - tan2 x)
3
= - sec x/tan
x - sec x/
.t
\
2
56. f(x) = . ~ 71
f’(x) = 2x5 + 2x3 + 2x2 - 2
(x2 + 1)2
(Form of answer may vary.)
48. y=-- sec X
x
y,= xsecxtanx-secx
2
0
57. g(O) - 1 - sin 0
x
sec x(x tan x - 1)
x2
49. y =-cscx-sinx
y’ = GSG x cot x - cos x
g’(O) = 1-sinO+OcosO
(1 - sin 0)2
(Form of answer may vary.)
COS X
= --- -- COS X
sin2 x
= COS X(CSC2 X --1)
= cos x cot2 x
© 2010 Brooks/Cole, Cengage Learning
138
Chapter 2 Differentiation
- 1 -sincos0 o
58.
f’(O)- 1 _ cosO-1
cos 0 - 1 (1 - cos O)z
59.
y-
(Form of answer may vary.)
1 + csc x
1 - csc x
y, = (1 - csc x)(=csc x cot x) - (1 + csc x)(csc x cot x)_ _ -2 csc x cot x
0 - csc x)~
0 - ~s~ x)~
60. f(x) = tan x cotx = 1
f(x) : (x + 3)(x2 - 2), (-2, 2)
64. (a)
f’(x) : o
f’(x) = (x + 3)(2x)+ (x2 - 2)(1)
f’(1) = 0
61.
= 3X2 ÷ 6x
h(t) - sect
t
h’(t) : t(sec t tan t) - (sec t)(1) : sec t(t tan t - 1)
t2
if(z) = sec z(z"7/.2tan z - 1)
62.
-2
f’(-2) : -2; Slope at (-2, 2)
t2
Tangent line:
y-2 =-2(x+2) => y =-2x-2
(b)
a
1
7/.2
-12 ~ 12
f(x) : sin x(sin x + cos x)
f’(x) : sin x(cos x -sin x)+ (sin x + cos x)cos x
= sin x cos x - sin2 x + sin x cos x + COS2 X
= sin 2x + cos 2x
f’
(c) Graphing utility confirms dy = -2 at @2, 2).
dr
/’
= sin m + cos- = 1
2 2
63. (a) f(x) : (x3 + 4x -1)(x- 2),
(1,
14)
= 4X3 --6X2 +8X--9
f’(1) = --3; Slope at (1,-4)
Tangent line: y + 4 = -3(x - 1) ~ y = -3x - 1
~
¯
(-5, 5)
f’(x) = (x + 4)(1)- x(1) _ 4
(x + 4)2
(x + 4)2
4 = 4; Slope at (-5, 5)
f’(-5) (-5 + 4)2
f’(x) = (X3 + 4X --1)(1)+ (X- 2)(3X2 + 4)
= X3 +4x--l+3x3 --6Xa +4X--8
(b)
x 4’
f(x)- x +
65. (a)
Tangent line: y - 5 = 4(x + 5) ~ y = 4x + 25
(b)
.¯
-8
I
I
I
(c) Graphing utility confirms dy " 4 at (-5, 5).
-6
(c) Graphing utility confirms dy = -3 at (1,-4).
dr
© 2010 Brooks/Cole, Cengage Learning
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 139
8
69. f(x) - x2 +4"; (2,1)
66. (a) f(x) - x + l’ 2,
~,(~) = (~ + 00) -(~ - 00) _ 2
(x+O~
(x + 1)2
-16x
f’(x) = (x2 + 4)(0) - 8(2x)
--
~’(2) = 9’2. Slope at (2, 31_/
~’(2) --16(2)_
(4 + 14)~ 2
Tangent line: y _1 2=2)~(~X=- 2
3
y
1
~x - 9-
1 2)
y -~ = -~(x-
(b)
1
y = ---x+2
2
2y+x-4=O
-3
-4
70.
67. (a)
f(x) = tan x,
27
2
x + 9; @3, 2.)
f’(x) = (x9 + 9)(0)- 27(2x)
-54x
--
f(x)
,1
f’(x) = sec2 x
~r’(-3) --54(-3)
(9 + 9)1 9 - 2
slo e I ,
Tangent line:
Y
y-l= 2(x - ~)
x+3)
1
y = -x+3
2
y-1 = 2x--2
4x-2y-a-+2 = 0
(b)
31
2
2y-x-6 = 0
71.
16x .
f(x) - x2 + 16’(-2,
256 - 16x2
f’(x) = (x2 + 16)(16)- 16x(2x)
(x ÷
68. (a)
x = see x,
,2
f’(x) = sec x tan x
f’(-2) = 256- 16(4)= 1:2
202
25
8
12
y+-=
+ 2)
~ ~(x
12
25
25y-12x+16 = 0
16
25
Tangent line:
6-,f~x - 3y + 6 - 2w/-~n" = 0
(b)
~
-2
(c) Graphing utility confirms
at (a")
-~xdY = 2-v/-~ ~,2.
© 2010 Brooks/Cole, Cengage Learning
140
Chapter 2 Differentiation
x-4
76. f(x) - ~2 : -7
72. f(x) : xZ + 6; 2,
24 - 4x2
f’(x) = (xz + 6)(4) - 4x(2x)
(; ÷6)~
f’(2)-
f’(x) = (x2 - 7)(1) -(x - 4)(2x)
X2 -- 7 -- 2X2 + 8X
24 - 16 2
102 - 25
x2 - 8x + 7
4 2=2)-~(xy --~
2 16
y =--x+m
25 25
25y-2x-16 = 0
73. f(x) - 2xx2
- 1- 2x-1
(x - 7)(x - 1)
1 s(7): ~1
f’(x) = 0 for x : 1, 7; f(1) : 7’
_ x-2
fhas horizontal tangents at 1,
and 7, ~ .
f’(x) =-2x-2 + 2x-3 - 2(-x
3 + 1)
x
f’(x) = 0when x = 1, and f(1) = 1.
(x -1)2
Horizontal tangent at (1, 1).
X2
74. f(x) = x2 + 1
f’(x) = (x2 + 1)(2x)- (x2)(2x)
(; ÷ 1)2
2x
1
1
2y+x = 6 ~ y = --x + 3; Slope: -2
2
-2
1
2
(x -1)
2
(x-l)2:4
x-1 = +2
x : -1, 3; f(-1) : O, f(3) : 2
f’(x) = 0when x = 0.
Horizontal tangent is at (0, 0).
X2
11
y-0=- ½(
x+ O~
Y 2 =--x-2
y-2=- x-
½( 3) ~
17
--~
2 +-
Y= 2
75. f(x) - x-1
f’(x) = (x- 1)(2x)- x2(1)
_ x2 - 2x _ x(x- 2)
(X - 1)2 (X - 1)2
f’(x) = Owhen x = Oor x = 2.
Horizontal tangents are at (0, 0)and (2, 4).
© 2010 Brooks/Cole, Cengage Learning
Section 2.3
Product and Quotient Rules and Higher-Order Derivatives 141
80. f’(x) = x(cos x - 3) -(sin x - 3x)(1) : x cos x - sin x
7a. i(x) : x-x 1
X2
-1
s’(x) = (x(x -1)x
_
(x-l)2
- 1)~
X2
g’(x) = x(cos x + 2)-(sin
x + 2x)(1)_ _ x cos xsin x
2
2
X
X
ggx~\ , - sin x + 2x _ sin x - 3x + 5x : -, ,t’(xl+ 5
Let(x, y): (x,d(x- 1))be apoint oftangency onthe
graph off.
x
x
fand g differ by a constant.
81. (a) p’(x): f’(x)g(x)+ f(x)g’(x)
if(l) : f’(1)g(1)+ f(1)g’(1): 1(4)
(b) q’(x) : g(x)f’(x)- f(x)g’(x)
g(x)
q’(4) : 3(-1)7(0) : _1
3~3
5 -(x/(x - 1)) _ -1
4x - 5
1
(x-1)(x+l) : (x-l)~
82. (a) p’(x) : f’(x)g(x) + f(x)g’(x)
p’(4) : -~(8) + 1(0) : 4
(4x-5)(x-1) : x+l
4x2-10x+4 = 0
(b) q’(x) : g(x)f’(x)- f(x)g’(x)
g(x)2
1
(x-2)(2x-1) = 0 => x = 7’2
Two tangent lines:
Y + 1 =-4/x-~) ~ y =-4x+l
q’(7) = 4(2)- 4(-1) 12 _ 3
42
16 4
83. Area : A(t) : (6t + 5)x/7 : 6ta12 + 5t1/2
A’(t) : 9t1/2 + 5-t-112 - 18t + 5
2 - 2-,,/7 em2/sec
y-2 =-l(x-2) ~ y =,x+4
79. f’(x) : (x + 2)3- 3x(1) : ~6
(. + :)~
(. + 2):
g’(x) = (x + 2)5- (5x + 4)(1) _ 6
(x + 2)2
(x + 2)2
5x + 4(x+2)
_ 3x+ (x+2~
2x -+f(x)+2
4
g(x)- (x+2)
fand g differ by a constant.
85.
C= iv-~ + -
x+30
, l<x
dC 100(_400+ 30
-g: ( x (x 7;0)
dC
(a) When x = 10: - -$38.13 thousand/100 components
dx
dC
(b) When x = 15: -- = -$10.37 thousand/100 components
(c) When x = 20: dC _ $3.80 thousand/100 components
dx
As the order size increases, the cost per item decreases.
© 2010 Brooks/Cole, Cengage Learning
142
Chapter 2 Differentiation
k
V
86.
dP k-dV
V2
l
P’(t) : 500[.(50 +(50 t2)(4)+ t2)2- (40(20
_ t2
P’(2) ~ 31.55bacteria/h
88.
F - Gmlm~
- Gmlm2d-~
d2
dF _ F’(d) - -2 Gm~m~
dd
d3
a9. (a)
sec x --
1
COS X
sin x
~x[Sec x] : -~
(b)
CSC X --
: --~s 2
cot x -
COS X COS X
1
=
sin x
= sec x tan x
COS X
COS X
1
sin x
~xx[CSC x] = -~x ~ :
(c)
:
cos x
1 cos x
=
sinxsinx
sinx sinx
(sin x)~
- csc x cot x
cos x
sin x
£[cot x] : d[cosx] : sinx(-sinx)- (cos x)(cos x)
sin2 x + cos2 x
sin2 x
1
sin2 x
-- CSC2 x
90. f(x) : secx
g(x) : csc x, [0, 2z)
S’(x) : g’(x)
sec x tan x
sec x tan x = -csc x cot x ~
csc x cot x
¯
1~
1 sin x
cos x cos x
=-1 ~
1
cos x
sin x sin x
sin3 x
COS3 X
--1 ~ tan3x =-1 ~ tanx =-1
© 2010 Brooks/Cole, Cengage Learning
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 143
91. (a) Using a graphing utility,
q(t) = -0.0546t3 + 2.529t2 - 36.89t + 186.6
v(t) = 0.0796t3 - 2.162t2 + 15.32t + 5.9.
(b) ao
95. f(x) = 4x3/2
f’(x) = 6x1/2
3
f"(x) = 3x-~/2 32
96. f(x) : x+ 2
X
16
0
f’(x) = 1- 6___4
3
x
v(t)
f"(x)- 192
4
x
x
~e[ ~
97. JkxY
- x- 1
:’(x) = (x -1)(1)- xO) _
(c) A = v(t) = 0.0796t3 - 2.162t2 + 15.32t + 5.9
q(t) -0.0546t3 + 2.529t~ - 36.89t + 186.6
(x- O:
f"(x) -
2
2
(x- 1)
2
98, :(x) = x + 2x - 1=x+2---
1
x
f’(x) = 1+ x-1 T
A represents the average value (in billions of dollars)
per one million personal computers.
(d) A’(t) represents the rate of change of the average
value per one million personal computers for the
given year t.
92. (a)
sin 0 -
2
99. f(x) = x sin x
f’(x) = x cos x + sin x
f"(x) = x(-sin x)+ cos x + cos x
= -x sin x + 2 cos x
r
r+h
r + h = rcsc0
h = rcscO-r = r(cscO-1)
(b)
:"(x) -
h’(O) = r(-csc 0. cot O)
100. f(x) = sec x
f’(x) = sec x tan x
S"(x) : sec x(sec2 x) + tan x(sec x tan x)
= sec x(sec2 x + tan~ x)
= -3960(2. ~/-~) = -7920~ mi/rad
93. f(x)
= x4 + 2x3- 3x2- x
f’(x) = 4x3 + 6x2 -6x-1
f"(x) = 12x2 + 12x- 6
94. f(x)
= 8x6 - 10x5
101. f’(x)= x:
f"(x) = 2x
102. f"(X) = 2- 2X-1
2
f’"(x) = 2x-2 = x---~-
+ 5x3
f’(x) = 48x5 - 50x4 -t- 15x2
f"(x) = 240x4 - 200x3 + 30x
103. f’"(x) = 2-,JTx
f(4)(x) = _~(2)x_l/2 _ 1
© 2010 Brooks/Cole, Cengage Learning
144
Chapter 2 Differentiation
104. f(4)(x) = 2x + 1
f(S)(x) = 2
110. The graph of a differentiable functionfsuch that
f > 0 and f’ < 0 for all real numbers x would, in
general, look like the graph below.
fl61(x) = o
105. f(x) = 2g(x) + h(x)
f’(x) = 2g’(x)+ h’(x)
f’(2) = 2g’(2) + h’(2)
f
= 2(-2)+ 4
=0
lO6. f(x) = 4f’(x) = -h’(x)
111.
It appears that f is cubic, so
f’ would be quadratic and
f" would be linear.
f’(2) = -h’(2) = -4
L2
107. f(x)- ~(x)
f’(x) = h(x)g’(x)- g(x)h’(x)
112.
It appears that f is quadratic
so f’ would be linear and
f" would be constant.
Y
f’(2) = h(2)g’(2)- g(2)h’(2)
(-1)(-2) -(3)(4)
/2~
= -10
108. f(x)= g(x)h(x)
f’(x) = g(x)h’(x) + h(x)g’(x)
113.
f’(2) = g(2)h’(2) + h(2)g’(2)
= (3)(4)+ (-1)(-2)
= 14
1 2345
L___
109. The graph of a differentiable functionfsuch that
f(2) = O,f’ < Ofor-oo < x < 2, and f’ > Ofor
2 < x < oo would, in general, look like the graph
below.
-3-
y
3
2
1
1
2
3
4
One such function is f(x) = (x - 2)2.
115.
Y
© 2010 Brooks/Cole, Cengage Learning
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 145
116.
Y
l~. "(0: ~oo~
2t + 15
1500 _ __
a(t) = v’(t)= (2t + 15)(100)- (lOOt)(2)
2
(2t + 15)
÷ 15)2
(a) a(5) = 1500 = 2.4 ft/sec2
~2(5) + 15]2
117. v(t) =36-t2,0_< t_< 6
(b) a(lO)
=
a(t) = v’(t)= -2t
v(3) = 27 m/sec
a(3) = -6 m/sec2
The speed of the object is decreasing.
(c) a(20)
=
1500 ~ 1.2
20o) +
flJsec2
1500 = 0.5 ff/sec2
+
119. s(t) = -8.25t2 + 66t
v(t) = s’(t) = 16.50t + 66
a(t) = v’(t) = -16.50
t(sec)
0
1
s(t) fit)
0
v(t) = s’(t) (ft/sec)
66
3
4
57.75 99
123.75
132
49.5
16.5
0
a(t) = v’(,) (ft/sec2) -16.5 -16.5
2
33
-16.5 -16.5
-16.5
Average velocity on:
[0,1] is 57.75 - 0 _ 57.75
1-0
[1,2]is 99 - 57.75 = 41.25
2-1
[2, 3] is 123.75 - 99 = 24.75
522
[3, 4] is 132
- 123.75
~2~
= 8.25
120.
16
12
8
s
t
_
s position function
v velocity function
a acceleration function
(b) The speed of the particle is the absolute value of its velocity. So, the particle’s speed is slowing
down on the intervals (0, 4/3)and (8/3, 4)and it speeds up on the intervals (4/3, 8/3)and (4, 6).
16!
1284-4-8-12 -16-
I4~
!
© 2010 Brooks/Cole, Cengage Learning
146
121.
Chapter 2 Differentiation
f(x) = x"
f(")(x) : n(n -1)(n - 2)... (2)(1) : n!
Note: n! : n(n - 1)... 3. 2. 1 (read "n factorial")
122.
f(.) : _
1
x
f(")(x) = (-1)"(n)(n-xn4:l
1)(n- 2)...(2)(1) _ (-1)"n!
xn+l
123. f(x): g(x)h(x)
(a) S’(x) : g(x)h’(x) + h(x)g’(z)
f"(x) : g(x)h"(x) + g’(x)h’(x) + h(x)g"(x) + h’(x)g’(x)
: g(x)h"(x) + 2g’(x)h’(x) + h(x)g"(x)
f’"(x) = g(x)h"’(x) + g’(x)h"(x) + 2g’(x)h"(x) + 2g"(x)h’(x) + h(x)g’"(x) + h’(x)g"(x)
: g(x)h’"(x)+ 3g’(x)h"(z)+ 3g"(x)h’(x)+ g’"(x)h(x)
f(4)(x) = g(x)h(4)(x) + g’(x)h’"(x) + 3g’(x)h’"(x) + 3g"(x)h"(x) + 3g"(x)h"(x) + 3g’"(x)h’(x)
+ g’"(x)h’(x) + g(4)(x)h(x)
= g(x)h(4)(x)+ 4g’(x)h"’(x)+ 6g"(x)h"(x)+ 4g’"(x)h’(x)+ g(4)(x)h(x)
n(n -1)(n - 2)... (2)(1) ,/(xhh(._O(xh + n(n -1)(n - 2)... (2)(1) o~,,(xhh(._Z)(x~
(b) f(")(x) = g(x)h(")(x)+ l[(n 1)(n 2) ~]~t , t, (2)~(; : ~ : ~::~(1)]~t,
t,
n(n - 1)(n - 2)... (2)(1)
::
+ [(, -1)(, - 2)... (2)(1)J(i)
:,,,(x~h(,_3)(x~ + ...
’’ ’’
+ g<")(x)Z(.)
: g(")~<")(") + it(, - 1~g’(")~’-’)(")) + 2t(, : 2)tg’’(x)~"-~)(") + ""
+ ~g("-i)(x)h’(x) + g(")(x)h(x)
Note: n[ = n(n - 1)...3 ¯ 2 ¯ 1 (read "n factorial")
124. [xf(x)]’ = xf’(x) + f(x)
[xf(x)]" = xf"(x) + f’(x) + f’(x) = xf"(x) + 2f’(x)
Vxf(x)]’" = xf’"(x)+ f"(x)+ 2f"(x)= xf’"(x)+ 3f"(x)
xf(x)
In general, [ ] = xf(")(x)+
125. f(x) : x" sin x
f’(x) = x" cos x + nx"-I sin x
When n = 1: f’(x) = x cos x + sin x
When n = 2:f’(x) = x2cosx+2sinx
When n = 3: f’(x) = x3 cos x + 3x2 sin x
When n = 4: f’(x) = x4 cos x + 4x3 sin x
For general n, f’(x) = x" cos x + nx"-I sin x.
© 2010 Brooks/Cole, Cengage Learning
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 147
126. f(x]
cos x _ x-" cos x
Xn
f’(x) = -x-’t sin x - r/x-n-l cos x
=-x ....’(xsinx+ cos x)
133. True
h’(c) = f(c)g’(c) + g(e)f’(e)
=
g(c)(o)
=0
xsinx+ ncosx
134. True
Xn+l
When n = l:f’(x) = xsinx+cosx
XE
When n = 2:f’(x) =
x sin x + 2 cos x
x3
When n = 3: f’(x) = x sin x +4 3 cos x
x
When n = 4: f’(x) = x sin x 5+ 4 cos x
x
For general
,ix~ =
x sin x + ncosx
Xn+l
127, y ..... 1 y,
1 y,,_ 2
X’
X2 ’ -- -- x3
x3y’’ + 2xEy’ = [_x3j + 2x2 _ =2-2=0
x r21
y = 2x3 - 6x + 10
128.
y’ = 6x2 - 6
y" = 12x
y’" = 12
-24xE
-y’" - xy"- 2y’= -12- x(12x)- 2(6x2-= 6)
129.
y = 2sinx + 3
y’ = 2 cos x
y" = -2 sin x
y"+y =-2sinx+(2sinx+3) =3
130.
y = 3cosx+sinx
y’ =-3sinx+ cosx
y" = -3 cos x - sin x
y"+y = (-3 cos x - sin x) + (3 cos x + sinx) = 0
131. False. If y = f(x)g(x), then
#Y =
135. True
136. True. If v(t) = c then a(t) = v’(t) = O.
137. f(x) : axE + bx + c
f’(x) = 2ax + b
x-intercept at (1, 0): 0 = a + b + c
(2, 7)on graph: 7 = 4a + 2b + c
Slope 10 at (2, 7):10 = 4a + b
Subtracting the third equation from the second,
-3 = b + c. Subtracting this equation from the first,
3 = a. Then, 10 = 4(3) + b ~ b = -2. Finally,
-3 = (-2)+c ~ c =-1.
f(x) = 3xE - 2x - 1
138. f(x) = ax3 + bxE + cx + d,a * 0
f’(x) = 3ax2 + 2bx + c
-2b + "x/4b2 - 12ac
6a
(a) No horizontal tangents: f’(x) , 0
f’(x) = 0 ~ x =
4b2 -12ac < 0
Example: a = c = 1, b = 0:
f(x) = x3 + x
(b) Exactly one horizontal tangent 4b2 - 12ac = 0
Example: a = 1, b = 3, c = 3:
f(x) = x3 + 3xE + 3x
(c) Exactly two horizontal tangents
4bE -12ac > 0
Example: b = 1, a = 1, c = 0:
f(x)
= x3 + X2
132. True. y is a fourth-degree polynomial.
d"y
n - = 0whenn > 4.
dx
© 2010 Brooks/Cole, Cengage Learning
148
139.
Chapter 2 Differentiation
f(x) = xx = Jx~’
~, if x_> 0
-x
ifx < 0
> °o}:
f’(x) = [-2x, ifx <
2, ifx > 0
f"(x) = -2, ifx < 0
f"(O) does not exist since the left and right derivatives are not equal.
140. (a) (fg’- f’g)’ = fg" + f’g’- f’g’- f"g
= fg"- f"g
(b) (fg)"
Tree
= (fg’+ fk)’
= fg"+ f’g’+ f’g’+ f"g
= fg" + 2fk’+ f~
~ fg" + f"g
False
~4~.
= ~[f(~)g(~)]~(~)+ f(~)g(~>’(~)
= Ef(x)g’(x)+ g(x)f’(x)Th(x) + f(x)g(x)h’(x)
= f’(x)g(x)h(x)+ f(x)g’(x)h(x)+ f(x)g(x)h’(x)
Section 2.4 The Chain Rule
y:
1o y : (Sx -- 8)4
3. y=~x3-7
u : g(x)
u = 5x-8
y=zt4
u=x+l
y = /d-l/2
Zt =X3-7
4, y =3tan(x-x2)
y = 3 tan u
5. y = csc3x
/A = CSC X
5X
6. y = sin2
5x
7. y = (4x- 1)3
y = sin u
9. g(x) : 3(4 - 9x)4
y’ : 3(4x -1)2(4) : 12(4x - 1)2
g’(x) : 12(4 - 9x)3(-9) : -108(4- 9x)3
10. f(t): (9t + 2)2/3
8. y = 2(6- X2)5
y’: 2(5)(6- X2)4(-2x) :
y =/23
-20x(6 - X2)4
6
f’(t) = 9t + 2)-1/’(9)- ~ + 2
© 2010 Brooks/Cole, Cengage Learning