Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
h
e 1
=1 .
1. (a) e is the number such that lim
h 0 h
(b)
x
x
x
(2.7 1)/x
x
(2.8 1)/x
0.001
0.9928
0.001
1.0291
0.0001
0.9932
0.0001
1.0296
0.001
0.9937
0.001
1.0301
0.0001
0.9933
0.0001
1.0297
h
h
2.7 1
2.8 1
From the tables (to two decimal places), lim
=0.99 and lim
=1.03 . Since
h
h
h 0
h 0
0.99<1<1.03 , 2.7<e<2.8 .
2. (a)
The function value at x=0 is 1 and the slope at x=0 is 1 .
x
e
(b) f (x)=e is an exponential function and g(x)=x is a power function.
d x x
d e
e
(e )=e and
(x )=ex
dx
dx
1
.
x
e
(c) f (x)=e grows more rapidly than g(x)=x when x is large.
/
3. f (x)=186.5 is a constant function, so its derivative is 0 , that is, f (x)=0 .
/
4. f (x)= 30 is a constant function, so its derivative is 0 , that is, f (x)=0 .
5. f (x)=5x 1
10
6. F(x)= 4x 2
7. f (x)=x +3x 4
/
f (x)=5 0=5
/
10 1
F (x)= 4(10x
/
)= 40x
9
2 1
f (x)=2x
+3 0=2x+3
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
8
5
8
5
8. g(x)=5x 2x +6
9. g(x)=5x 2x +6
10. f (t)=
11. y=x
y =
x
3/5
/
8 1
5 1
7
4
x
=
x
/
7
1/3
1
x
2
2
21. g(x)=x +
5x
7/5
3/5) 1
= 54t
= 3t
8/5
10
8
7 10
R (x)= 7 10 x =
x
18. y= x =x 20. f (t)= t =
/
= 10 x x
19. F(x)=
2
7/5
10
1/2
/
x
17. G(x)= x 2e =x 2e 3
)+0=40x 10x
x
3 (
t
5
Y (t)=6( 9)t
7
2 x
5
2/5) 1
/
10
) 2(5x
)+0=40x 10x
1 5
3
5
3
(6t ) 3(4t )+1=3t 12t +1
2
/
R (t)=5
9
) 2(5x
4
/
2
2
V (r)= (3r )=4 r
3
15. Y (t)=6t 16. R(x)=
4
y =5(e )+0=5e
4 3
13. V (r)= r 3
14. R(t)=5t
7
f (t)=
/
12. y=5e +3
5 1
2 (
x
5
/
8 1
g (x)=5(8x
1 6 4
t 3t +t 2
2/5
/
g (x)=5(8x
/
y =
5
=
1 x
3
2
x
2
2
=x +x 1/2
1 x
2
1/2
1
x
2e =
2 x
2e
x
1
2/3
=
2/3
3x
5 5
1
2
1 1/2 =t t
t
1
G (x)=
8
x=
1 5
x
32
/
f (t)=
/
1 t
2
/
F (x)=
1/2
3
1
5 4
4
(5x )=
x
32
32
1 t
2
g (x)=2x+( 2)x =2x
3/2
=
1
2 t
+
1
2t t
2
x
3
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
22. y= x (x 1)=x
3/2
/
1/2
x y =
3 1/2 1 x x
2
2
1/2
=
1 x
2
1/2
(3x 1) [ factor out
1 x
2
1/2
]
3x 1
.
2 x
/
or y =
2
x +4x+3 3/2
1/2
1/2
23. y=
=x +4x +3x x
1
1
/ 3 1/2
1/2
y = x +4
x +3 2
2
2
3/2
2/2
x
3/2
3
2
3
x+
2
x 2x x
=
1/2
note that x =x x =x x
2
x 2 x
24. y=
=x 2x
x
2
25. y=4
1/2
/
/
2
y =0 since 4
v
4
t
/
3
2 3/4
2
A
31. z=
3
2/3
10
y
/
3/2
y
2
v
/
u =
3
4
2 t
3
t
3
7/4
3
=2t+
4t
1/3
+2
11
/
z = 10Ay
3
2
y
+Be =
y
32. y=e
v
b
= 2 + 3 /(2 u )
2
2c
v
v =2t
+Be +Be =Ay
10
1/2
y =ae bv 2cv =ae /
t +2 t =t +2t
30. u=
1 u
2
y =2ax+b
=t t
3
=1+1/(x x )
g (u)= 2 (1)+ 3
b c
v
1
2
+ =ae +bv +cv 2
v
v
1
2
29. v=t 3/2
/
27. y=ax +bx+c
28. y=ae +
x
is a constant.
26. g(u)= 2 u+ 3u = 2 u+ 3 u 2
1
2
y =1 2
t
7/4
=
2
3
3
=2t+
4
4t
1/2
v
t
3
+3 t
3
3 t
10A
11
+Be
y
y
x+1
x 1
x
+1=e e +1=e e +1
x
33. f (x)=e 5x
/
/
x
y =e e =e
x+1
x
f (x)=e 5 .
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
/
Notice that f (x)=0 when f has a horizontal tangent, f
negative when f is decreasing.
5
3
/
34. f (x)=3x 20x +50x
4
/
is positive when f is increasing, and f
2
/
15
3
/
14
/
is an even function while f is an
2
/
1
/
f (x)=45x 15x .
Notice that f (x)=0 when f has a horizontal tangent, f
negative when f is decreasing.
36. f (x)=x+1/x=x+x is
f (x)=15x 60x +50 .
Notice that f (x)=0 when f has a horizontal tangent and that f
odd function.
35. f (x)=3x 5x +3
/
/
2
/
is positive when f is increasing, and f
/
is
2
f (x)=1 x =1 1/x .
Notice that f (x)=0 when f has a horizontal tangent, f
/
is positive when f is increasing, and f
/
is
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
negative when f is decreasing.
/
2
3
37. To graphically estimate the value of f (1) for f (x)=3x x , we’ll graph f in the viewing
rectangle [1 0.1,1+0.1] by [ f (0.9),f(1.1)] , as shown in the figure. If we have sufficiently zoomed in
on the graph of f , we should obtain a graph that looks like a diagonal line; if not, graph again with
1 0.01 and 1+0.01 , etc.
2.299 1.701 0.589
/
=
=2.99 .
Estimated value: f (1)
1.1 0.9
0.2
2
3
/
Exact value: f (x)=3x x 2
/
f (x)=6x 3x , so f (1)=6 3=3 .
38. See the previous exercise. Since f is a decreasing function, assign Y (3.9) to Y
1
Y
min
and Y (4.1) to
1
.
0.49386 0.50637 0.01251
=
= 0.06255 .
4.1 3.9
0.2
1 3/2
1 3/2
1
/
/
f (x)=
x
, so f (4)=
(4 )=
2
2
2
/
Estimated value: f (4)
Exact value: f (x)=x
4
x
39. y=x +2e y=2x+2.
/
1/2
3
x
2
2
3
=
1
= 0.0625 .
16
/
y =4x +2e . At (0, 2) , y =2 and an equation of the tangent line is y 2=2(x 0) or
2
/
40. y=(1+2x) =1+4x+4x y 9=12(x 1) or y=12x 3.
41. y=3x x y=3x 1 .
1
8
/
/
y =4+8x. At (1, 9), y =12 and an equation of the tangent line is
2
/
y =6x 3x . At (1, 2) , y =6 3=3 , so an equation of the tangent line is y 2=3(x 1) , or
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
3 1/2
/ 3
x . At (4, 8) , y = (2)=3, so an equation of the tangent line is
2
2
y 8=3(x 4), or y=3x 4 .
42. y=x x =x
3/2
/
y =
43. (a)
(b)
From the graph in part (a), it appears that f
negative (so f
/
(x , x ) and (x , 1
2
3
4
3
is negative) on ( /
is zero at x 1.25 , x 0.5 , and x 3 . The slopes are
1
2
3
, x ) and (x , x ). The slopes are positive (so f
1
2
3
/
is positive) on
).
2
(c) f (x)=x 3x 6x +7x+30
/
3
2
f (x)=4x 9x 12x+7
44. (a)
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
(b)
From the graph in part (a), it appears that f
(so f
/
is positive) on ( x
2
(c) g(x)=e 3x , x ) and (x , 1
/
/
is zero at x 0.2 and x 2.8 . The slopes are positive
1
2
) . The slopes are negative (so f
2
/
is negative) on (x , x ) .
1
2
x
g (x)=e 6x
3
2
/
2
45. The curve y=2x +3x 12x+1 has a horizontal tangent when y =6x +6x 12=0
2
6(x +x 2)=0 and (1, 6) .
3
6(x+2)(x 1)=0 x= 2 or x=1 . The points on the curve are ( 2, 21)
2
/
2
46. f (x)=x +3x +x+3 has a horizontal tangent when f (x)=3x +6x+1=0
6
36 12
1
6 .
x=
= 1
6
3
3
47. y=6x +5x 3
/
2
2
m=y =18x +5 , but x 0 for all x , so m 5 for all x .
x
/
x
48. The slope of y=1+2e 3x is given by m=y =2e 3 .
The slope of 3x y=5 y=3x 5 is 3 .
m=3
x
2e 3=3
x
e =3
x=ln 3 . This occurs at the point (ln 3, 7 3ln 3) (1.1, 3.7) .
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
49.
2
Let (a, a ) be a point on the parabola at which the tangent line passes through the point (0, 4). The
tangent line has slope 2a and equation y ( 4)=2a(x 0)
2
2
y=2ax 4 . Since (a, a ) also lies on the line,
2
a =2a(a) 4, or a =4 . So a=
2 and the points are (2, 4) and ( 2, 4).
2
/
2
50. If y=x +x , then y =2x+1 . If the point at which a tangent meets the parabola is (a, a +a), then
the slope of the tangent is 2a+1 . But since it passes through (2, 3), the slope must also be
2
y a +a+3
=
.
x
a 2
2
a +a+3
2
2
Therefore, 2a+1=
. Solving this equation for a we get a +a+3=2a 3a 2
a 2
2
a 4a 5=(a 5)(a+1)=0 a=5 or 1 . If a= 1 , the point is ( 1, 0) and the slope is 1, so the equation
is y 0=( 1)(x+1) or y= x 1. If a=5, the point is (5, 30) and the slope is 11, so the equation is
y 30=11(x 5) or y=11x 25.
2
/
/
51. y= f (x)=1 x f (x)= 2x, so the tangent line at (2, 3) has slope f (2)= 4. The normal line has
1 1
1
1
7
slope = and equation y+3= (x 2) or y= x .
4 4
4
4
2
2
52. y= f (x)=x x /
/
f (x)=1 2x. So f (1)= 1, and the slope of the normal line is the negative
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
reciprocal of that of the tangent line, that is, 1/( 1)=1. So the equation of the normal line at (1, 0) is
2
y 0=1(x 1) y=x 1. Substituting this into the equation of the parabola, we obtain x 1=x x x=
1. The solution x= 1 is the one we require. Substituting x= 1 into the equation of the parabola
to find the y coordinate, we have y= 2. So the point of intersection is ( 1, 2), as shown in the
sketch.
53.
1
1
x+h x
x (x+h)
=lim
h
h 0 hx(x+h)
/
f (x) =lim f (x+h) f (x) =lim
h
h 0
h 0
=lim
h
0
h
1
1
=lim
=
2
hx(x+h) h 0 x(x+h)
x
2
54. Substituting x=1 and y=1 into y=ax +bx gives us a+b=1 (1) . The slope of the tangent line
/
/
y=3x 2 is 3 and the slope of the tangent to the parabola at (x, y) is y =2ax+b . At x=1 , y =3
3=2a+b (2) . Subtracting (1) from (2) gives us 2=a and it follows that b= 1 . The parabola has
2
equation y=2x x .
2
55. f (x)=2 x if x 1 and f (x)=x 2x+2 if x>1 . Now we compute the right and left hand
derivatives defined in Exercise :
f (1+h) f (1)
2 (1+h) 1
h
/
f (1)=lim
=lim
=lim
=lim 1= 1 and
h
h
h
h
0
h
0
h
0
h
0
2
2
f (1+h) f (1)
(1+h) 2(1+h)+2 1
h
f (1)=lim
=lim
=lim
=lim h=0 .
+
h
h
+
+
+ h
+
/
h
0
h
0
/
h
/
Thus, f (1) does not exist since f (1)
/
0
h
0
/
f (1) , so f is not differentiable at 1 . But f (x)= 1 for x<1
+
/
and f (x)=2x 2 if x>1 .
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
56. g(x)=
lim
h
0
1 2x if x< 1
2
if 1 x 1
x
if x>1
x
{
g( 1+h) g( 1)
=lim
h
h
0
1 2( 1+h) 1
2h
=lim
=lim ( 2 ) = 2 and
h
h
h
0
h
2
lim
h
+
0
0
2
g( 1+h) g( 1)
( 1+h) 1
2h+h
=lim
=lim
=lim ( 2+h)= 2,
h
h
h
+
+
+
h
0
h
0
h
0
/
so g is differentiable at 1 and g ( 1)= 2.
2
2
g(1) g(1)
(1) 1
2h+h
lim
=lim
=lim
=lim (2+h)=2 and
h
h
h
h
0
h
0
h
0
h
0
g(1) g(1)
(1) 1
h
/
lim
=lim
=lim
=lim 1=1 , so g (1) does not exist.
h
h
+
+
+ h
+
h
0
h
0
h
0
h
0
/
Thus, g is differentiable except when x=1 , and g (x)=
2
2
57. (a) Note that x 9<0 for x <9
f (x) =
{
2
x 9
if x 3
x +9
if 3<x<3
x 9
ifx 3
2
2
{
2 if x< 1
2x if 1 x<1
1 if x>1
| x|<3 3<x<3 . So
/
f (x) =
{
2x
if x< 3
2x if 3<x<3
2x
if x>3
=
{
2x
if | x|>3
2x if | x|<3
To show that
10
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
/
f (3) does not exist we investigate lim
h
0
f (3+h) f (3)
by computing the left and right hand
h
derivatives defined in Exercise .
2
f (3+h) f (3)
[ (3+h) +9] 0
f (3)=lim
=lim
=lim ( 6 h)= 6 and
h
h
/
h
0
h
0
h
0
2
2
f (3+h) f (3)
(3+h) 9 0
6h+h
f (3)=lim
=lim
=lim
=lim (6+h)=6 .
+
h
h
h
+
+
+
+
/
h
0
h
0
h
Since the left and right limits are different, lim
h
0
0
h
0
f (3+h) f (3)
/
does not exist, that is, f (3) does not
h
/
exist. Similarly, f ( 3) does not exist. Therefore, f is not differentiable at 3 or at 3 .
(b)
58. If x 1 , then h(x)=| x 1|+| x+2|=x 1+x+2=2x+1.
If 2<x<1 , then h(x)= (x 1)+x+2=3.
If x 2, then h(x)= (x 1) (x+2)= 2x 1. Therefore,
h(x)=
/
To see that h (1)=lim
x
1
{
2x 1 if x 2
3
if 2<x<1
2x+1 if x 1
/
h (x)=
2 if x< 2
0 if 2<x<1
2 if x>1
{
h(x) h(1)
h(x) h(1)
3 3
does not exist, observe that lim
=lim
=0 but
x 1
x 1
3 1
x
1
x
1
h(x) h(1)
2x 2
/
lim
=lim
=2 . Similarly, h ( 2) does not exist.
x 1
+
+ x 1
x
1
x
1
11
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
2
/
59. y= f (x)=ax f (x)=2ax. So the slope of the tangent to the parabola at x=2 is m=2a(2)=4a. The
1
slope of the given line, 2x+y=b y= 2x+b, is seen to be 2, so we must have 4a= 2 a= . So
2
1 2
when x=2, the point in question has y coordinate 2 = 2. Now we simply require that the given
2
line, whose equation is 2x+y=b, pass through the point (2, 2): 2(2)+( 2)=b b=2. So we must have
1
a=
and b=2.
2
/
/
60. f is clearly differentiable for x<2 and for x>2. For x<2, f (x)=2x, so f (2)=4. For x>2,
/
/
/
/
f (x)=m, so f (2)=m. For f to be differentiable at x=2, we need 4= f (2)= f (2)=m. So
+
+
f (x)=4x+b. We must also have continuity at x=2, so 4= f (2)=lim f (x)=lim (4x+b)=8+b. Hence,
x
+
2
x
+
2
b= 4.
3
2
/
2
61. y= f (x)=ax +bx +cx+d f (x)=3ax +2bx+c. The point ( 2, 6) is on f , so f ( 2)=6
8a+4b 2c+d=6 (1) . The point (2, 0) is on f , so f (2)=0 8a+4b+2c+d=0 (2) . Since there are
/
/
horizontal tangents at ( 2, 6) and (2, 0), f (
2)=0. f ( 2)=0
12a 4b+c=0 (3) and
/
f (2)=0 12a+4b+c=0 (4) . Subtracting equation (3) from (4) gives 8b=0 b=0. Adding (1) and (2)
gives 8b+2d=6, so d=3 since b=0. From (3) we have c= 12a, so (2) becomes 8a+4(0)+2( 12a)+3=0
3
3
9
3=16a a=
. Now c= 12a= 12
=
and the desired cubic function is
16
16
4
3 3 9
y=
x
x+3.
16
4
62. (a) xy=c
c
c
y= . Let P= a,
x
a
/
. The slope of the tangent line at x=a is y (a)=
c
2
. Its
a
c
2c
2c
c
c
equation is y
=
(x a) or y=
x+ . so its y intercept is
. Setting y=0 gives x=2a, so
2
2
a
a
a
a
a
2c
the x intercept is 2a. The midpoint of the line segment joining 0,
and (2a, 0) is
a
c
a,
=P.
a
(b) We know the x and y intercepts of the tangent line from part (a), so the area of the triangle
1
1
1
bounded by the axes and the tangent is (base)(height)= xy= (2a)(2c/a)=2c, a constant.
2
2
2
12
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions
1000
63. Solution 1: Let f (x)=x
. Then, by the definition of a derivative,
1000
/
f (1)=lim
x
1
f (x) f (1)
x
1
. But this is just the limit we want to find, and we know (from
=lim
x 1
x 1
x 1
/
the Power Rule) that f (x)=1000x
999
/
1000
999
, so f (1)=1000(1)
=1000 . So lim
x
1000
1)=(x 1)(x
Solution 2: Note that (x
1000
lim
x
1
x
1
(x 1)(x
= lim
x 1
x 1
=
lim (x
x
999
+x
999
+x
998
+x
998
999
+x
998
+x
997
997
1
x
1
=1000 .
x 1
2
+ +x +x+1). So
2
+x + +x +x+1)
x 1
997
2
+ +x +x+1) = 1+1+1+...+1+1+1
1
1000 ones
= 1000, as above.
64.
In order for the two tangents to intersect on the y axis, the points of tangency must be at equal
2
distances from the y axis, since the parabola y=x is symmetric about the y axis. Say the points of
2
2
2
tangency are (a, a ) and ( a, a ), for some a>0. Then since the derivative of y=x is dy/dx=2x, the
2
2
left hand tangent has slope 2a and equation y a = 2a(x+a), or y= 2ax a , and similarly the right
hand tangent line has equation
2
2
2
y a =2a(x a), or y=2ax a . So the two lines intersect at (0, a ). Now if the lines are perpendicular,
1
2 1
then the product of their slopes is 1, so ( 2a)(2a)= 1 a = a= . So the lines intersect at
4
2
1
0, .
4
13
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
dV dV dx
2 dx
=
=3x
dt
dx dt
dt
3
1. V =x dA dA dr
dr
=
=2 r
dt dr dt
dt
2
2. (a) A= r dA
dr
2
=2 r
=2 (30m)(1m/s)=60 m / s
dt
dt
(b)
dy dy dx
dy
2
2
=
=(3x +2)(5)=5(3x +2) . When x=2 ,
=5(14)=70 .
dt dx dt
dt
3
3. y=x +2x
2
2
4. x +y =25
2x
2
dx
dy
+2y
=0
dt
dt
2
When y=4 , x +4 =25
2
2
dz
dx
dy
dz 1
dx
dy
2 2
2
=2x
+2y
=
x
+y
. When x=5 and y=12 , z =5 +12 dt
dt
dt
dt z
dt
dt
dx
dy
dz
1
46
z= 13 . For
=2 and
=3, =
(5 2+12 3)=
.
dt
dt
dt 13
13
2
5. z =x +y 2
z =169
dx
dy
dx
y dy
= y
=
.
dt
dt
dt
x dt
dy
dx
4
x= 3 . For
=6 ,
=
(6)= 8 .
dt
dt
3
x
3
2z
6. y= 1+x we have 4=
dy dy dx 1
3
=
= (1+x )
dt dx dt 2
3(4) dx
2(3) dt
dx
(3x ) =
dt
1/2
2
2
3x
2 1+x
3
dx
dy
. With
=4 when x=2 and y=3 ,
dt
dt
dx
=2 cm / s.
dt
7. (a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mi / h passes
directly over a radar station. If we let t be time (in hours) and x be the horizontal distance traveled by
the plane (in mi), then we are given that dx/dt=500 mi / h.
(b) Unknown: the rate at which the distance from the plane to the station is increasing when it is 2 mi
from the station. If we let y be the distance from the plane to the station, then we want to find dy/dt
when y=2 mi.
(c)
2
2
(d) By the Pythagorean Theorem, y =x +1 2y(dy/dt)=2x(dx/dt) .
3
dy x dx x
dy
2 2
(e)
=
= (500) . Since y =x +1 , when y=2 , x= 3 , so
=
(500)=250 3 433 mi / h.
dt y dt y
dt
2
8. (a) Given: the rate of decrease of the surface area is 1 cm
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
2
2
/ min. If we let t be time (in minutes) and S be the surface area (in cm ), then we are given that
2
dS/dt= 1 cm / s.
(b) Unknown: the rate of decrease of the diameter when the diameter is 10 cm. If we let x be the
diameter, then we want to find dx/dt when x=10 cm.
(c)
(d) If the radius is r and the diameter x=2r , then r=
1
2
x and S=4 r =4
2
dS dS dx
dx
=
=2 x
.
dt dx dt
dt
dS
dx
dx
1
dx
1
(e) 1=
=2 x
=
. When x=10 ,
=
dt
dt
dt
2 x
dt
20
min.
1
x
2
2
2
= x . So the rate of decrease is
1
20
cm /
9. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15 ft tall pole at a rate of 5
ft / s. If we let t be time (in s) and x be the distance from the pole to the man (in ft), then we are given
that dx/dt=5 ft / s.
(b) Unknown: the rate at which the tip of his shadow is moving when he is 40 ft from the pole. If we
let y be the distance from the man to the tip of
d
his shadow (in ft), then we want to find
(x+y) when x=40 ft.
dt
(c)
(d) By similar triangles,
15 x+y
=
6
y
15y=6x+6y
(e) The tip of the shadow moves at a rate of
9y=6x
d
d
(x+y)=
dt
dt
2
x.
3
2
5 dx 5
25
x+ x =
= (5)=
ft / s.
3
3 dt 3
3
y=
10. (a) Given: at noon, ship A is 150 km west of ship B; ship A is sailing east at 35 km / h, and ship B
is sailing north at 25 km / h. If we let t be time (in hours), x be the distance traveled by ship A (in
km), and y be the distance traveled by ship B (in km), then we are given that dx/dt=35 km / h and
dy/dt=25 km / h.
(b) Unknown: the rate at which the distance between the ships is changing at 4:00 P.M. If we let z be
the distance between the ships, then we want to find dz/dt when t=4 h.
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
(c)
2
2
2
(d) z =(150 x) +y 2z
dz
=2(150 x)
dt
dx
dt
+2y
dy
dt
2
2
(e) At 4:00 P.M., x=4(35)=140 and y=4(25)=100 z= (150 140) +100 = 10,100 . So
dz 1
dx
dy
10(35)+100(25)
215
=
(x 150) +y
=
=
21.4 km / h.
dt z
dt
dt
10,100
101
11.
dx
dy
2 2 2
=60 mi / h and
=25 mi / h. z =x +y dt
dt
dz
dx
dy
dz 1
dx
dy
z =x
+y
=
x
+y
.
dt
dt
dt
dt z
dt
dt
We are given that
2
2z
dz
dx
dy
=2x
+2y
dt
dt
dt
2
After 2 hours, x=2(60)=120 and y=2(25)=50 z= 120 +50 =130 , so
dz 1
dx
dy
120(60)+50(25)
=
x
+y
=
=65 mi / h.
dt z
dt
dt
130
12.
We are given that
When x=8 ,
dx
y 2
=1.6 m / s. By similar triangles,
= dt
12 x
y=
24
x
dy
24 dx
24
=
=
(1.6) .
2 dt
2
dt
x
x
dy
24(1.6)
=
= 0.6 m / s, so the shadow is decreasing at a rate of 0.6 m / s.
dt
64
13.
We are given that
dx
dy
2
2
2
=4 ft / s and
=5 ft / s. z =(x+y) +500 dt
dt
2z
dz
=2(x+y)
dt
dx dy
+
dt dt
. 15
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
minutes after the woman starts, we have x=(4ft/s)(20min)(60s/min)=4800 ft and y=5 15 60=4500
2
2
z= (4800+4500) +500 = 86,740,000 , so
dz x+y
dx dy
4800+4500
837
=
+
=
(4+5)=
8.99 ft / s .
dt
z
dt dt
86,740,000
8674
14. We are given that
dx
=24 ft / s.
dt
(a)
dx
.
dt
dy 90 x
2
2
When x=45 , y= 45 +90 =45 5 , so
=
dt
y
2
2
2
y =(90 x) +90 2y
dy
=2(90 x)
dt
dx
dt
so the distance from second base is decreasing at a rate of
=
45
24
( 24)=
,
45 5
5
24
10.7 ft / s.
5
(b) Due to the symmetric nature of the problem in part (a), we expect to get the same answer and we
dz
dx
dz
45
24
2 2
2
. When x=45 , z=45 5 , so
=
(24)=
10.7 ft / s.
do. z =x +90 2z =2x
dt
dt
dt 45 5
5
1
dh
dA
bh , where b is the base and h is the altitude. We are given that
=1 cm / min and
=2
2
dt
dt
dh
db
dA 1
2
=
b
+h
cm / min. Using the Product Rule, we have
. When h=10 and A=100 , we
dt 2
dt
dt
1
1
1
db
db
4=20+10
have 100= b(10) b=10 b=20 , so 2=
20 1+10
2
2
2
dt
dt
db 4 20
=
= 1.6 cm / min.
dt
10
15. A=
16.
dy
dx
2 2
= 1 m / s, find
when x=8 m. y =x +1
dt
dt
, y= 65 , so
Given
2y
dy
dx
=2x
dt
dt
dx y dy
y
=
=
. When x=8
dt x dt
x
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
dx
=
dt
65
. Thus, the boat approaches the dock at
8
65
1.01 m / s.
8
17.
We are given that
dx
dy
2
2
2
=35 km / h and
=25 km / h. z = ( x+y ) +100 dt
dt
2
2z
dz
=2(x+y)
dt
dx dy
+
dt dt
.
2
At 4:00 P.M., x=4(35)=140 and y=4(25)=100 z= (140+100) +100 = 67,600 =260 , so
dz x+y
dx dy
140+100
720
=
+
=
(35+25)=
55.4 km / h.
dt
z
dt dt
260
13
18. Let D denote the distance from the origin (0,0) to the point on the curve y= x .
dD 1 2 1/2
dx
2x+1 dx
2
2
2
2
2
D= (x 0) +(y 0) = x +( x ) = x +x = (x +x) (2x+1) =
. With
dt 2
dt
dt
2
2 x +x
dx
dD
9
27
=3 when x=4 ,
=
(3)=
3.02 cm / s.
dt
dt 2 20
4 5
19.
dV
1 2
=C 10 , 000 , where V = r h is the volume at
dt
3
2
r h
1
1
1
dV 3
2 dh
time t . By similar triangles, = r= h V = h h=
h
= h
.
2 6
3
3
3
27
dt 9
dt
dh
800,000
2
When h=200cm ,
=20cm/min , so C 10 , 000= (200) (20) C=10 , 000+
289 ,
dt
9
9
If C= the rate at which water is pumped in, then
3
253 cm / min.
20.
By similar triangles,
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
3 b
1
2
= , so b=3h . The trough has volume V = bh(10)=5(3h)h=15h 1 h
2
1 dh
2
4
. When h= ,
=
= ft / min.
2 dt
1 5
5
2
12=
dV
dh
=30h
dt
dt
dh 2
=
dt 5h
21.
1
( base + base )( height ) , and the
1
2
2
volume V of the 10 meter long trough is 10A . Thus, the volume of the trapezoid with height h is
1
a 0.25 1
V =(10) [0.3+(0.3+2a)]h . By similar triangles, =
= , so
2
h 0.5 2
dV dV dh
dh
dh
0.2
2
2a=h V =5(0.6+h)h=3h+5h . Now
=
0.2=(3+10h)
=
. When h=0.3
dt
dh dt
dt
dt 3+10h
dh
0.2
0.2
1
10
,
=
=
m / min =
m / min or
cm / min.
dt 3+10(0.3) 6
30
3
The figure is labeled in meters. The area A of a trapezoid is
22.
1
(b+12)h(20)=10(b+12)h and, from similar triangles,
2
x 6
y 16 8
8h
11h
= and =
= , so b=x+12+y=h+12+
=12+
. Thus,
h 6
h 6 3
3
3
The figure is drawn without the top 3 feet. V =
2
11h
110h
dV
220
V =10 24+
h=240h+
and so 0.8=
= 240+
h
3
3
dt
3
dh
0.8
3
=
=
0.00132 ft / min.
dt 240+5(220/3) 2275
dh
. When h=5 ,
dt
23.
dV
1 2 1
3
We are given that
=30 ft / min. V = r h= dt
3
3
h
2
2
3
h
h=
12
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
dV dV dh
=
dt
dh dt
2
h dh
30=
4 dt
dh 120
dh 120
6
=
. When h=10 ft,
= 2 =
0.38 ft / min.
2
dt
dt
5
10 h
24.
x
We are given dx/dt=8 ft / s. cot =
100
100 1
When y=200 , sin =
= 200 2
1
rad / s.
50
x=100cot dx
2 d
= 100csc dt
dt
2
d
sin =
8.
dt
100
2
d
(1/2)
1
rad / s. The angle is decreasing at a rate of
=
8=
dt
100
50
25.
1
h
1
bh , but b=5 m and sin = h=4sin , so A= (5)(4sin )=10sin . We are given
2
4
2
d
dA dA d
=0.06 rad / s, so
=
=(10cos )(0.06)=0.6cos . When =
,
dt
dt d dt
3
dA
1
2
=0.6 cos
=(0.6)
=0.3 m / s.
dt
3
2
A=
26.
rad / min. By the Law of Cosines,
90
dx
d
dx 180sin 2
2
2
x =12 +15 2(12)(15)cos =369 360cos 2x
=360sin =
dt
dt
dt
x
We are given d /dt=2 / min =
dx 180sin 60
=60 , x= 369 360cos 60 = 189 =3 21 , so
=
dt
3 21
min.
d
. When
dt
3
7
=
=
0.396 m /
90 3 21
21
27. Differentiating both sides of PV =C with respect to t and using the Product Rule gives us
dV
dP
dV
V dP
dP
P
+V
=0
=
. When V =600 , P=150 and
=20 , so we have
dt
dt
dt
P dt
dt
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
dV
600
3
=
(20)= 80 . Thus, the volume is decreasing at a rate of 80 cm / min.
dt
150
28. PV
1.4
=C P 1.4V
0.4
dV
=
dt
dV
1.4 dP
+V
=0
dt
dt
V
1.4
0.4
dP
V dP
=
. When V =400 ,
dt
1.4P dt
P 1.4V
dP
dV
400
250
P=80 and
= 10 , so we have
=
( 10)=
. Thus, the volume is increasing at a
dt
dt
1.4(80)
7
250
3
rate of
36 cm / min.
7
29. With R =80 and R =100 ,
1
2
1 1
1
1
180
9
400
1
=
+
=
+
=
=
, so R=
. Differentiating
R R
9
R 80 100 8000 400
1
2
dR
dR
1
2
1 1
1 dR
1
1
1
=
+
with respect to t , we have =
2 dt
2 dt
2 dt
R R
R
R
R
R
1
2
1
dR 2
=R
dt
1
2
R
dR
1
dt
dR 400
=
2
dt
9
2
R
1
2
+
1
2
dR
2
dt
. When R =80 and R =100 ,
1
2
2
1
2
(0.3)+
80
1
2
100
(0.2) =
107
0.132
/ s.
810
dB
2/3
2.53
when L=18 using B=0.007W and W =0.12L
.
dt
dB
dB dW dL
2 1/3
20 15
1.53
=
= 0.007 W
(0.12 2.53 L )
dt
dW dL dt
3
10,000,000
2
2.53 1/3
1.53
8
5
(0.12 2.53 18 )
1.045 10 g/yr
= 0.007 3 (0.12 18 )
7
10
30. We want to find
31.
dx
x
x=10sin =2 ft / s. sin =
dt
10
2
d
d
2
2=10cos
=
=
rad / s.
4 dt
dt 10(1/ 2 )
5
We are given that
dx
=10cos dt
d
. When =
,
dt
4
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
32.
Using Q for the origin, we are given
dx
dy
= 2 ft / s and need to find
when x= 5 . Using the
dt
dt
2
Pythagorean Theorem twice, we have
2
2
2
x +12 + y +12 =39 , the total length of the rope.
dx
dy
x
y
+
=0 , so
Differentiating with respect to t , we get
2
2 dt
2
2 dt
x +12
y +12
dy
x
=
dt
y
2
2
y +12
2
2
x +12
2
dx
2
2
2
2
2
2
. Now when x= 5 , 39= ( 5) +12 + y +12 =13+ y +12 dt
2
, and y= 26 12 = 532 . So when x= 5 ,
dy
=
dt
( 5)(26)
( 2)=
532 (13)
2
2
y +12 =26
10
0.87 ft / s. So cart B is
133
moving towards Q at about 0.87 ft / s.
33. (a)
2
2
2
By the Pythagorean Theorem, 4000 +y = . Differentiating with respect to t , we obtain
dy
d
dy
2y =2
. We know that
=600 ft / s, so when y=3000 ft,
dt
dt
dt
d
y dy 3000
1800
2
2
=
=
(600)=
=360 ft / s.
= 4000 +3000 = 25,000,000 =5000 ft and
dt dt 5000
5
y
(b) Here tan =
4000
d
d
(tan )=
dt
dt
y
4000
2
sec d
1 dy
=
dt 4000 dt
2
d
cos dy
=
.
dt
4000 dt
2
dy
4000 4000 4
d
(4/5)
When y=3000 ft,
=600 ft / s, =5000 and cos =
=
= , so
=
(600)=0.096
dt
5000 5
dt 4000
rad / s.
34.
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
We are given that
2
sec =1+
1
3
d
dx
1
2 d
=4(2 )=8 rad / min. x=3tan =3sec . When x=1 , tan = , so
dt
dt
dt
3
2 10
dx
10
80
=
and
=3
(8 )=
83.8 km / min.
9
dt
9
3
35.
dx
=300 km / h. By the Law of Cosines,
dt
1
dy
dx dx
dy 2x+1 dx
2 2 2
2
2
y =x +1 2(1)(x)cos 120 =x +1 2x =x +x+1 , so 2y =2x
+
=
. After
2
dt
dt dt
dt
2y dt
1 minute,
300
dy 2(5)+1
1650
2
x=
=5 km y= 5 +5+1 = 31 km =
(300)=
296 km / h.
60
dt 2 31
31
We are given that
36.
dx
dy
=3 mi / h and
=2 mi / h. By the Law of Cosines,
dt
dt
dz
dx
dy
dy
dx
2 2 2
2 2
z =x +y 2xycos 45 =x +y 2 xy 2z =2x
+2y
2x
2y
. After 15 minutes
dt
dt
dt
dt
dt
1
= h ,
4
13 6 2
3
2 1
3 2
2 2
3
2
2
we have x= and y= = z =
z=
and
+
2
4
4
4 2
4
4
4
4
13 6 2
2
3
1
3
1
2
dz
=
2
3+2
2 2
2 2
3 =
= 13 6 2
4
2
4
2
2
dt
13 6 2
13 6 2
2.125mi/h.
We are given that
37.
Let the distance between the runner and the friend be . Then by the Law of Cosines,
10
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates
2
2
2
=200 +100 2 200 100 cos =50,000 40,000cos (*). Differentiating implicitly with respect to t
d
d
, we obtain 2
= 40,000( sin )
. Now if D is the distance run when
dt
dt
the angle is radians, then by the formula for the length of an arc on a circle, s=r , we have
1
d
1 dD
7
d
D=100 , so =
D
=
=
. To substitute into the expression for
, we must
100
dt 100 dt 100
dt
1
2
know sin at the time when =200 , which we find from (*): 200 =50,000 40,000cos cos =
4
sin =
1
1
4
2
=
15
15
d
. Substituting, we get 2(200) =40,000
4
dt
4
7
100
7 15
6.78 m / s. Whether the distance between them is increasing or decreasing depends on
4
the direction in which the runner is running.
d /dt=
38.
2
= rad / h.
12 6
The minute hand goes around once an hour, or at the rate of 2 rad / h. So the angle between them
(measuring clockwise from the minute hand to the hour hand) is changing at the rate of
11
d /dt= 2 =
rad / h. Now, to relate to , we use the Law of Cosines:
6
6
The hour hand of a clock goes around once every 12 hours or, in radians per hour,
2
2
2
=4 +8 2 4 8 cos =80 64cos (*).
d
d
= 64( sin )
. At 1:00, the angle
dt
dt
2
between the two hands is one twelfth of the circle, that is,
= radians. We use (*) to find at
12 6
d
11
1:00: = 80 64cos
= 80 32 3 . Substituting, we get 2
=64sin
6
dt
6
6
1
11
64
d
2
6
88
=
=
18.6 . So at 1:00, the distance between the tips of the
dt
2 80 32 3
3 80 32 3
hands is decreasing at a rate of 18.6 mm / h 0.005 mm / s.
Differentiating implicitly with respect to t , we get 2
11
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
1. As in Example 1, T (0)=185 , T (10)=172 , T (20)=160 , and
T (10) T (20) 172 160
/
/
T (20)
=
= 1.2 F / min. T (30) T (20)+T (20)(30 20) 160 1.2(10)=148
10 20
10
F.
We would expect the temperature of the turkey to get closer to 75 F
as time increases. Since the temperature decreased 13 F in the first 10 minutes and 12 F in the
second 10 minutes, we can assume that the slopes of the tangent line are increasing through negative
values: 1.3, 1.2,... . Hence, the tangent lines are under the curve and 148 F
is an underestimate. From the figure, we estimate the slope of the tangent line at t=20 to be
184 147
37
=
.
0 30
30
37
2
/
Then the linear approximation becomes T (30) T (20)+T (20) 10 160
(10)=147 147.7 .
30
3
/
2. P (2)
P(1) P(2) 87.1 74.9
=
= 12.2 kilopascals / km.
1 2
1
/
P(3) P(2)+P (2)(3 2) 74.9 12.2(1)=62.7 kPa.
From the figure, we estimate the slope of the tangent line at h=2 to be
/
approximation becomes P(3) P(2)+P (2) 1 74.9
98 63
35
=
. Then the linear
0 3
3
35
63.23 kPa.
3
3. Extend the tangent line at the point (2030,21) to the t axis. Answers will vary based on this
approximation we’ll use t=1900 as our t intercept. The linearization is then
/
P(t)
P(2030)+P (2030)(t 2030)
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
21
(t 2030)
130
21
P(2040)=21+
(2040 2030) 22.6%
130
21
P(2050)=21+
(2050 2030) 24.2%
130
21+
These predictions are probably too high since the tangent line lies above the graph at t=2030.
N(1980) N(1985) 15.0 17.0
N(1990) N(1985) 19.3 17.0
=
=0.4 and B=
=
=0.46 . Then
1980 1985
5
1990 1985
5
N(t) N(1985) A+B
/
N (1985)= lim
=0.43 million / year. So
t 1985
2
t 1985
4. Let A=
/
N(1984) N(1985)+N (1985)(1984 1985) 17.0+0.43( 1)=16.57 million.
N(1995) N(2000) 22.0 24.9
/
N (2000)
=
=0.58 million / year.
1995 2000
5
/
N(2006) N(2000)+N (2000)(2006 2000) 24.9+0.58(6)=28.38 million.
3
/
5. f (x)=x 2
/
/
f (x)=3x , so f (1)=1 and f (1)=3 . With a=1 , L(x)= f (a)+ f (a)(x a) becomes
/
L(x)= f (1)+ f (1)(x 1)=1+3(x 1)=3x 2 .
/
6. f (x)=ln x /
7. f (x)=cos x L(x)= f
2
3
/
/
f (x)=1/x , so f (1)=0 and f (1)=1 . Thus, L(x)= f (1)+ f (1)(x 1)=0+1(x 1)=x 1.
f (x)= sin x , so f
+f
1/3
8. f (x)= x =x /
/
x
2
/
f (x)=
1 x
3
L(x)= f ( 8)+ f ( 8)(x+8)= 2+
2
2/3
2
=0 1
=0 and f
x
2
/
2
= x+
/
= 1 . Thus,
2
, so f ( 8)= 2 and f ( 8)=
.
1
. Thus,
12
1
1
4
.
(x+8)=
x
12
12
3
9. f (x)= 1 x 2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
1
1
/
, so f (0)=1 and f (0)=
. Therefore,
2
2 1 x
1 x = f (x) f (0)+ f /(0)(x 0)
1
1
= 1+ (x 0)=1
x
2
2
1
1
So 0.9 = 1 0.1 1
(0.1)=0.95 and 0.99 = 1 0.01 1
(0.01)=0.995 .
2
2
/
f (x)=
1/3
3
10. g(x)= 1+x =(1+x) /
g (x)=
1
(1+x)
3
2/3
,
1
1
/
3
. Therefore, 1+x =g(x) g(0)+g (0)(x 0)=1+ x . So
3
3
1
1
3
3
3
0.95= 1+ ( 0.05) 1+ ( 0.05) =0.983 , and 1.1 = 1+0.1 1+ (0.1)=1.03 .
3
3
/
so g(0)=1 and g (0)=
3
1/3
3
11. f (x)= 1 x =(1 x) f (x)
/
f (0)+ f (0)(x 0)=1
1.204<x<0.706 .
/
/
f (x)=
1
(1 x)
3
1
x . We need
3
2
3
2/3
/
, so f (0)=1 and f (0)=
1 x 0.1<1
/
1
. Thus,
3
1 3
x< 1 x +0.1 , which is true when
3
/
12. f (x)=tan x f (x)=sec x , so f (0)=0 and f (0)=1 . Thus, f (x) f (0)+ f (0)(x 0)=0+1(x 0)=x .
We need tan x 0.1<x<tan x+0.1 , which is true when 0.63<x<0.63 .
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
4
1
13. f (x)=
4
5
/
=(1+2x) f (x)= 4(1+2x) (2)=
(1+2x)
f (x)
8
/
5
, so f (0)=1 and f (0)= 8 . Thus,
(1+2x)
/
f (0)+ f (0)(x 0)=1+( 8)(x 0)=1 8x .
4
4
We need 1/(1+2x) 0.1<1 8x<1/(1+2x) +0.1 , which is true when 0.045<x<0.055 .
x
/
14. f (x)=e x
/
f (x)=e , so f (0)=1 and f (0)=1 . Thus, f (x)
x
/
f (0)+ f (0)(x 0)=1+1(x 0)=1+x .
x
We need e 0.1<1+x<e +0.1 , which is true when 0.483<x<0.416.
/
4
15. If y= f (x) , then the differential dy is equal to f (x)dx . y=x +5x
16. y=cos x
17. y=xln x
dy= sin x dx= sin xdx
dy= x
2
18. y= 1+t 19. y=
3
dy=(4x +5)dx .
u+1
u 1
dy=
dy=
1
+ln x 1 dx=(1+ln x)dx
x
1
2
(1+t )
2
1/2
(2t)dt=
4
dt
1+t
(u 1)(1) (u+1)(1)
2
(u 1)
20. y=(1+2r) t
5
du=
2
2
2
du
(u 1)
5
dy= 4(1+2r) 2dr= 8(1+2r) dr
2
21. (a) y=x +2x dy=(2x+2)dx
(b) When x=3 and
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
dx=
1
, dy=[2(3)+2]
2
x/4
22. (a) y=e dy=
1
2
=4 .
1 x/4
e dx
4
1 0
e
4
(b) When x=0 and dx=0.1 , dy=
1
(4+5x)
2
(0.1)=0.025 .
5
dx
2 4+5x
5
5 1
1
(b) When x=0 and dx=0.04 , dy=
(0.04)= =
=0.05 .
4 25 20
2 4
23. (a) y= 4+5x dy=
24. (a) y=1/(x+1) dy=
1
2
1/2
5dx=
dx
(x+1)
(b) When x=1 and dx= 0.01 , dy=
25. (a) y=tan x 1 1
1
( 0.01)= =
=0.0025 .
2
4 100 400
2
1
2
dy=sec xdx
2
2
(b) When x= /4 and dx= 0.1 , dy=[sec ( /4)] ( 0.1)=( 2 ) ( 0.1)= 0.2 .
26. (a) y=cos x dy= sin xdx
(b) When x= /3 and dx=0.05 , dy= sin ( /3)(0.05)= 0.5 3 (0.05)= 0.025 3 0.043 .
2
2
2
27. y=x , x=1 , x=0.5
y=(1.5) 1 =1.25 . dy=2xdx=2(1)(0.5)=1
28. y= x , x=1 , x=1
1
1
dy=
dx= (1)=0.5
2
2 x
y= 2 1 = 2 1 0.414
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
2
29. y=6 x , x= 2 , x=0.4
dy= 2xdx= 2( 2)(0.4)=1.6
30. y=
16
, x=4 , x= 1
x
5
31. y= f (x)=x 2
2
y=(6 ( 1.6) ) (6 ( 2) )=1.44
y=
16 16 4
= . dy=
3
4 3
4
16
2
dx=
x
16
2
( 1)=1
4
4
dy=5x dx . When x=2 and dx=0.001 , dy=5(2) (0.001)=0.08 , so
5
(2.001) = f (2.001)
32. y= f (x)= x f (2)+dy=32+0.08=32.08 .
dy=
1
dx . When x=100 and dx= 0.2 , dy=
1
( 0.2)= 0.01 , so
2 100
2 x
99.8 = f (99.8) f (100)+dy=10 0.01=9.99 .
2
2
2/3
33. y= f (x)=x dy= 3 dx . When x=8 and dx=0.06 , dy= 3 (0.06)=0.02 , so
3 x
3 8
2/3
(8.06) = f (8.06)
f (8)+dy=4+0.02=4.02 .
2
2
34. y= f (x)=1/x dy=( 1/x )dx . When x=1000 and dx=2,dy=[ 1/(1000) ](2)= 0.000002 , so
1/1002= f (1002) f (1000)+dy=1/1000 0.000002=0.000998
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
35. y= f (x)=tan x
2
2
dy=sec xdx . When x=45 and dx= 1 ,
2
dy=sec 45 ( /180)=( 2 ) ( /180)= /90 , so tan 44 = f (44 )
f (45 )+dy=1 /90 0.965 .
1
1
dx . When x=1 and dx=0.07 , dy= (0.07)=0.07 , so
x
1
f (1)+dy=0+0.07=0.07.
36. y= f (x)=ln x
dy=
ln 1.07= f (1.07)
/
37. y= f (x)=sec x
/
f (x)=sec xtan x , so f (0)=1 and f (0)=1 0=0 . The linear approximation of f at
/
0 is f (0)+ f (0)(x 0)=1+0(x)=1 . Since 0.08 is close to 0 , approximating sec 0.08 with 1 is
reasonable.
6
/
5
38. If y=x , y =6x and the tangent line approximation at (1,1) has slope 6 . If the change in x is
6
0.01 , the change in y on the tangent line is 0.06 , and approximating (1.01) with 1.06 is reasonable.
/
39. y= f (x)=ln x
/
f (x)=1/x , so f (1)=0 and f (1)=1 . The linear approximation of f at 1 is
/
f (1)+ f (1)(x 1)=0+1(x 1)=x 1 . Now f (1.05)=ln 1.05 1.05 1=0.05 , so the approximation is
reasonable.
2
/
40. (a) f (x)=(x 1) /
f (x)=2(x 1) , so f (0)=1 and f (0)= 2 .
/
Thus, f (x) L (x)= f (0)+ f (0)(x 0)=1 2x .
2x
g(x)=e
f
/
2x
g (x)= 2e
/
, so g(0)=1 and g (0)= 2 .
/
Thus, g(x) L (x)=g(0)+g (0)(x 0)=1 2x .
g
h(x)=1+ln (1 2x)
/
h (x)=
2
/
, so h(0)=1 and h (0)= 2 .
1 2x
/
Thus, h(x) L (x)=h(0)+h (0)(x 0)=1 2x .
h
Notice that L =L =L . This happens because f , g , and h have the same function values and the
f
g
h
same derivative values at a=0 .
(b)
The linear approximation appears to be the best for the function f since it is closer to f for a larger
domain than it is to g and h . The approximation looks worst for h since h moves away from L faster
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
than f and g do.
3
41. (a) If x is the edge length, then V =x 2
2
dV =3x dx . When x=30 and dx=0.1 , dV =3(30) (0.1)=270
3
, so the maximum possible error in computing the volume of the cube is about 270 cm . The relative
error is calculated by dividing the change in V , V , by V . We approximate V with dV .
2
V
Relative error =
V
dV 3x dx
dx
0.1
=
=3 =3
=0.01 .
3
V
x
30
x
Percentage error = relative error 100%=0.01 100%=1% .
2
(b) S=6x dS=12xdx . When x=30 and dx=0.1 , dS=12(30)(0.1)=36 , so the maximum possible
2
error in computing the surface area of the cube is about 36 cm .
S dS 12xdx
dx
0.1
=
=2 =2
=0.006 .
Relative error =
2
S
S
x
30
6x
Percentage error = relative error 100%=0.006 100%=0.6% .
2
42. (a) A= r dA=2 r dr . When r=24 and dr=0.2 , dA=2 (24)(0.2)=9.6
, so the maximum
2
possible error in the calculated area of the disk is about 9.6 30 cm .
A dA 2 rdr 2dr 2(0.2) 0.2 1
(b) Relative error =
=
=
=
=
=
=0.016 .
2
A
A
r
24
12 60
r
Percentage error = relative error 100%=0.016 100%=1.6% .
2
43. (a) For a sphere of radius r , the circumference is C=2 r and the surface area is S=4 r , so
2
84
2
2
r=C/(2 ) S=4 (C/2 ) =C / dS=(2/ )C dC . When C=84 and dC=0.5 , dS= (84)(0.5)=
,
84
dS 84/
1
2
so the maximum error is about
27 cm . Relative error = 2 =
0.012
S
84
84 /
4 3 4
(b) V = r = 3
3
1
dV =
2
2
2
(84) (0.5)=
C
2
1764
2
3
C
=
6
3
2
1
dV =
2
2
2
C dC . When C=84 and dC=0.5 ,
, so the maximum error is about
dV
1764/
=
approximately
3
V
(84) /(6
1764
2
2
)
=
2
3
179 cm . The relative error is
1
0.018 .
56
44. For a hemispherical dome,
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
2 3
V= r 3
1
(50)=25 m and
2
5
5
2
dr=0.05 cm =0.0005 m, dV =2 (25) (0.0005)=
, so the amount of paint needed is about
2
8
8
2
dV =2 r dr . When r=
3
m .
2
V dV =2 rhdr=2 rh r
45. (a) V = r h
(b) The error is
V dV =[ (r+ r)2h r 2h] 2 rh r= r 2h+2 rh r+ ( r)2h r 2h 2 rh r
2
= ( r) h
3
dF 4kR dR
dR
46. F=kR dF=4kR dR
=
=4
. Thus, the relative change in F is about 4
4
F
R
kR
times the relative change in R . So a 5% increase in the radius corresponds to a 20% increase in blood
flow.
4
3
dc
dx=0dx=0
dx
d
du
(b) d(cu)= (cu)dx=c
dx=cdu
dx
dx
d
du dv
(c) d(u+v)= (u+v)dx=
+
dx
dx dx
d
dv
du
(d) d(uv)= (uv)dx= u
+v
dx
dx
dx
du
v
u
u
d
u
dx
(e) d
=
dx=
2
v
dx
v
v
d n
n
n 1
(f) d(x )= (x )dx=nx dx
dx
47. (a) dc=
48. (a) f (x)=sin x
f (x)
(b)
/
du
dv
dx+
dx=du+dv
dx
dx
dv
du
dx=u
dx+v
dx=udv+vdu
dx
dx
dv
du
dv
v
dx u
dx
dx
dx
dx
vdu udv
dx=
=
2
2
v
v
dx=
/
f (x)=cos x , so f (0)=0 and f (0)=1 . Thus,
/
f (0)+ f (0)(x 0)=0+1(x 0)=x .
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
We want to know the values of x for which y=x approximates y=sin x with less than a 2% difference;
that is, the values of x for which
x sin x
x sin x
<0.02 0.02<
<0.02 sin x
sin x
{
0.02sin x<x sin x<0.02sin x
0.02sin x>x sin x>0.02sin x
{
if sin x>0
if sin x<0
0.98sin x<x<1.02sin x
1.02sin x<x<0.98sin x
if sin x>0
if sin x<0
In the first figure, we see that the graphs are very close to each other near x=0 . Changing the viewing
rectangle and using an intersect feature (see the second figure) we find that y=x intersects y=1.02sin x
at x 0.344 . By symmetry, they also intersect at x 0.344 (see the third figure.). Converting 0.344
180
radians to degrees, we get 0.344
/
19.7 20 , which verifies the statement.
/
49. (a) The graph shows that f (1)=2 , so L(x)= f (1)+ f (1)(x 1)=5+2(x 1)=2x+3 .
f (0.9) L(0.9)=4.8 and f (1.1) L(1.1)=5.2 .
/
(b) From the graph, we see that f (x) is positive and decreasing. This means that the slopes of the
tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the
curve. Thus, the estimates in part (a) are too large.
10
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials
/
2
50. (a) g (x)= x +5 /
/
g (2)= 9 =3 . g(1.95) g(2)+g (2)(1.95 2)= 4+3( 0.05)= 4.15 .
/
g(2.05) g(2)+g (2)(2.05 2)= 4+3(0.05)= 3.85 .
/
2
/
(b) The formula g (x)= x +5 shows that g (x) is positive and increasing. This means that the slopes
of the tangent lines are positive and the tangents are getting steeper. So the tangent lines lie below the
graph of g . Hence, the estimates in part (a) are too small.
11
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
1. Product Rule:
2
3
y=(x +1)(x +1)
/
2
2
3
4
2
4
4
2
y =(x +1)(3x )+(x +1)(2x)=3x +3x +2x +2x=5x +3x +2x .
2
3
5
3
2
/
2. Quotient Rule: F(x)=
1/2
/
F (x) =
x
x 3x x
x
=
x 3x
2
3/2
1/2
x
1 x
2
9 1/2
x
2
( x 3x3/2)
( x1/2) 2
1
4
y =5x +3x +2x (equivalent).
Multiplying first: y=(x +1)(x +1)=x +x +x +1
1/2
9
1 1/2 3
1 1/2
x
x + x
x 3x
=
2
2
2
2
1 1/2
=
= x 3
x
x
2
x 3x x
1 1/2
1/2
/
= x 3x=x 3x F (x)= x 3 (equivalent).
Simplifying first: F(x)=
2
x
For this problem, simplifying first seems to be the better method.
1/2
x f (x)=x
d x x d 2 2 x x
x
(e )+e
(x )=x e +e (2x)=xe (x+2) .
dx
dx
1/2 x
/
/
2 x
3. By the Product Rule, f (x)=x e x
2
4. By the Product Rule, g(x)= x e =x e 5. By the Quotient Rule, y=
2
x
e
2
x
/
y =
d x x d 2
(e ) e
(x )
dx
dx
22
x
e
6. By the Quotient Rule, y=
1+x
7. g(x)=
8. f (t)=
3x 1
2x+1
QR
2t
4+t
2
1 x
2
x
/
2 x
=
y =
x
x
2
x
=
=
x
e +xe e
2
1 x
2
x
x
(1+x)e e (1)
1/2
x (e ) e (2x)
(x )
x
QR
1/2 x
g (x)=x (e )+e
x
=
4
1/2 x
e (2x+1) .
x
=
xe (x 2)
x
xe
4
x
=
e (x 2)
x
3
x
2
.
(1+x)
(x+1)
(x+1)
/
(2x+1)(3) (3x 1)(2) 6x+3 6x+2
5
g (x)=
=
=
2
2
2
(2x+1)
(2x+1)
(2x+1)
/
f (t)=
2
(4+t )(2) (2t)(2t)
22
(4+t )
2
=
8+2t 4t
22
(4+t )
2
=
8 2t
2
22
(4+t )
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
3
4
9. V (x)=(2x +3)(x 2x) /
3
PR
3
4
2
6
3
6
2
3
4
3
6
3
V (x)=(2x +3)(4x 2)+(x 2x)(6x )=(8x +8x 6)+(6x 12x )=14x 4x 6
2
3
5
2
10. Y (u)=(u +u )(u 2u )
2
/
3
PR
4
5
Y (u) =(u +u )(5u 4u)+(u 2u )( 2u 3u )
1
2
2
1
2
2
2
=(5u 4u +5u 4u )+( 2u 3u+4u +6u )=3u +2u+2u
1
11. F(y)=
2
3
4
y
2
(
3
4
(y+5y )= y 3y
) ( y+5y3) 2
PR
y
( 2 4) ( 1+15y2) + ( y+5y3) ( 2y 3+12y 5)
2
4
2
2
4
2
= ( y +15 3y 45y ) +( 2y +12y 10+60y )
/
F (y) = y 3y
2
4
2
4
=5+14y +9y or 5+14/y +9/y
t
12. R(t)=(t+e )(3 t )=
1 1/2
t
t
/
R (t) =(t+e )( 2 t )+(3 t )(1+e )
1 1/2 1 1/2 t
t
= t t e +(3+3e t 2
2
13. y=
t
2
y =
2
3t 2t+1
2
(3t 2t+1)(2t) t (6t 2)
2
2
2
2
2
2
=
2
4
t 2
QR
/
2
2
(3t 2t+1)
4
y =
2
2t(1 t)
=
(3t 2t+1)
14. y=
t
(3t 2t+1)
2t (3t 2t+1 3t +t)
3
t
t e e /(2 t )
2t[3t 2t+1 t(3t 1)]
(3t 2t+1)
t +t
3
t
2
2
=
t
QR
2
/
t
t e )=3+3e 2
3
3
(t 2)(3t +1) (t +t)(4t )
4
2
6
=
(t 2)
6
=
4
2
t 3t 6t 2
4
2
(t 2)
2
6
4
4
2
(t 2)
6
=
4
(3t +t 6t 2) (4t +4t )
4
2
4
2
t +3t +6t +2
(t 2)
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
2
r
/
2
r
r
r
2
r
2
15. y=(r 2r)e = y =(r 2r)(e )+e (2r 2)=e (r 2r+2r 2)=e (r 2)
16. y=
1
s
s+ke
s
s
(s+ke )(0) (1)(1+ke )
/
=y =
s2
1+ke
=
s
s2
(s+ke )
(s+ke )
3
v 2v v 2
2
1/2
17. y=
=v 2 v =v 2v v
1
2
/
y =2v 2
We can change the form of the answer as follows: 2v v
3/2
w
5/2
18. z=w (w+ce )=w +cw
19. y=
1
4
2
4
/
y =
e x 1
x +1
2
4
23. y=
y =
1/2
2v v 1 2v 1
1
=
=
v
v
v
3/2
3/2
=2v
w
w
3 1/2
5 3/2 1 1/2 w
w
= w + cw e (2w+3)
2
2
2
=
2x(2x +1)
4
2
2
(x +x +1)
1
2 x
=
2
1
1
1
1
+
+
2 2 x 2 2 x
2
( x +1)
/
( x +1)
1
2
x ( x +1)
2
2
(cx+d)(a) (ax+b)(c)
2
(cx+d)
/
=
x
2cx
2c/x
(x+c/x)(1) x(1 c/x ) x+c/x x+c/x
=
=
f (x)=
=
2
2
2
2
2
2
2
c 2
x (x +c)
(x +c)
x +c
x+
x
2
x
x
/
f (x)=
y =
.
2
( x 1)
2 x
/
(x+1)(2) (2x)(1)
2
(x+1)
1
1
1
is y 1= (x 1) , or y= x+ .
2
2
2
24. y=
2
1
( x +1)
ax+b
cx+d
2x
x+1
2
1/2
=2v v
w e +e 3
(x +x +1)
x
21. f (x)=
x+c/x
22. f (x)=
z =
(x +x +1)(0) 1(4x +2x)
x +x +1
20. y=
5 3/2
w +c
2
/
3/2 w
1/2
v
=
=
acx+ad acx bc
2
(cx+d)
2
/
2
(x+1)
. At (1,1) , y =
=
ad bc
2
(cx+d)
1
, and an equation of the tangent line
2
x
x+1
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
1
(x+1)
2 x
/
y =
x (1)
=
2
(x+1) (2x)
1 x
=
2
/
. At (4,0.4) , y =
2
(x+1)
2 x (x+1) 2 x (x+1)
equation of the tangent line is y 0.4= 0.03(x 4) , or y= 0.03x+0.52 .
x
/
x
x
x
/
3
= 0.03 , and an
100
0
25. y=2xe y =2(x e +e 1)=2e (x+1) . At (0, 0) , y =2e (0+1)=2 1 1=2 , and an equation of the
tangent line is y 0=2(x 0) , or y=2x .
x
e
26. y= x
x
x
x e e 1
/
y =
2
x
=
e (x 1)
2
x
/
. At (1, e) , y =0 , and an equation of the tangent line is
x
y e=0(x 1) , or y=e .
1
27. (a) y= f (x)=
2
/
2
f (x)=
(1+x )(0) 1(2x)
22
1+x
point
=
2x
22
. So the slope of the tangent line at the
(1+x )
(1+x )
1
1 1
1
/
2
is f ( 1)= = and its equation is y
= (x+1) or y= x+1 .
2 2
2 2
2
2
1
1,
2
(b)
x
28. (a) y= f (x)=
2
1+x
2
/
f (x)=
(1+x )1 x(2x)
22
2
=
(1+x )
/
point (3,0.3) is f (3)=
1 x
22
. So the slope of the tangent line at the
(1+x )
8
and its equation is y 0.3= 0.08(x 3) or y= 0.08x+0.54 .
100
(b)
29. (a) f (x)=
e
x
x
3
/
f (x)=
3
x
x
2
x (e ) e (3x )
32
(x )
2 x
=
x e (x 3)
x
6
x
=
e (x 3)
x
4
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
(b)
/
f =0 when f has a horizontal tangent line, f
when f is increasing.
30. f (x)=
x
2
x 1
2
(x 1)1 x(2x)
/
f (x)=
2
2
/
is negative when f is decreasing, and f
2
2
(x 1)
is positive
2
x 1
=
/
=
2
(x 1)
x +1
2
2
Notice that the slopes of all tangents to
(x 1)
/
f are negative and f (x)<0 always.
/
/
31. We are given that f (5)=1 , f (5)=6 , g(5)= 3 , and g (5)=2 .
/
/
/
(a) ( fg) (5)= f (5)g (5)+g(5) f (5)=(1)(2)+( 3)(6)=2 18= 16
(b)
(c)
f
g
/
g
f
/
/
(5)=
/
g(5) f (5) f (5)g (5)
2
=
( 3)(6) (1)(2)
2
[g(5)]
( 3)
/
(5)=
=
/
f (5)g (5) g(5) f (5)
2
=
(1)(2) ( 3)(6)
2
[ f (5)]
20
9
=20
(1)
/
/
32. We are given that f (3)=4 , g(3)=2 , f (3)= 6 , and g (3)=5 .
/
/
/
(a) ( f +g ) (3)= f (3)+g (3)= 6+5= 1
/
/
/
(b) ( fg ) (3)= f (3)g (3)+g(3) f (3)=(4)(5)+(2)( 6)=20 12=8
(c)
f
g
/
/
(3)=
/
g(3) f (3) f (3)g (3)
2
[g(3)]
=
(2)( 6) (4)(5)
2
(2)
=
32
= 8
4
(d)
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
/
f
f g
/
(3) =
/
2
[ f (3) g(3)]
(4 2)( 6) 4( 6 5) 12+44
=
=
=8
2
2
(4 2)
2
x
/
33. f (x)=e g(x)
/
/
[ f (3) g(3)] f (3) f (3)[ f (3) g (3)]
0
x
/
x
x
/
f (x)=e g (x)+g(x)e =e g (x)+g(x) .
/
f (0)=e g (0)+g(0) =1(5+2)=7
d
34.
dx
/
h(x)
x
xh (x) h(x) 1
=
2
x
d
dx
h(x)
x
/
=
2h (2) h(2)
2
x=2
2
=
2( 3) (4) 10
=
= 2.5
4
4
35. (a) From the graphs of f and g , we obtain the following values: f (1)=2 since the point (1,2) is on
/
the graph of f ; g(1)=1 since the point (1,1) is on the graph of g ; f (1)=2 since the slope of the line
4 0
/
segment between (0,0) and (2,4) is
=2 ; g (1)= 1 since the slope of the line segment between
2 0
0 4
/
/
/
( 2,4) and (2,0) is
= 1 . Now u(x)= f (x)g(x) , so u (1)= f (1)g (1)+g(1) f (1)=2 ( 1)+1 2=0 .
2 ( 2)
8
1
2
/
/
2 3
3
2
/
g(5) f (5) f (5)g (5)
3
3
(b) v(x)= f (x)/g(x) , so v (5)=
=
=
=
2
2
4
3
[g(5)]
2
/
/
/
2
3
+2 0= .
4
2
1
2
1 5 4
3
36. (a) P(x)=F(x)G(x) , so P (2)=F(2)G (2)+G(2)F (2)=3
/
/
(b) Q(x)=F(x)/G(x) , so Q (7)=
/
G(7)F (7) F(7)G (7)
2
[G(7)]
/
37. (a) y=xg(x) x
(b) y=
g(x)
g(x)
(c) y=
x
/
=
2
1
=
1 10 43
+
=
4 3 12
/
y =xg (x)+g(x) 1=xg (x)+g(x)
/
y =
/
y =
/
g(x) 1 xg (x)
2
/
=
g(x) xg (x)
2
[g(x)]
[g(x)]
xg (x) g(x) 1
xg (x) g(x)
/
2
(x)
/
=
2
x
38. (a)
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
2
/
y=x f (x)
(b) y=
2
/
/
x f (x) f (x)(2x)
y =x f (x)+ f (x)(2x)
f (x)
2
2
y =
/
22
x
2
x
2
/
y =
3
/
f (x)(2x) x f (x)
2
[ f (x)]
1+xf (x)
x
/
x[xf (x)+ f (x)] [1+xf (x)]
/
=
(x )
x
(c) y=
f (x)
(d) y=
/
xf (x) 2 f (x)
y =
1
2 x
2
( x)
x
3/2
/
1/2
f (x)+x
1 x
2
f (x)
=
1/2
x
1 1/2
x f (x)
1/2
2 /
2
2x
xf (x)+2x f (x) 1
=
1/2
3/2
2x
2x
39. If P(t) denotes the population at time t and A(t) the average annual income, then T (t)=P(t)A(t) is
/
/
/
the total personal income. The rate at which T (t) is rising is given by T (t)=P(t)A (t)+A(t)P (t)
/
/
/
T (1999) =P(1999)A (1999)+A(1999)P (1999)=( 961,400 )( $1400 / yr )+( $30,593 )( 9200 / yr )
=$ 1,345,960,000 / yr +$ 281,455,600 / yr =$ 1,627,415,600 / yr
So the total personal income was rising by about $ 1.627 billion per year in 1999.
/
The term P(t)A (t) $ 1.346 billion represents the portion of the rate of change of total income due to
/
the existing population’s increasing income. The term A(t)P (t) $ 281 million represents the portion
of the rate of change of total income due to increasing population.
40. (a) f (20)=10 , 000 means that when the price of the fabric is $20/ yard, 10 , 000 yards will be
sold.
/
f (20)= 350 means that as the price of the fabric increases past $20/ yard, the amount of fabric
which will be sold is decreasing at a rate of 350 yards per (dollar per yard).
/
/
/
/
(b) R( p)= pf ( p) R ( p)= pf ( p)+ f ( p) 1 R (20)=20 f (20)+ f (20) 1=20( 350)+10,000=3000.
This means that as the price of the fabric increases past $20/ yard, the total revenue is increasing at
$3000/($/yard). Note that the Product Rule indicates that we will lose $7000/($/yard) due to selling
less fabric, but that that loss is more than made up for by the additional revenue due to the increase in
price.
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
x
/
(x+1)(1) x(1)
1
, then f (x)=
=
. When x=a , the equation of the tangent
2
2
x+1
(x+1)
(x+1)
a
a
1
1
line is y
=
(x a) . This line passes through (1,2) when 2
=
(1 a)
2
2
a+1
a+1
(a+1)
(a+1)
41. If y= f (x)=
2
2
2(a+1) a(a+1)=1 a 2
2a +4a+2 a a 1+a=0
2
a +4a+1=0 .
The quadratic formula gives the roots of this equation as a=
2
4 4(1)(1) 4 12
=
= 2
2(1)
2
4
3 ,
so there are two such tangent lines. Since
f ( 2
2
3) =
2
3
3 +1
=
2
3
1
3
1
3
1
3
2 2 3 3 3 1 3 1 3
=
=
,
1 3
2
2
1 3
the lines touch the curve at A 2+ 3,
( 0.27, 0.37) and
2
1+ 3
B 2 3,
( 3.73,1.37) .
2
x 1
/ (x+1)(1) (x 1)(1)
2
42. y=
y =
=
. If the tangent intersects the curve when x=a ,
2
2
x+1
(x+1)
(x+1)
1
2
then its slope is 2/(a+1) . But if the tangent is parallel to x 2y=2 , that is, y= x 1 , then its slope is
2
1
1
2
2
. Thus,
= (a+1) =4 a+1= 2 a=1 or 3 . When a=1 , y=0 and the equation of the
2 2
2
(a+1)
1
1
1
tangent is y 0= (x 1) or y= x
.
2
2
2
1
1
7
When a= 3 , y=2 and the equation of the tangent is y 2= (x+3) or y= x+ .
2
2
2
=
/
/
/
/
/
/
/
/
/
43. (a) ( fgh) =[( fg)h] =( fg) h+( fg)h =( f g+ fg )h+( fg)h = f gh+ fg h+ fgh
/
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules
(b) Putting f =g=h in part (a) , we have
d
3
/
/
/
/
/
2 /
[ f (x)] = ( fff ) = f ff + ff f + fff =3 fff =3[ f (x)] f (x).
dx
(c)
d 3x d x 3
x2 x
2x x
3x
(e )= (e ) =3(e ) e =3e e =3e
dx
dx
44. (a)
d
dx
d
d
(1) 1
[g(x)]
dx
dx
g(x)
1
g(x)
=
[Quotient Rule]
2
[g(x)]
/
=
g(x) 0 1 g (x)
2
/
=
[g(x)]
(b) y=
1
4
2
2
y =
2
2
(x +x +1)
1
x
n
=
2
[g(x)]
2
4x +2x
4
/
g (x)
[g(x)]
3
/
x +x +1
d n d
(x )=
(c)
dx
dx
0 g (x)
or
2x(2x +1)
4
2
2
(x +x +1)
n /
=
(x )
n2
(x )
[by the Reciprocal Rule] =
nx
n 1
2n
= nx
n 1 2n
= nx
n 1
x
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
2
/
1. (a) s= f (t)=t 10t+12 v(t)= f (t)=2t 10
(b) v(3)=2(3) 10= 4 ft / s
(c) The particle is at rest when v(t)=0 2t 10=0 t=5 s.
(d) The particle is moving in the positive direction when v(t)>0 2t 10>0 2t>10 t>5 .
(e) Since the particle is moving in the positive direction and in the negative direction, we need to
calculate the distance traveled in the intervals [0,5] and [5,8] separately. | f (5) f (0)|=| 13 12|=25 ft
and | f (8) f (5)|=|{ 4 ( 13)|=9 ft. The total distance traveled during the first 8 s is 25+9=34 ft.
(f)
3
2
2. (a) s= f (t)=t 9t +15t+10
/
2
v(t)= f (t)=3t 18t+15=3(t 1)(t 5)
(b) v(3)=3(2)( 2)= 12 ft / s
(c) v(t)=0 t=1 s or 5 s
(d) v(t)>0 0 t<1 or t>5
(e) | f (1) f (0)|=|17 10|=7 , | f (5) f (1)|=| 15 17|=32 , and | f (8) f (5)|=|66 ( 15)|=81 . Total distance
=7+32+81=120 ft.
(f)
3
2
2
/
3. (a) s= f (t)=t 12t +36t v(t)= f (t)=3t 24t+36
(b) v(3)=27 72+36= 9 ft / s
2
(c) The particle is at rest when v(t)=0 . 3t 24t+36=0 3(t 2)(t 6)=0 t=2 s or 6 s.
(d) The particle is moving in the positive direction when v(t)>0 . 3(t 2)(t 6)>0 0 t<2 or t>6 .
(e) Since the particle is moving in the positive direction and in the negative direction, we need to
calculate the distance traveled in the intervals (0,2) , (2,6) , and [6,8] separately.
| f (2) f (0)|=|32 0|=32 .
| f (6) f (2)|=|0 32|=32 .
| f (8) f (6)|=|32 0|=32 .
The total distance is 32+32+32=96 ft.
(f)
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
4
/
4. (a) s= f (t)=t 4t+1
3
v(t)= f (t)=4t 4
3
(b) v(3)=4(3) 4=104 ft / s
3
2
(c) It is at rest when v(t)=4(t 1)=4(t 1)(t +t+1)=0 t=1 s.
3
(d) It moves in the positive direction when 4(t 1)>0 t>1 .
(e) Distance in positive direction =| f (8) f (1)|=|4065 ( 2)|=4067 ft
Distance in negative direction =| f (1) f (0)|=| 2 1|=3 ft
Total distance traveled =4067+3=4070 ft
(f)
t
5. (a) s=
/
2
2
v(t)=s (t)=
(t +1)(1) t(2t)
2
t +1
(t +1)
2
(b) v(3)=
1 (3)
2
2
2
(3 +1)
=
1 9
2
10
=
=
1 t
2
2
2
(t +1)
8
2
=
ft / s
100
25
2
(c) It is at rest when v=0 1 t =0 t=1 s [ t 1 since t 0 ].
2
2
(d) It moves in the positive direction when v>0 1 t >0 t <1 0 t<1 .
1
1
(e) Distance in positive direction =|s(1) s(0)|=
0 = ft
2
2
8 1
49
Distance in negative direction =|s(8) s(1)|=
=
ft
65 2
130
1 49 57
Total distance traveled = +
=
ft
2 130 65
(f)
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
2
5/2
3/2
1/2
6. (a) s= t (3t 35t+90)=3t 35t +90t 15 3/2 105 1/2
15
/
1/2 15 1/2 2
v(t)=s (t)=
t t +45t =
t (t 7t+6)=
(t 1)(t 6)
2
2
2
2 t
15
(b) v(3)=
(2)( 3)= 15 3 ft / s
2 3
(c) It is at rest when v=0 t=1 s or 6 s.
(d) It moves in the positive direction when v>0 (t 1)(t 6)>0 0 t<1 or t>6.
(e)
Distance in positive direction =|s(1) s(0)|+|s(8) s(6)|=|58 0|+|4 2 ( 12 6 )
=58+4 2 +12 6 93.05 ft
Distance in negative direction =|s(6) s(1)|=| 12 6 58|=58+12 6 87.39 ft
Total distance traveled =58+4 2 +12 6 +58+12 6 =116+4 2 +24 6 180.44 ft
(f)
3
2
/
2
2
7. s(t)=t 4.5t 7t v(t)=s (t)=3t 9t 7=5 3t 9t 12=0 , the particle reaches a velocity of 5 m / s at t=4 s.
t=4 or 1 . Since t 0
ds
=5+6t , so v(2)=5+6(2)=17 m / s.
dt
5+6t=35 6t=30 t=5 s.
2
8. (a) s=5t+3t (b) v(t)=35
3(t 4)(t+1)=0
v(t)=
dh
=10 1.66t , so v(3)=10 1.66(3)=5.02 m / s.
dt
10
17
2
2
(b) h=25 10t 0.83t =25 0.83t 10t+25=0 t=
3.54 or 8.51 .
1.66
The value t =(10 17 )/1.66 corresponds to the time it takes for the stone to rise 25 m and
2
9. (a) h=10t 0.83t v(t)=
1
t =(10+ 17 )/1.66 corresponds to the time when the stone is 25 m high on the way down. Thus,
2
v(t )=10 1.66[(10
1
17 )/1.66]= 17 4.12 m / s.
/
10. (a) At maximum height the velocity of the ball is 0 ft / s. v(t)=s (t)=80 32t=0
32t=80 t=
5
.
2
So the maximum height is
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
s
5
2
=80
5
2
16
2
5
2
2
=200 100=100 ft.
2
2
(b) s(t)=80t 16t =96 16t 80t+96=0 16(t 5t+6)=0 16(t 3)(t 2)=0 .
So the ball has a height of 96 ft on the way up at t=2 and on the way down at t=3 . At these times the
velocities are v(2)=80 32(2)=16 ft / s and v(3)=80 32(3)= 16 ft / s, respectively.
2
/
/
2
11. (a) A(x)=x A (x)=2x . A (15)=30 mm / mm is the rate at which the area is increasing with
respect to the side length as x reaches 15 mm.
1
1
(4x)= P(x) . The figure suggests that if x is small,
2
2
then the change in the area of the square is approximately half of its perimeter ( 2 of the 4 sides)
/
(b) The perimeter is P(x)=4x , so A (x)=2x=
2
times x . From the figure, A=2x( x)+( x) . If x is small, then A 2x( x) and so
A/ x 2x .
dV
dV
2
=3x .
dx
dx
increasing as x increases past 3 mm.
3
12. (a) V (x)=x 2
3
=3(3) =27 mm / mm is the rate at which the volume is
x=3
1
2 1
(6x )= S(x) . The figure suggests that if x is
2
2
small, then the change in the volume of the cube is approximately half of its surface area (the area of
2
/
2
(b) The surface area is S(x)=6x , so V (x)=3x =
2
2
3
3 of the 6 faces) times x . From the figure, V =3x ( x)+3x( x) +( x) . If x is small, then
2
2
V 3x ( x) and so V / x 3x .
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
13. (a)
(i)
A(3) A(2) 9 4
=
=5
3 2
1
(ii) A(2.5) A(2) 6.25 4
=
=4.5
2.5 2
0.5
(iii) A(2.1) A(2) 4.41 4
=
=4.1
2.1 2
0.1
2
/
(b) A(r)= r /
A (r)=2 r , so A (2)=4
.
/
(c) The circumference is C(r)=2 r=A (r) . The figure suggests that if r is small, then the change in
the area of the circle (a ring around the outside) is approximately equal to its circumference times r
. Straightening out this ring gives us a shape that is approximately rectangular with length 2 r and
2
2
2
width r , so A 2 r( r) . Algebraically, A=A(r+ r) A(r)= (r+ r) r =2 r( r)+ ( r) .
So we see that if r is small, then A 2 r( r) and therefore, A/ r 2 r .
/
2
14. (a) A (1)=7200
cm / s
/
2
/
2
(b) A (3)=21 , 600 cm / s
(c) A (5)=36 , 000 cm / s
/
15. (a) S (1)=8
/
(b) S (2)=16
/
(c) S (3)=24
2
ft / ft
2
ft / ft
2
ft / ft
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
16. (a)
(a)
4
4
(512) (125)
V (8) V (5) 3
3
=
=172 8 5
3
(b)
(c)
(d)
(e)
(f)
4
4
(512) (125)
V (8) V (5) 3
3
=
=172 8 5
3
3
m /
3
m /
4
4
(216) (125)
V (6) V (5) 3
3
=
=121.3 6 5
1
4
4
(216) (125)
V (6) V (5) 3
3
=
=121.3 6 5
1
3
/
3
m /
4
3 4
3
(5.1) (5)
V (5.1) V (5) 3
3
=
=102.013 5.1 5
0.1
2
m
m /
4
3 4
3
(5.1) (5)
V (5.1) V (5) 3
3
=
=102.013 5.1 5
0.1
/
m
m
m
3
m /
3
m /
m
m
3
(b) V (r)=4 r , so V (5)=100 m / m.
4 3
/
2
(c) V (r)= r V (r)=4 r =S(r) . By analogy with Exercise 13(c) , we can say that the change in
3
the volume of the spherical shell, V , is approximately equal to its thickness, r , times the surface
2
2
area of the inner sphere. Thus, V 4 r ( r) and so V / r 4 r .
17. (a) (1)=6 kg / m
(b) (2)=12 kg / m
(c) (3)=18 kg / m
5
= 218.75 gal / min
40
10
/
(b) V (10)= 250 1
= 187.5 gal / min
40
20
/
(c) V (20)= 250 1
= 125 gal / min
40
(d)
/
18. (a) V (5)= 250
1
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
/
V (40)= 250
1
/
40
40
=0 gal / min
2
19. (a) Q (0.5)=3(0.5) 4(0.5)+6=4.75 A
/
2
(b) Q (1)=3(1) 4(1)+6=5 A
2
dF
3
2GmM
= 2(GmM)r =
, which is the rate of change of the
2
3
dr
r
r
force with respect to the distance between the bodies. The minus sign indicates that as the distance r
between the bodies increases, the magnitude of the force F exerted by the body of mass m on the
body of mass M is decreasing.
/
/
3
2GmM
GmM=20 , 000 .
(b) Given F (20 , 000)= 2 , find F (10 , 000) . 2=
3
20,000
20. (a) F=
/
GmM
F (10,000)=
=(GmM)r (
3
2 20,000
3
) = 2 23= 16 N / km
10,000
21. (a) To find the rate of change of volume with respect to pressure, we first solve for V in terms of
P.
C
dV
C
PV =C V = =
.
2
P
dP
P
(b) From the formula for dV /dP in part (a), we see that as P increases, the absolute value of dV /dP
decreases. Thus, the volume is decreasing more rapidly at the beginning.
1 dV
1
C
C
1
C
(c) =
=
=
=
=
2
V dP
V
(PV )P CP P
P
22. (a)
.
C(6) C(2)
0.0295 0.0570
=
6 2
4
(i)
.
= 0.006875 ( moles/L ) / min
C(4) C(2)
0.0408 0.0570
=
4 2
2
= 0.008 ( moles/L ) / min
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
.
C(2) C(0)
0.0570 0.0800
=
2 0
2
(iii)
= 0.0115 ( moles/L ) / min
(b) Slope =
C
0.077
0.01 ( moles/L ) / min
t
7.8
1860 1750 110
2070 1860 210
=
=11 , m =
=
=21 ,
1 1920 1910
2 1930 1920
10
10
(m +m )2=(11+21)/2=16 million / year
23. (a) 1920: m =
1
2
4450 3710 740
5280 4450 830
=
=74 , m =
=
=83 ,
1 1980 1970
2 1990 1980
10
10
(m +m )2=(74+83)/2=78.5 million / year
1980: m =
1
2
3
2
3
2
(b) P(t)=at +bt +ct+d (in millions of people), where a 0.0012937063 , b 7.061421911 ,
c 12,822.97902 , and d 7,743,770.396 .
/
(c) P(t)=at +bt +ct+d
(d)
2
P (t)=3at +2bt+c (in millions of people per year)
/
2
P (1920) =3(0.0012937063)(1920) +2( 7.061421911)(1920)+12,822.97902
14.48 million / year
/
P (1980) 75.29 million / year (smaller, but close)
/
(e) P (1985) 81.62 million / year, so the rate of growth in 1985 was about 81.62 million / year.
4
3
6
2
24. (a) A(t)=at +bt +ct +dt+e , where a= 5.8275058275396 10 , b=0.0460458430461 ,
c= 136.43277039706 , d=179 , 661.02676871 , and e= 88 , 717 , 597.060767 .
4
3
2
(b) A(t)=at +bt +ct +dt+e
(c)
/
3
2
A (t)=4at +3bt +2ct+d
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
/
A (1990) 0.0833 years of age per year
(d)
2
a kt
25. (a) C =
akt+1
rate of reaction
2
2
2
2
d C
(akt+1)(a k) (a kt)(ak) a k(akt+1 akt)
ak
=
=
=
=
2
2
2
dt
(akt+1)
(akt+1)
(akt+1)
2
2
2
a kt
a kt+a a kt
a
(b) If x= C , then a x=a
=
=
.
akt+1
akt+1
akt+1
2
So k(a x) =k
a
akt+1
2
=
2
ak
2
(akt+1)
=
d C
dt
=
dx
.
dt
2
26. (a) After an hour the population is n(1)=3 500 ; after two hours it is n(2)=3(3 500)=3 500 ; after
2
3
4
three hours, n(3)=3(3 500)=3 500 ; after four hours, n(4)=3 500 . From this pattern, we see that the
t
t
population after t hours is n(t)=3 500=500 3 .
d x
t
x
(b) From (5) in Section 3.1, we have
(3 ) (1.10)3 . Thus, for n(t)=500 3 ,
dx
dn
d t
dn
t
6
=500 (3 ) 500(1.10)3 500(1.10)3 400 , 950 bacteria / hour.
dt
dt
dt t=6
27. (a) Using v=
P
2 2
(R r with R=0.01 , l=3 , P=3000 , and =0.027 , we have v as a function of r :
4 l
3000
2 2
(0.01 r ) . v(0)=0.925 cm / s, v(0.005)=0.694 cm / s, v(0.01)=0 .
4(0.027)3
P
P
Pr
2 2
/
(R r ) v (r)=
( 2r)=
. When l=3 , P=3000 , and =0.027 , we have
(b) v(r)=
4 l
4 l
2 l
v(r)=
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
3000r
/
/
/
. v (0)=0 , v (0.005)= 92.592(cm/s)/ cm, and v (0.01)= 185.185(cm/s)/ cm.
2(0.027)3
(c) The velocity is greatest where r=0 (at the center) and the velocity is changing most where
r=R=0.01 cm (at the edge).
/
v (r)=
28. (a)
(a)
T
=
1
2
T
L 1
2L
T
=
1
2
T
L f=
1
2L
T
=
1
2L T
f=
1
2L
T
=
1
2L T
f=
1
2L
T
=
T
2L
1
2L
T
=
T
2L
(b)
f=
(c)
(d)
(e)
(f)
f=
(b)
(i)
1
1
2L
f=
1
2
df
=
dL
1
2
T
L =
df
=
dL
1
2
T
L =
2
df 1
=
dT 2
1
2L T
1/2
df 1
=
dT 2
1
2L T
1/2
df
1
=
d
2
T
2L
df
1
=
d
2
T
2L
1/2
1
4L T
1/2
1
4L T
=
=
3/2
4L
3/2
4L
higher note
(ii)
df
>0 and T is increasing dT
f is increasing higher note
(iii)
df
<0 and is increasing d
f is decreasing lower note
3
/
3/2
T
=
f is increasing 2
T
=
df
<0 and L is decreasing dL
29. (a) C(x)=2000+3x+0.01x +0.0002x 1
2L
2
T
2L
1/2
1/2
2
T
1
/
3/2
2
C (x)=3+0.02x+0.0006x
/
(b) C (100)=3+0.02(100)+0.0006(10 , 000)=3+2+6=$11/ pair. C (100) is the rate at which the cost
is increasing as the 100 th pair of jeans is produced. It predicts the cost of the 101 st pair.
(c) The cost of manufacturing the 101 st pair of jeans is
10
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
C(101) C(100) = (2000+303+102.01+206.0602) (2000+300+100+200)
= 11.0702 $11.07
2
3
/
2
30. (a) C(x)=84+0.16x 0.0006x +0.000003x C (x)=0.16 0.0012x+0.000009x This is the rate at which the cost is increasing as the 100 th item is produced.
(b) C(101) C(100)=97.13030299 97 $0.13 .
p(x)
31. (a) A(x)=
x
/
/
A (x)=
xp (x) p(x) 1
2
/
=
xp (x) p(x)
2
/
. A (x)>0
/
C (100)=0.13 .
A(x) is increasing; that is,
x
x
the average productivity increases as the size of the workforce increases.
/
(b) p (x) is greater than the average productivity /
xp (x)> p(x)
/
/
xp (x) p(x)>0
xp (x) p(x)
2
>0
/
p (x)>A(x)
/
p (x)>
p(x)
x
/
A (x)>0 .
x
32. (a)
(
1+4x
dR
=
S=
dx
=
9.6x
0.4
0.6
) ( 9.6x 0.6) ( 40+24x0.4) ( 1.6x 0.6)
( 1+4x0.4) 2
+38.4x
0.2
64x
0.6
0.4 2
( 1+4x )
38.4x
0.2
=
54.4x
0.6
( 1+4x0.4) 2
(b)
At low levels of brightness, R is quite large and is quickly decreasing, that is, S is negative with large
absolute value. This is to be expected: at low levels of brightness, the eye is more sensitive to slight
changes than it is at higher levels of brightness.
PV
PV
1
=
=
(PV ) . Using the Product Rule, we have
nR (10)(0.0821) 0.821
dT
1
1
/
/
=
[P(t)V (t)+V (t)P (t)]=
[(8)( 0.15)+(10)(0.10)] 0.2436 K / min.
dt 0.821
0.821
33. PV =nRT T=
11
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences
34. (a) If dP/dt=0 , the population is stable (it is constant).
dP
P
P
P
(b)
=0 P=r
1
P
=1
=1
0
dt
P
r
P
P
r
c
0
c
c
P=P
0
(c) If =0.05 , then P=10 , 000
1
5
5
c
0
If P =10 , 000 , r =5%=0.05 , and =4%=0.04 , then P=10 , 000
c
r
1
1
4
5
.
0
=2000 .
=0 . There is no stable population.
35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is,
dC
dW
=0 and
=0 .
dt
dt
(b) ‘‘The caribou go extinct’’ means that the population is zero, or mathematically, C=0 .
dC
dW
(c) We have the equations
=aC bCW and
= cW +dCW . Let dC/dt=dW /dt=0 , a=0.05 ,
dt
dt
b=0.001 , c=0.05 , and d=0.0001 to obtain 0.05C 0.001CW =0 (1) and 0 .05W +0.0001CW =0 (2) .
Adding 10 times (2) to (1) eliminates the CW terms and gives us 0.05C 0.5W =0 C=10W .
Substituting C=10W into (1) results in
2
2
0.05(10W ) 0.001(10W )W =0 0.5W 0.01W =0 50W W =0 W (50 W )=0 W =0 or 50 . Since
C=10W , C=0 or 500 . Thus, the population pairs (C,W) that lead to stable populations are (0,0) and
(500,50) . So it is possible for the two species to live in harmony.
12
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
/
f (x)=1 3cos x
1. f (x)=x 3sin x
/
2. f (x)=xsin x
f (x)=x cos x+(sin x) 1=xcos x+sin x
/
3. y=sin x+10tan x
/
4. y=2 x+5cos x
3
y = 2 xcot x 5sin x
/
5. g(t)=t cos t 2
y =cos x+10sec x
3
2
2
3
2
g (t)=t ( sin t)+(cos t) 3t =3t cos t t sin t or t (3cos t tsin t)
/
6. g(t)=4sec t+tan t 2
g (t)=4sec ttan t+sec t
7. h( )=csc +e cot /
2
2
h ( )= csc cot +e ( csc )+(cot )e = csc cot +e (cot csc )
u
/
8. y=e (cos u+cu)
9. y=
x
cos x
y =
u
/
y =
(cos x)(1) (x)( sin x)
2
=
cos x+xsin x
2
cos x
(cos x)
(x+cos x)(cos x) (1+sin x)(1 sin x)
2
(x+cos x)
2
=
u
1+sin x
x+cos x
10. y=
/
u
y =e ( sin u+c)+(cos u+cu)e =e (cos u sin u+cu+c)
2
(x+cos x)
=
2
2
(x+cos x)
2
xcos x+cos x (cos x)
2
xcos x+cos x (1 sin x)
=
xcos x
2
(x+cos x)
sec 1+sec (1+sec )(sec tan ) (sec )(sec tan )
11. f ( )=
/
f ( )
12. y=
=
2
(1+sec )
=
(sec tan )[(1+sec ) sec ]
2
(1+sec )
=
sec tan 2
(1+sec )
tan x 1
sec x
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
(
2
2
)
2
dy sec xsec x (tan x 1)sec xtan x sec x sec x tan x+tan x
1+tan x
=
=
=
2
2
dx
sec x
sec x
sec x
/
y =cos x+sin x .
Another method: Simplify y first: y=sin x cos x
13. y=
sin x
2
2
/
y =
x cos x (sin x)(2x)
( x2) 2
x
=
x(xcos x 2sin x)
x
4
=
xcos x 2sin x
x
3
14. y= ( +cot )
/
2
2
2
y = (1 csc )+( +cot )( cot )= (1 csc cot cot )
2
2
2
= ( cot cot cot )
2
{1+cot =csc }
2
= ( cot 2cot )= cot ( +2cot )
15. y=sec tan /
2
2
2
y =sec (sec )+tan (sec tan )=sec (sec +tan )
2
2
Using the identity 1+tan =sec 2
, we can write alternative forms of the answer as
2
sec (1+2tan ) or sec (2sec 1)
/
/
/
/
16. Recall that if y= fgh , then y = f gh+ fg h+ fgh . y=xsin xcos x
dy
2
2
=sin xcos x+xcos xcos x+xsin x( sin x)=sin xcos x+xcos x xsin x
dx
1
sin x
=
d
d
( sec x ) =
dx
dx
1
cos x
=
d
d
19.
( cot x ) =
dx
dx
cos x
sin x
17.
18.
d
d
( csc(x) ) =
dx
dx
(sin x)(0) 1(cos x)
2
=
sin x
2
=
1
cos x
= xcot x
sin x sin x
=
1
sin x
=sec xtan x
cos x cos x
sin x
(cos x)(0) 1( sin x)
2
cos x
=
cos x
=
sin x
2
cos x
(sin x)( sin x) (cos x)(cos x)
2
sin x
2
=
2
sin x+cos x
2
=
sin x
1
2
=
2
x
sin x
20. f (x)=cos x
f (x+h) f (x)
cos (x+h) cos x
cos xcos h sin xsin h cos x
/
=
lim
=
lim
=
lim
f (x)
h
h
h
h 0
h 0
h 0
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
= lim
h
cos x
0
cos h 1
sin h
sin x
h
h
=cos x lim
h
0
cos h 1
sin h
sin x lim
h
h 0 h
=(cos x)(0) (sin x)(1)= sin x
21. y=tan x
/
2
y =sec x
equation of the tangent line is y 1=2
x
22. y=e cos x
/
the slope of the tangent line at
x
x
4
x
or y=2x+1
x
y =e ( sin x)+(cos x)e =e (cos x sin x) 4
2
,1
is sec
2
2
4
=( 2 ) =2 and an
.
the slope of the tangent line at (0, 1) is
0
e (cos 0 sin 0)=1(1 0)=1 and an equation is y 1=1(x 0) or y=x+1 .
23. y=x+cos x
y=x+1 .
24. y=
/
/
y =1 sin x . At ( 0,1 ) , y =1 , and an equation of the tangent line is y 1=1(x 0) , or
1
sin x+cos x
cos x sin x
/
y =
/
2
[ Reciprocal Rule]. At ( 0,1 ) , y =
(sin x+cos x)
equation of the tangent line is y 1= 1(x 0) , or y= x+1 .
1 0
2
= 1 , and an
(0+1)
/
25. (a) y=xcos x y =x( sin x)+cos x(1)=cos x xsin x . So the slope of the tangent at the point
( , ) is cos sin = 1 (0)= 1 , and an equation is y+ = (x ) or y= x .
(b)
26. (a) y=sec x 2cos x
/
y =sec xtan x+2sin x
the slope of the tangent line at
,1 is sec tan +2sin =2
3
3
3
3
equation is y 1=3 3 x
or y=3 3 x+1 3 .
3
(b)
3 +2
3
=3 3 and an
2
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
27. (a) f (x)=2x+cot x
/
2
f (x)=2 csc x
(b)
/
Notice that f (x)=0 when f has a horizontal tangent.
/
f is positive when f is increasing and f
negative when the graph of f is steep.
28. (a) f (x)= x sin x
/
/
is negative when f is decreasing. Also, f (x) is large
/
f (x)= x cos x+(sin x)
1 x
2
1/2
= x cos x+
sin x
2 x
(b)
/
Notice that f (x)=0 when f has a horizontal tangent.
f
/
is positive when f is increasing and f
/
is negative when f is decreasing.
/
29. f (x)=x+2sin x has a horizontal tangent when f (x)=0 1+2cos x=0
cos x=
1
2
2
4
4
2
+2 n or
+2 n , where n is an integer. Note that
and
are units from 3
3
3
3
3
This allows us to write the solutions in the more compact equivalent form (2n+1) , n an
3
integer.
x=
cos x
30. y=
2+sin x
/
y =
(2+sin x)( sin x) cos xcos x
2
(2+sin x)
2
=
2
2sin x sin x cos x
2
(2+sin x)
=
2sin x 1
2
.
=0 when
(2+sin x)
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
11
7
+2 n or x=
+2 n , n an integer. So y=
6
6
11
1
the points on the curve with horizontal tangents are:
+2 n,
,
6
3
n an integer.
2sin x 1=0 sin x=
1
2
31. (a) x(t)=8sin t
x=
v(t)=x
2
(b) The mass at time t=
3
2
2
v
=8cos
=8 3
3
32. (a) s(t)=2cos t+3sin t
(b)
(c) s=0
/
1
or y=
3
7
+2 n,
6
1
and
3
1
,
3
(t)=8cos t
has position x
1
2
2
3
= 4 . Since v
=8sin
2
3
2
=8
3
3
2
=4 3 and velocity
<0 , the particle is moving to the left.
v(t)= 2sin t+3cos t
t 2.55 . So the mass passes through the equilibrium position for the first time when
2
t 2.55 s.
(d) v=0 t 0.98 , s(t ) 3.61 cm. So the mass travels a maximum of about 3.6 cm (upward and
1
1
downward) from its equilibrium position.
(e) The speed |v| is greatest when s=0 ; that is, when t=t +n , n a positive integer.
2
33.
From the diagram we can see that sin =x/10 x=10sin . We want to find the rate of change of x
with respect to ; that is, dx/d . Taking the derivative of the above expression, dx/d =10(cos ) .
dx
1
So when =
,
=10cos
=10
=5 ft/rad.
3 d
3
2
34. (a) F=
W
sin +cos dF ( sin +cos )(0) W ( cos sin ) W (sin cos )
=
=
2
2
d
( sin +cos )
( sin +cos )
(b)
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
dF
=0
d
W (sin cos )=0
sin = cos 1
tan = =tan (c)
From the graph of F=
0.6(50)
0.6sin +cos for 0 1 , we see that
1
dF
=0 0.54 . Checking this
d
with part (b) and =0.6 , we calculate =tan 0.6 0.54 . So the value from the graph is consistent
with the value in part (b).
35.
sin 3x
3sin 3x
=lim
[ multiply numerator and denominator by 3 ]
x
3x
0
x 0
lim
x
sin 3x
[ as x
0 3x
=3 lim
3x
sin 0 0 , 3x
=3 lim
[ let =3x ]
=3(1)
=3
[ Equation 2]
0]
36.
sin 4x
=lim
0 sin 6x x 0
sin 4x
x
x
sin 6x
lim
x
=lim
x
0
4sin 4x
6x
lim
4x
x 0 6sin 6x
sin 4x 1
6x
1
2
lim
=4(1) (1)=
6 x 0 sin 6x
6
3
0 4x
=4lim
x
37.
tan 6t
=lim
0 sin 2t t 0
sin 6t
1
t
t
cos 6t sin 2t
lim
t
=lim
t
0
6sin 6t
1
2t
lim
lim
6t
t 0 cos 6t t 0 2sin 2t
sin 6t
1
1
2t
1 1
lim
lim
=6(1) (1)=3
1 2
0 6t
t 0 cos 6t 2 t 0 sin 2t
=6lim
t
38.
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
cos 1
cos 1
lim
cos 1
0
0
lim
= lim
=
= =0
sin sin 1
0 sin 0
lim
0 sin (cos )
39. lim
=
sec 0
( lim (cos )) = sin 1 =sin 1
sin
0
lim (sec )
1
0
40.
2
lim
t
sin 3t
0
t
2
sin 3t sin 3t
t
t
=lim
t
=
0
sin 3t
t
0
lim
t
sin 3t
sin 3t
lim
t
t
0
t 0
=lim
t
2
sin 3t
0 3t
= 3lim
t
2
2
=(3 1) =9
41.
cot 2x
cos2x sinx
lim
=lim
=lim cos 2x
x
sin2x
x 0
x 0
x 0
=1
lim [(sin x)/x]
(sin x)/x
(sin 2x)/x
=lim
x
cos2x
0
x
0
2 lim [(sin 2x)/2x]
x
0
1
1
=
2 1 2
42.
sin x cos x
sin x cos x
sin x cos x
= lim
=
lim
2
2
cos 2x
x /4 cos x sin x x /4 (cos x+sin x)(cos x sin x)
/4
1
1
1
= lim
=
=
2
x /4 cos x+sin x cos +sin 4
4
lim
x 43. Divide numerator and denominator by . ( sin( ) also works.)
sin sin lim
sin 0 1
1
lim
= lim
=
=
=
1
sin 1
1+1 1 2
0 +tan 0 1+ sin 1+ lim
lim
cos 0 0 cos 44.
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
lim
x
1
sin (x 1)
2
x +x 2
sin (x 1)
1
sin (x 1) 1
1
=lim
lim
= 1=
x 1
3
3
1 (x+2)(x 1) x 1 x+2 x 1
=lim
x
d
d sin x
45. (a)
tan x=
dx
dx cos x
d
d
1
sec x=
(b)
dx
dx cos x
(c)
2
sec x=
sec xtan x=
cos xcos x sin x( sin x)
2
cos x
(cos x)(0) 1( sin x)
2
2
=
2
cos x+sin x
2
cos x
2
.
cos x
sin x
. So sec xtan x=
cos x
1
2
. So sec x=
2
.
cos x
d
d 1+cot x
(sin x+cos x)=
dx
dx
x
2
cos x sin x =
x( csc x) (1+cot x)( xcot x)
2
2
x[ csc x+(1+cot x)cot x]
=
2
csc x
2
csc x
2
csc x+cot x+cot x 1+cot x
=
=
x
x
So cos x sin x=
cot x 1
.
x
46. Let PR =x . Then we get the following formulas for r and h in terms of h
1
r=xsin
and cos = h=xcos
. Now A ( ) = 2
2 x
2
2
1 2
A ( )
r
2
1
r 1
xsin
lim
= lim = lim
+ B ( ) = lim
rh
2
2
0
+
+ h
+ xcos
1
= 2
0
0
0
and x : sin
2
=
r
x
1
r and B ( ) = (2r)h=rh . So
2
2
( /2 )
( /2 )
lim tan ( /2 ) =0 .
+
0
47. By the definition of radian measure, s=r , where r is the radius of the circle.
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions
By drawing the bisector of the angle , we can see that sin
=
d/2
r
2
s
r
2 ( /2 )
/2
So lim = lim
= lim
= lim
=1 .
+ d
+ 2rsin ( /2 )
+ 2sin ( /2 ) 0 sin ( /2 )
0
0
d=2rsin
2
.
0
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
1. Let u=g(x)=4x and y= f (u)=sin u . Then
2. Let u=g(x)=4+3x and y= f (u)= u =u
2
1/2
dy dy du
=
=(cos u)(4)=4cos 4x .
dx du dx
dy dy du 1 =
= u
dx du dx 2
. Then
10
3. Let u=g(x)=1 x and y= f (u)=u . Then
1/2
(3)=
3
3
=
.
2 u 2 4+3x
dy dy du
9
29
=
=(10u )( 2x)= 20x(1 x ) .
dx du dx
4. Let u=g(x)=sin x and y= f (u)=tan u . Then
dy dy du
2
2
=
=(sec u)(cos x)=(sec u)(sin x) cos x , or
dx du dx
2
equivalently, sec (sin x) cos x .
5. Let u=g(x)=sin x and y= f (u)= u . Then
x
6. Let u=g(x)=e and y= f (u)=sin u . Then
3
7
/
7. F(x)=(x +4x) 2
3
6
3
/
2
4
2
3
2
6
2
1
2
3 3/4
(2+3x )
4(1+2x+x )
2+3x
=
4(1+2x+x )
4
4
33
(1+2x+x )
2
4
f (x)= (1+x )
3
4 2/3
/
10. f (x)=(1+x ) 4
(t +1)
2
2
3 3/4
3
6
[ or 7x (x +4) (3x +4) ]
3 1/4
2+3x
4
cos x
cos x
=
.
2 u 2 sin x
2
1+2x+x =(1+2x+x ) 1
3
3 3/4 d
/
(1+2x+x )=
= (1+2x+x ) F (x) 4
dx
11. g(t)=
cos x=
F (x)=3(x x+1) (2x 1)
9. F(x)=
1
1/2
dy dy du
x
x
x
=
=(cos u)(e )=e cos e .
dx du dx
F (x)=7(x +4x) (3x +4)
8. F(x)=(x x+1) =
dy dy du 1 =
= u
dx du dx 2
3
=(t +1) /
1/3
8x
3
(4x )=
3
1+x
3
4
4
3
3
4
3 4
4
g (t)= 3(t +1) (4t )= 12t (t +1) =
12t
4
3
4
(t +1)
12.
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
1
f (t)= (1+tan t)
3
1/3
3
/
f (t)= 1+tan t =(1+tan t) 3
3
/
13. y=cos (a +x )
3
3
mx
2
sec t=
3
2
3
(1+tan t)
3
2
3
3
/
2
3
2
y =3(cos x) ( sin x) [ a is just a constant] = 3sin xcos x
mx
/
3
2
sec t
2
y = sin (a +x ) 3x [ a is just a constant] = 3x sin (a +x )
14. y=a +cos x
15. y=e
3
2/3
d
( mx)=e
dx
y =e
mx
mx
( m)= me
/
16. y=4sec 5x
y =4sec 5xtan 5x(5)=20sec 5xtan 5x
5
28
17. g(x)=(1+4x) (3+x x ) /
5
27
28
4
g (x) =(1+4x) 8(3+x x ) (1 2x)+(3+x x ) 5(1+4x) 4
4
27
2
4
27
2
4
27
2
=4(1+4x) (3+x x ) [2(1+4x)(1 2x)+5(3+x x )]
2
=4(1+4x) (3+x x ) [(2+4x 16x )+(15+5x 5x )]
=4(1+4x) (3+x x ) (17+9x 21x )
4
3 3
4
18. h(t)=(t 1) (t +1) /
4
3
3
3
2
3
4
4
2
3
h (t) =(t 1) 4(t +1) (3t )+(t +1) 3(t 1) (4t )
2 4
2 3
3
4
3
2 4
2 3
3
4
2
2/3
=12t (t 1) (t +1) [(t 1)+t(t +1)]=12t (t 1) (t +1) (2t +t 1)
4
2
3
19. y=(2x 5) (8x 5) /
3
3
2
4
4
2
y =4(2x 5) (2)(8x 5) +(2x 5) ( 3)(8x 5) (16x)
3
3
2
2
4
1/3
2
4
=8(2x 5) (8x 5) 48x(2x 5) (8x 5)
2
2
1/3
/
20. y=(x +1)(x +2) 2
x
21. y=xe
/
2
y =2x(x +2) +(x +1)
2
x
2
x
2
x
y =xe ( 2x)+e 1=e
2
1
3
(x +2)
1/3
(2x)=2x(x +2)
2
1+
x +1
2
3(x +2)
2
x
( 2x +1)=e
2
2
(1 2x )
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
5x
22. y=e
23. y=e
xcos x
y =e
/
5x
/
cos 3x
y =e
5x
( 3sin 3x)+(cos 3x)( 5e
5x
)= e
d
xcos x
xcos x
(xcos x)=e
x( sin x)+(cos x) 1 =e
(cos x xsin x)
dx
xcos x
2
1 x
24. Using Formula 5 and the Chain Rule, y=10
z 1
=
z+1
25. F(z)=
1
=
F (z) 2
1/2
z 1
z+1
1/2
z 1
z+1
/
(3sin 3x+5cos 3x)
d
dz
2
/
1 x
y =10
d
2
1
(ln 10)
(1 x )= 2x(ln 10)10
dx
2
x
.
z 1
z+1
1
=
2
1/2
1/2
z+1
z 1
(z+1)(1) (z 1)(1)
2
(z+1)
1/2
1 (z+1)
z+1 z+1 1 (z+1)
2
1
=
=
=
1/2
2
1/2
2
1/2
3/2
2
2
(z 1)
(z+1)
(z 1)
(z+1) (z 1) ( z+1 )
4
(y 1)
26. G(y)=
2
5
(y +2y)
2
/
G (y) =
5
3
4
2
4
(y +2y) 4(y 1) 1 (y 1) 5(y +2y) (2y+2)
2
52
[(y +2y) ]
2
=
4
3
2
2(y +2y) (y 1) [2(y +2y) 5(y 1)(y+1)]
2
10
(y +2y)
3
=
2
2
2(y 1) [(2y +4y)+( 5y +5)]
2
6
3
=
2
(y +2y)
27. y=
r
2
2
2(y 1) ( 3y +4y+5)
6
(y +2y)
r +1
y
2
r +1 (1) r
/
=
(
1 2 (r +1)
2
r +1 )
2
2
r
2
1/2
r +1 (2r)
=
r +1 )
2
2
2
r +1 r
r +1
2
r +1
(
2
2
2
2
=
r +1
(
r +1 )
2
2
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
2
(r +1) r
=
(
2
3
r +1 )
2
1
=
3/2
2
2
or (r +1)
3/2
(r +1)
2
1/2
Another solution: Write y as a product and make use of the Product Rule. y=r (r +1)
1 2 3/2
2
1/2
/
y =r 2 (r +1) (2r)+r +1) 1
3/2
2
=(r +1)
2
2
1
2
3/2
[ r +(r +1) ]=(r +1)
2
3/2
(1)=(r +1)
2
3/2
The step that students usually have trouble with is factoring out (r +1)
2
2
. But this is no different
5
than factoring out x from x +x ; that is, we are just factoring out a factor with the smallest exponent
3
1
that appears on it. In this case, is smaller than .
2
2
2u
e
28. y=
u
u
e +e
u
u
/
y =
2u
2u
u
u
2u
(e +e )(e 2) e (e e )
u
u
u
u
2u
e (2e +2e e +e )
u
u
e (e +3e )
=
=
u u2
u u2
u u2
(e +e )
(e +e )
(e +e )
Another solution: Eliminate negative exponents by first changing the form of y .
2u
e
y=
u
u
u
e
u
e +e
/
y =
=
3u
2u
e +1
e
2u
e
3u
3u
2
2u
3u
2u
(e +1)(3e ) e (2e )
=
(e +1)
2u
2u
e (3e +3 2e )
2u
2
(e +1)
/
3u
=
2u
e (e +3)
2u
2
(e +1)
2
2
y =sec (cos x) ( sin x)= sin x sec (cos x)
29. y=tan (cos x)
2
sin x
30. y=
cos x
2
/
y =
cos x(2sin x cos x) sin x( sin x)
2
cos x
2
=
2
sin x(2cos x+sin x)
2
cos x
2
=
sin x(1+cos x)
2
cos x
2
=sin x(1+sec x)
Another method: y=tan x sin x
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
/
2
2
y =sec x sin x+tan x cos x=sec x sin x+sin x
sin x
31. Using Formula 5 and the Chain Rule, y=2
d
/ sin x
sin x
sin x
y =2
(ln 2)
(sin x)=2
(ln 2) cos x =2
( ln 2)cos x
dx
2
2 6
34. y=xsin
1
x
2
5
2
5
y =6(1+cos x) 2cos x( sin x)= 12cos xsin x(1+cos x)
/
y =sin
2
y =2(tan 3 )
/
33. y=(1+cos x) d
2
2
(tan 3 )=2tan 3 sec 3 3=6tan 3 sec 3
d
/
2
32. y=tan (3 )=(tan 3 ) 1
1
+xcos
x
x
2
2
1
2
=sin
x
1 1
1
cos
x x
x
2
35. y=sec x+tan x=(sec x) +(tan x) /
2
2
2
2
y =2(sec x) (sec xtan x)+2(tan x)(sec x)=2sec x tan x+2sec x tan x=4sec x tan x
ktan
36. y=e
x
ktan
/
y =e
2
x
ktan
d
(ktan x )=e
dx
x
ksec
2
1 x
x
2
1/2
=
ksec
2
2 x
x
ktan
e
x
2
37. y=cot (sin )=[cot (sin )] d
/
2
2
y =2[cot (sin )]
[cot (sin )]=2cot (sin ) [ csc (sin ) cos ]= 2cos cot (sin ) csc (sin )
d
d
/
38. y=sin (sin (sin x)) y =cos (sin (sin x)) (sin (sin x))=cos (sin (sin x)) cos (sin x) cos x
dx
39. y= x+ x /
y =
40. y= x+ x+ x 1
(x+ x )
2
/
y =
1/2
1+
1 x
2
1
x+ x+ x
2
(
1/2
) 1/2
=
1
2 x+ x
1
1+ (x+ x )
2
1+
1/2
1
2 x
1 1+ x
2
1/2
41. y=sin (tan sin x )
d
d
2
1/2
/
y =cos (tan sin x ) dx (tan sin x )=cos (tan sin x ) sec sin x dx (sin x)
1
2
1/2
=cos (tan sin x )sec sin x (sin x) cos x
2
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
(
=cos (tan sin x ) sec
2
x
2
2
x
1
2 sin x
)
sin x
2
x
( )
2
(cos x)
2
d
x
3
x
3 =2 (ln 2)3 (ln 3)(2x)
y =2 (ln 2)
dx
3
/
42. y=2 3
10
/
9
9
/
9
43. y=(1+2x) y =10(1+2x) 2=20(1+2x) . At (0,1) , y =20(1+0) =20 , and an equation of the
tangent line is y 1=20(x 0) , or y=20x+1 .
2
/
/
44. y=sin x+sin x y =cos x+2sin xcos x . At (0,0) , y =1 , and an equation of the tangent line is
y 0=1(x 0) , or y=x .
/
/
45. y=sin (sin x) y =cos (sin x) cos x . At ( ,0) , y =cos (sin ) cos =cos (0) ( 1)=1( 1)= 1 ,
and an equation of the tangent line is y 0= 1(x ) , or y= x+ .
2 x
/
46. y=x e x
2
x
x
2 x
y =x ( e )+e (2x)=2xe x e
47. (a) y=
x
x
/
y =
1+e
/
At (0, 1) , y =
1,
1
e
/
1
1
, y =2e e =
1
. So an equation of
e
1 1
1
= (x 1) or y= x .
e e
e
the tangent line is y
2
. At
x
(1+e )(0) 2( e )
x2
x
=
(1+e )
0
2e
02
(1+e )
=
2(1)
2
(1+1)
=
2
2
2
=
2e
x2
.
(1+e )
1
.
2
So an equation of the tangent line is y 1=
1
1
(x 0) or y= x+1 .
2
2
(b)
48. (a) For x>0 , |x|=x , and y= f (x)=
x
2
2 x
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
/
f (x) =
(
2
=
2 x )
2 3/2
( 2x)
2 1/2
(2 x )
2
2 1/2
(2 x )
2
=
(2 x )
(2 x )
2
2
(2 x )+x
2 1/2
1
2
2
2 x (1) x
2 3/2
(2 x )
/
So at (1,1) , the slope of the tangent line is f (1)=2 and its equation is y 1=2(x 1) or y=2x 1 .
(b)
2
49. (a) f (x)=
1
(1 x
2
x
/
1 x
x
f (x) =
1/2
2
( 2x)
1 x (1)
2
x
2
=
2
x (1 x )
2
x
2
1 x
2
1 x
2
1 x
1
=
2
x
2
1 x
(b)
/
Notice that all tangents to the graph of f have negative slopes and f (x)<0 always.
50. (a)
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
From the graph of f , we see that there are 5 horizontal tangents, so there must be 5 zeros on the
graph of f
/
. From the symmetry of the graph of f , we must have the graph of f
it is low at x= . The intervals of increase and decrease as well as the signs of f
figure.
/
/
as high at x=0 as
are indicated in the
(b)
f (x) =sin (x+sin 2x)
d
/
f (x) =cos (x+sin 2x) dx (x+sin 2x)
=cos (x+sin 2x)(1+2cos 2x)
/
2
/
51. For the tangent line to be horizontal, f (x)=0 . f (x)=2sin x+sin x f (x)=2cos x+2sin xcos x=0
3
2cos x ( 1+sin x ) =0 cos x=0 or sin x= 1 , so x= +2n or
+2n , where n is any integer.
2
2
3
Now f
=3 and f
= 1 , so the points on the curve with a horizontal tangent are
2
2
3
+2n ,3 and
+2n , 1 , where n is any integer.
2
2
52. f (x)=sin 2x 2sin x
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
/
2
2
f (x)=2cos 2x 2cos x=4cos x 2cos x 2 , and 4cos x 2cos x 2=0 (cos x 1)(4cos x+2)=0
1
cos x=1 or cos x=
. So x=2n or (2n+1) , n any integer.
2
3
53. F(x)= f (g(x))
/
/
/
/
/
/
F (x)= f (g(x)) g (x) ,
/
/
/
so F (3)= f (g(3)) g (3)= f (6) g (3)=7 4=28 . Notice that we did not use f (3)=2 .
/
/
w(x)=u(v(x)) 54. w=u v
/
/
/
/
w ( x ) =u (v(x)) v (x) , so
/
/
/
w (0)=u (v(0)) v (0)=u (2) v (0)=4 5=20 . The other pieces of information, u(0)=1 , u (0)=3 ,
/
and v (2)=6 , were not needed.
/
55. (a) h(x)= f (g(x))
(b) H(x)=g( f (x))
/
/
/
/
/
/
/
/
/
/
/
/
H (x)=g ( f (x)) f (x) , so H (1)=g ( f (1)) f (1)=g (3) 4=9 4=36 .
/
56. (a) F(x)= f ( f (x))
(b) G(x)=g(g(x))
/
h (x)= f (g(x)) g (x) , so h (1)= f (g(1)) g (1)= f (2) 6=5 6=30 .
/
/
/
/
/
/
F (x)= f ( f (x)) f (x) , so F (2)= f ( f (2)) f (2)= f (1) 5=4 5=20 .
/
/
/
/
/
/
/
G (x)=g (g(x)) g (x) , so G (3)=g (g(3)) g (3)=g (2) 9=7 9=63 .
/
/
/
/
/
/
/
/
/
u (x)= f (g(x))g (x) . So u (1)= f (g(1))g (1)= f (3)g (1) . To find f (3) ,
3 4
1
/
note that f is linear from (2,4) to (6,3) , so its slope is
=
. To find g (1) , note that g is linear
6 2
4
0 6
1
3
/
/
from (0,6) to (2,0) , so its slope is
= 3 . Thus, f (3)g (1)= ( 3)= .
2 0
4
4
57. (a) u(x)= f (g(x))
(b) v(x)=g( f (x))
/
/
/
/
/
/
/
/
/
/
/
v (x)=g ( f (x)) f (x) . So v (1)=g ( f (1)) f (1)=g (2) f (1) , which does not
/
exist since g (2) does not exist.
/
(c) w(x)=g(g(x))
/
/
/
/
/
w (x)=g (g(x))g (x) . So w (1)=g (g(1))g (1)=g (3)g (1) . To find g (3) ,
2 0 2
2
/
/
note that g is linear from (2,0) to (5,2) , so its slope is
= . Thus, g (3) g (1)=
( 3)= 2
5 2 3
3
.
/
/
/
/
/
h (x)= f ( f (x)) f (x) .
58. (a) h(x)= f ( f (x))
/
/
/
So h (2)= f ( f (2)) f (2)= f (1) f (2) ( 1)( 1)=1 .
d 2
2
/
/ 2
/
/
/ 2
/ 2
(b) g(x)= f (x ) g (x)= f (x )
(x )= f (x )(2x) . So g (2)= f (2 )(2 2)=4 f (4) 4(1.5)=6 .
dx
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
/
/
/
/
/
/
/
/
59. h(x)= f (g(x)) h (x)= f (g(x))g (x) . So h (0.5)= f (g(0.5))g (0.5)= f (0.1)g (0.5) . We can
estimate the derivatives by taking the average of two secant slopes.
m +m
1
2 22+36
14.8 12.6
18.4 14.8
/
/
For f ( 0.1 ) : m =
=22 , m =
=36 . So f (0.1)
=
=29 .
1
2
2
0.1 0
0.2 0.1
2
m +m
1
2
0.10 0.17
0.05 0.10
/
/
= 0.6 .
For g (0.5) : m =
= 0.7 , m =
= 0.5 . So g (0.5)
1
2
2
0.5 0.4
0.6 0.5
/
/
/
Hence, h (0.5)= f (0.1)g (0.5) (29)( 0.6)= 17.4 .
/
60. g(x)= f ( f (x))
/
/
/
/
/
/
/
g (x)= f ( f (x)) f (x) . So g (1)= f ( f (1)) f (1)= f (2) f (1) .
m +m
1
2
3.1 2.4
4.4 3.1
/
/
For f (2) : m =
=1.4 , m =
=2.6 . So f (2)
=2 .
1 2.0 1.5
2 2.5 2.0
2
m +m
1
2
2.0 1.8
2.4 2.0
/
/
For f (1) : m =
=0.4 , m =
=0.8 . So f (1)
=0.6 .
1 1.0 0.5
2 1.5 1.0
2
/
/
/
Hence, g (1)= f (2) f (1) (2)(0.6)=1.2 .
/
x
(b) G(x)=e
f (x)
/
G (x)=e
x
d x
/ x x
(e )= f (e )e
dx
d
f (x) /
f (x)=e f (x)
dx
/
F (x)= f (x )
/
(b) G(x)= f (x) G (x)=
4
f (x)
/
63. (a) f (x)=L(x )
/
d / (x )= f (x ) x
dx
1
4
1
/
f (x)
3
4
3
f (x)=L (x ) 4x =(1/x ) 4x =4/x for x>0 .
/
(b) g(x)=L(4x)
f (x)
/
62. (a) F(x)= f (x )
/
F (x)= f (e )
61. (a) F(x)= f (e )
/
g (x)=L (4x) 4=(1/(4x)) 4=1/x for x>0 .
4
/
3
/
3
3
(c) F(x)=[L(x)] F (x)=4[L(x)] L (x)=4[L(x)] (1/x)=4[L(x)] /x
(d) G(x)=L(1/x)
G (x)=L (1/x) ( 1/x )=(1/(1/x)) ( 1/x )=x ( 1/x )= 1/x for x>0 .
64. r(x)= f (g(h(x)))
/
/
/
/
/
2
/
2
/
2
/
r (x)= f (g(h(x))) g (h(x)) h (x) , so
/
/
/
/
/
r (1)= f (g(h(1))) g (h(1)) h (1)= f (g(2)) g (2) 4= f (3) 5 4=6 5 4=120
65. s(t)=10+
1
sin (10 t)
4
the velocity after t seconds is
10
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
/
v(t)=s (t)=
1
5
cos (10 t)(10 )=
cos (10 t) cm / s.
4
2
/
velocity =s = 66. (a) s=Acos ( t+ )
Asin ( t+ ) .
/
(b) If A
0 and 0 , then s =0 sin ( t+ )=0 t+ =n t=
2 t
dB
2 t
= 0.35cos
5.4
dt
5.4
dB 7
2
(b) At t=1 ,
=
cos
0.16 .
dt 54
5.4
2
5.4
67. (a) B(t)=4.0+0.35sin
2
(t 80) 365
68. L(t)=12+2.8sin
=
2
(t 80)
365
/
L (t)=2.8cos
n , n an integer.
0.7
2 t 7
2 t
cos
=
cos
5.4
5.4 54
5.4
2
365
/
.
/
On March 21, t=80 , and L (80) 0.0482 hours per day. On May 21, t=141 , and L (141) 0.02398 ,
/
which is approximately one half of L (80) .
1.5t
69. s(t)=2e
sin 2 t
1.5t
/
v(t)=s (t)=2 e
1.5t
(cos 2 t)(2 )+(sin 2 t)e
70. (a) lim p(t)=lim
t
t
kt 1
(b) p(t)=(1+ae ) 1
kt
1+ae
=
1.5t
( 1.5) =2e
1
=1 , since k>0
1+a 0
dp
kt 2
kt
= (1+ae ) ( kae )=
dt
kt
(2 cos 2 t 1.5sin 2 t)
kt
e 0.
kt
kae
kt 2
(1+ae )
(c)
11
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
0.5t 1
) , it seems that p(t)=0.8 (indicating that 80% of the population
From the graph of p(t)=(1+10e
has heard the rumor) when t 7.4 hours.
t
71. (a) Using a calculator or CAS, we obtain the model Q=ab with a=100.0124369 and
t
t lnb
b=0.000045145933 . We can change this model to one with base e and exponent ln b [ b =e
precalculus mathematics or from Section 7.3]: Q=ae
/
from
10.005531t
tln b
=100.012437e
.
t
x
(b) Use Q (t)=ab ln b or the calculator command nDeriv(Y , X, .04) with Y =ab to get
1
1
/
Q (0.04) 670.63 A. The result of Example 2 in Section 2.1 was 670 A.
20
t
72. (a) P=ab with a=4.502714 10 and b=1.029953851 ,
where P is measured in thousands of people. The fit appears to be very good.
(b) For 1800: m =
1
5308 3929
7240 5308
=137.9 , m =
=193.2 .
2 1810 1800
1800 1790
/
So P (1800) (m +m )/2=165.55 thousand people / year.
1
For 1850: m =
1
2
23,192 17,063
31,443 23,192
=612.9 , m =
=825.1 .
2
1850 1840
1860 1850
/
So P (1850) (m +m )/2=719 thousand people / year.
1
2
x
(c) Use the calculator command nDeriv(Y , X, .04) with Y =ab to get
1
/
1
/
P (1800) 156.85 and P (1850) 686.07. These estimates are somewhat less than the ones in part
(b).
(d) P(1870) 41,946.56. The difference of 3.4 million people is most likely due to the Civil War
(1861 1865).
12
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
8
/
73. (a) Derive gives g (t)=
45(t 2)
10
without simplifying. With either Maple or Mathematica, we first
(2t+1)
8
9
(t 2)
/
get g (t)=9
9
(t 2)
18
(2t+1)
10
, and the simplification command results in the above expression.
(2t+1)
/
3
3
4
3
2
(b) Derive gives y =2(x x+1) (2x+1) (17x +6x 9x+3) without simplifying.
/
4
3
4
5
3
3
2
With either Maple or Mathematica, we first get y =10(2x+1) (x x+1) +4(2x+1) (x x+1) (3x 1) .
If we use Mathematica’s Factor or Simplify , or Maple’s factor , we get the above expression, but
Maple’s simplify gives the polynomial expansion instead. For locating horizontal tangents, the
factored form is the most helpful.
4
4
74. (a) f (x)=
x x+1
(3x 1)
1/2
4
x +x+1
/
. Derive gives f (x)=
4
4
x x+1
x +x+1
4
whereas either Maple or
4
(x +x+1)(x x+1)
4
3x 1
/
Mathematica give f ( x ) =
4
x x+1
4
after simplification.
4
2
(x +x+1)
x +x+1
/
(b) f (x)=0
/
4
3x 1=0
x=
4
1
0.7598 .
3
(c) f (x)=0 where f has horizontal tangents. f
inflection points.
/
has two maxima and one minimum where f has
75. (a) If f is even, then f (x)= f ( x) . Using the Chain Rule to differentiate this equation, we get
d
/
/
/
/
/
/
f (x)= f ( x)
( x)= f ( x) . Thus, f ( x)= f (x) , so f is odd.
dx
/
/
/
(b) If f is odd, then f (x)= f ( x) . Differentiating this equation, we get f (x)= f ( x)( 1)= f ( x) ,
so f
/
is even.
13
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
76.
/
f (x)
g(x)
=
{ f (x) g(x) 1} /= f /(x) g(x) 1+( 1) g(x) 2g /(x) f (x)
/
/
/
/
f (x) f (x)g (x) f (x)g(x) f (x)g (x)
=
=
2
2
g(x)
g(x)
g(x)
77. (a)
d
n
sin xcos nx =nsin n 1 xcos xcos nx+sin nx( nsin nx) [Product Rule]
dx
(
)
=nsin
=nsin
=nsin
n 1
n 1
n 1
n 1
x(cos nxcos x sin nxsin x)
[factor out nsin
x]
xcos (nx+x)
[Addition Formula for cosine]
xcos (n+1)x
[factor out x ]
(b)
d
n
cos xcos nx =ncos n 1 x( sin x)cos nx+cos nx( nsin nx) [Product Rule]
dx
(
)
= ncos
= ncos
= ncos
n 1
n 1
n 1
n 1
x(cos nxsin x+sin nxcos x)
[factor out ncos
x]
xsin (nx+x)
[Addition Formula for sine]
xsin (n+1)x
[factor out x ]
5
78. ‘‘The rate of change of y with respect to x is eighty times the rate of change of y with respect to
d 5
dy
dy
dy
4
4 dy
x ’’ y =80
5y
=80
5y =80 (Note that
0 since the curve never has a
dx
dx
dx
dx
dx
4
horizontal tangent) 79. Since d
d
y =16
y=2 (since y>0 for all x )
rad, we have
180
d
=
sin
=
cos
=
cos d
180
180
180
180
=
( sin )
2
2 1/2
80. (a) f (x)=|x|= x =(x ) /
f (x)=
1 2
(x )
2
1/2
(2x)=
x
=
2
x
.
x
for x
0 .
|x|
f is not differentiable at x=0 .
14
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule
2
(b) f (x)=|sin x|= sin x 1
sin x
/
2 1/2
f (x)= (sin x) 2sin x cos x=
cos x=
2
|sin x|
{
cos x if sin x>0
cos x if sin x<0
f is not differentiable when x=n
, n an integer.
x
x
/
g (x)=cos |x|
= cos x=
|x| |x|
2
x (c) g(x)=sin |x|=sin
{
cos x if x>0
cos x if x<0
g is not differentiable at 0 .
81. First note that products and differences of polynomials are polynomials and that the derivative of
a polynomial is also a polynomial. When n=1,
A (x)
/ Q(x)P /(x) P(x)Q /(x)
P(x)
(1)
/
/
1
f (x)=
=
=
, where A (x)=Q(x)P (x) P(x)Q (x).
1
2
1+1
Q(x)
[Q(x)]
[Q(x)]
A (x)
(k)
k
Suppose the result is true for n=k, where k 1. Then f (x)=
, so
k+1
[Q(x)]
f
(k+1)
k
(x) =
k+1
/
A (x)
[Q(x)]
/
k+1 2
[Q(x)]
k+1
[Q(x)]
=
/
k
k
=
k+1
k
A (x) A (x) (k+1)[Q(x)] Q (x)
{[Q(x)] }
/
k
/
A (x) (k+1)A (x)[Q(x)] Q (x)
k
k
2k+2
[Q(x)]
k1
/
/
/
[Q(x)] { A (x) (k+1)A (x)Q (x)}
k
k
=
k
k+2
[Q(x)] [Q(x)]
=A
k+2
(x)/[Q(x)]
k+1
k
k
=
k+2
/
Q(x)A (x) (k+1)A (x)Q (x)
[Q(x)]
, where A
/
/
(x)=Q(x)A (x) (k+1)A (x)Q (x) .
k+1
k
k
We have shown that the formula holds for n=1, and that when it holds for n=k it also holds for
n=k+1. Thus, by mathematical induction, the formula holds for all positive integers n.
15
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
d
d
2
(xy+2x+3x )= (4)
dx
dx
y+2
/
y = 6
.
x
/
1. (a)
/
(x y +y 1)+2+6x=0
xy = y 2 6x
/
y =
y 2 6x
or
x
2
4 2x 3x
4
/
4
(b) xy+2x+3x =4 xy=4 2x 3x y=
= 2 3x , so y =
3.
2
x
x
x
/ y 2 6x (4/x 2 3x) 2 6x 4/x 3x
4
(c) From part (a), y =
=
=
=
3.
2
x
x
x
x
2
2
d
d
2
2
(4x +9y )= (36)
dx
dx
2. (a)
2
2
2
(b) 4x +9y =36
2 1
2
(9 x )
3 2
/
y =
y=
1/2
2x
( 2x)=
2
1 1
+
x y
1 1
+ =1
x y
(c) y =
y
2
=
/
(c) y =
5.
y
x
1
2
x
=
y=(4
4
x
x
d 2 2 d
(x +y )=
(1)
dx
dx
1
/
y =
2
1
2
/
y =
2
y
2
y
y
x
x
x
/ (x 1)(1) (x)(1)
1
y=
, so y =
=
.
2
2
x 1
(x 1)
(x 1)
2
2
1
2 x
2
1
=
x (x 1)
+
2
(x 1)
1
/
2 y
x ) =16 8 x +x
=
y =0
2
=
2
3 9 x
/
2
.
2
9 x
1
2x
=
2
x
x
x
y =4
2
3
9 2
d
d
( x + y )=
(4)
dx
dx
4. (a)
4x
d
= (1)
dx
[x/(x 1)]
x
(b)
4x
=
9y
1
1 x 1
=1
=
y
x
x
2
/
y=
8x
4x
=
18y
9y
2
2
9 x , so
3
3 9 x
/
(b)
/
y =
4
2
(9 x )
9
2
2
9y =36 4x (c) From part (a), y =
d
3. (a)
dx
/
8x+18y y =0
y =0
/
y =
y
/
y =
x
4
+1
x
4
+1
x
/
2x+2yy =0
/
2yy = 2x
/
y =
x
y
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
6.
d 2 2 d
(x y )=
(1)
dx
dx
7.
d 3 2
d
2
(x +x y+4y )=
(6)
dx
dx
2
/
2
/
2
2
/
2
3x +2xy
y =
d 2
d
3
(x 2xy+y )=
(c)
dx
dx
/ 2x 2y
y =
2
2x 3y
/
2
2
2
2
/
/
2
2
x +8y
2
/
/
/
/
2
/
/
2x 2y=2xy 3y y 2x 2(xy +y 1)+3y y =0
2
2
(x y +y 2x)+(x 2yy +y 1)=3
/
2
2x 2y=y (2x 3y )
/
2
x y +2xyy =3 2xy y 2
/
y (x +2xy)=3 2xy y 2
x y +8yy = 3x 2xy
x(3x+2y)
=
/
8.
d 2
d
2
(x y+xy )=
(3x)
dx
dx
x
y
/
x +8y
9.
y =
3x +(x y +y 2x)+8yy =0
/
(x +8y) y = 3x 2xy
/
/
2x=2yy 2x 2yy =0
y =
3 2xy y
2
x +2xy
10.
d 5 2 3 d
4
(y +x y )= (1+x y)
dx
dx
/
4
2 2
4
3
4
3
/
/
2
2
/
3
y =
3
4
/
3
4
3
2 2
5y +3x y x
d 2 2
d
(x y +xsin y)=
(4) dx
dx
2
2
4x y 2xy
/
y (5y +3x y x )=4x y 2xy 11.
/
5y y +x 3y y +y 2x=0+x y +y 4x 2
/
4
2
/
x 2yy +y 2x+xcos y y +sin y 1=0
2x yy +xcos y y = 2xy sin y
2
/
2
(2x y+xcos y)y = 2xy sin y
/
y =
2
2xy sin y
2
2x y+xcos y
12.
d
d
2
(1+x)= [sin (xy )]
dx
dx
2
2
2
2
/
2
1=[cos (xy )](x 2yy +y 1)
/
1 y cos (xy )=2xycos (xy )y /
y =
2
2
/
2
2
1=2xycos (xy )y +y cos (xy )
2
1 y cos (xy )
2
2xycos (xy )
13.
d
d
(4cos xsin y)= (1)
dx
dx
/
4[cos x cos y y +sin y ( sin x)]=0
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
/
14.
y =
d
d
2
2
[ysin (x )]= [xsin (y )]
dx
dx
/
2
4sin xsin y
=tan xtan y
4cos xcos y
/
y (4cos xcos y)=4sin xsin y 2
2
2
/
2
/
2
ycos (x ) 2x+sin (x ) y =xcos (y ) 2yy +sin (y ) 1
2
2
/
y [sin (x ) 2xycos (y )]=sin (y ) 2xycos (x )
y =
2
2
2
2
sin (y ) 2xycos (x )
sin (x ) 2xycos (y )
( )
2
d
x
15.
e
dx
y
2
d
=
(x+y)
dx
x y 2
/
2
/
2 x y
e (x y +y 2x)=1+y xe
2
/
y +2xye
x y
/
=1+y 2
2
2 x y
xe
/
2
/
y y =1 2xye
x y
2
/
2
2 x y
1)=1 2xye
y (x e
x y
/
y =
1 2xye
2
2 x y
1
xe
16.
1
(x+y)
2
d
d
2 2
( x+y )=
(1+x y )
dx
dx
/
1
y
2
/
2
+
=2x yy +2xy 2 x+y 2 x+y
1/2
x y
/
2
/
2
(1+y )=x 2yy +y 2x
/
2
/
2
x+y 1+y =4x y x+y y +4xy
2
/
2
/
2
y 4x y x+y y =4xy
/
x+y 1
2
2
y (1 4x y x+y )=4xy
/
x+y 1
y =
x+y 1
4xy
2
1 4x y x+y
1
(xy)
2
2
17. xy =1+x y
y
1/2
/
x
y
2
x =2xy
2 xy
2 xy
/
y
18. tan (x y)=
2
2
2
y
x 2x
/
xy
=
2 xy
2
x
y
/
2 /
y +
=x y +2xy
2 xy
2 xy
/
(xy +y 1)=0+x y +y 2x
2
4xy xy y
2 xy
2
/
y =
4xy xy y
2
x 2x
/
xy
/
(1+x )sec (x y) (1 y )+tan (x y) 2x=y (1+x )tan (x y)=y
1+x
2
2
2
2
2
2
/
/
(1+x )sec (x y) (1+x )sec (x y) y +2xtan (x y)=y 2
2
/
(1+x )sec (x y)+2xtan (x y)=[1+(1+x )]sec (x y) y /
y =
2
2
(1+x )sec (x y)+2xtan (x y)
2
2
1+(1+x )sec (x y)
19. xy=cot (xy)
/
2
/
y+xy = csc (xy)(y+xy )
/
2
(y+xy )[1+csc (xy)]=0
/
y+xy =0
/
y = y/x
20. sin x+cos y=sin xcos y
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
/
/
/
cos x sin y y =sin x( sin y y )+cos ycos x
/ cos x(cos y 1)
y =
sin y(sin x 1)
21.
d
d
2
3
{1+ f (x)+x [ f (x)] }= (0)
dx
dx
/
2
2
/
/
2
d
d 2
= (x )
dx dx
2
/
3
f (x)+x 3[ f (x)] f (x)+[ f (x)] 2x=0 . If x=1 , we have
3
/
f (1)+1 3[ f (1)] f (1)+[ f (1)] 2(1)=0
16
/
/
13 f (1)= 16 f (1)=
.
13
22.
(sin xsin y sin y) y =cos xcos y cos x
/
2
/
3
/
f (1)+1 3 2 f (1)+2 2=0
/
f (1)+12 f (1)= 16
/
g (x)+xcos g(x) g (x)+sin g(x) 1=2x . If x=1 , we have
/
/
/
g (1)+1cos g(1) g (1)+sin g(1)=2(1)
/
/
/
dx 4
+x 1 =1
dy
2xy
g (1)+cos 0 g (1)+sin 0=2
g (1)+g (1)=2
/
2g (1)=2
/
g (1)=1 .
4
2 2
3
4
2
22
2
2
24. (x +y ) =ax y
2
2
25. x +xy+y =3
3
2(x +y )
/
/
2
/
2x+xy +y 1+2yy =0
/
+ y 4x
3
2
4
2
/
xy +2yy = 2x y
2
2
dx ax 4y(x +y )
=
2 2
dy
4x(x +y ) 2axy
dx
dx
2
2x
+2y =2ayx
+ax dy
dy
2
x=1 and y=1 , we have y =
dx
dy
dx 1 4y 2x y x
=
2
3
dy
2xy +4x y
dx
3 dx
3
2
4
+4x y
=1 4y 2x y x dy
dy
2
2
4y + x 2y+y 2x
23. y +x y +yx =y+1
/
y (x+2y)= 2x y
/
y =
2x y
. When
x+2y
2 1 3
= = 1 , so an equation of the tangent line is y 1= 1(x 1) or
1+2 1 3
y= x+2 .
2
2
/
/
/
/
/
26. x +2xy y +x=2 2x+2(xy +y 1) 2yy +1=0 2xy 2yy = 2x 2y 1 y (2x 2y)= 2x 2y 1
/ 2x 2y 1
/ 2 4 1 7 7
y =
. When x=1 and y=2 , we have y =
=
= , so an equation of the tangent
2x 2y
2 4
2 2
7
7
3
line is y 2= (x 1) or y= x
.
2
2
2
2
2
2
2
2
27. x +y =(2x +2y x) /
2
2
/
2x+2yy =2(2x +2y x)(4x+4yy 1) . When x=0 and y=
1
, we have
2
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
1
2
/
0+y =2
y=x+
/
/
(2y 1)
/
/
y =2y 1
y =1 , so an equation of the tangent line is y
1
=1(x 0) or
2
1
.
2
2/3
2 x
3
2/3
28. x +y =4
1/3
y =0
1/3
( 3 3 )
=
=
3 3
( 3 3 )
1
/
2/3
1
/
have y =
2 + y
3
1/3
3
+
x
y
3
/
=0
3
/
y =
3
y
y
. When x= 3 3 and y=1 , we
x
3
1
, so an equation of the tangent line is
=
3 3
3
1
1
(x+3 3 ) or y=
x+4 .
3
3
y 1=
2
22
/
2
2
2
2
29. 2(x +y ) =25(x y )
2
/
/
4(x +y )(2x+2yy )=25(2x 2yy )
2
/
/
4(x+yy )(x +y )=25(x yy ) 2
2
/
2
2
4yy (x +y )+25yy =25x 4x(x +y )
/
y =
2
2
2
2
25x 4x(x +y )
. When
25y+4y(x +y )
75 120
45
9
=
=
, so an equation of the tangent line is
25+40
65
13
9
9
40
y 1=
(x 3) or y=
x+
.
13
13
13
/
x=3 and y=1 , we have y =
2 2
2 2
30. y (y 4)=x (x 5)
/
/
/
32y +16y =0
4
2
4
16y =0
2
31. (a) y =5x x 2
4
2
y 4y =x 5x 3
/
/
3
4y y 8yy =4x 10x . When x=0 and y= 2 , we have
/
y =0 , so an equation of the tangent line is y+2=0(x 0) or y= 2 .
3
/
2yy =5(4x ) 2x
3
10x x
y =
. So at the point (1,2) we have
y
/
3
10(1) 1 9
9
9
5
y =
= , and an equation of the tangent line is y 2= (x 1) or y= x
.
2
2
2
2
2
/
(b)
2
3
2
32. (a) y =x +3x /
2
2yy =3x +3(2x)
2
3x +6x
y =
. So at the point (1, 2) we have
2y
/
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
2
3(1) +6(1)
9
9
9
1
y =
=
, and an equation of the tangent line is y+2=
(x 1) or y=
x+ .
2( 2)
4
4
4
4
/
/
(b) The curve has a horizontal tangent where y =0
2
3x +6x=0
3x(x+2)=0
x=0 or x= 2 . But note
2
2
3
that at x=0 , y=0 also, so the derivative does not exist. At x= 2 , y =( 2) +3( 2) = 8+12=4 , so y= 2
. So the two points at which the curve has a horizontal tangent are ( 2, 2) and ( 2,2) .
(c)
33. (a)
There are eight points with horizontal tangents: four at x 1.57735 and four at x 0.42265 .
2
3x 6x+2
/
(b) y =
3
2
2(2y 3y y+1)
/
/
y = 1 at (0, 1) and y =
Equations of the tangent lines are y= x+1 and y=
/
(c) y =0
2
3x 6x+2=0
x=1
1
at (0, 2)
3
1
x+2 .
3
1
3
3
(d)
By multiplying the right side of the equation by x 3 , we obtain the
first graph.
By modifying the equation in other ways, we can generate the other
graphs.
2
y(y 1)(y 2)=x(x 1)(x 2)(x 3)
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
2
y(y 4)(y 2)=x(x 1)(x 2)
2
y(y+1)(y 1)(y 2)=(x 1)(x 2)
2
(y+1)(y 1)(y 2)=x(x 1)(x 2)
2
x(y+1)(y 1)(y 2)=y(x 1)(x 2)
2
2
y(y +1)(y 2)=x(x 1)(x 2)
2
2
y(y+1)(y 2)=x(x 1)(x 2)
34. (a)
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
1
, and 3 at x=1 . The three
2
horizontal tangents along the top of the wagon are hard to find, but by limiting the y range of the
graph (to [1.6,1.7] , for example) they are distinguishable.
(b) There are 9 points with horizontal tangents: 3 at x=0 , 3 at x=
/
2
2
35. From Exercise 29, a tangent to the lemniscate will be horizontal if y =0 25x 4x(x +y )=0
2 2
2 2 25
x[25 4(x +y )]=0 x +y =
( 1 ). (Note that when x is 0 , y is also 0 , and there is no horizontal
4
25
2 2
tangent at the origin.) Substituting
for x +y in the equation of the lemniscate,
4
2 22
2 2
2 2 25
2 75
2 25
2(x +y ) =25(x y ) , we get x y =
( 2 ). Solving ( 1 ) and ( 2 ), we have x =
and y =
, so
8
16
16
5 3
5
the four points are ,
.
4
4
2
x
36.
2
2
y
+
2
a
2x
=1
+
2
b
/
2yy
=0
2
a
2
/
y =
b x
2
b
2
b x
y y =
0
0
x x
0
the ellipse, we have
2
37.
2
2
2
a
2x
=1
2
b
a
2
y y
+
a
y
0
2
x
y y =
0
0
2
a y
2
=
b
2yy
2
b
/
=0
0
2
0
2
2
x x
0
=
2
x
+
a
b
0
2
. Since (x ,y ) lies on
0 0
a
2
+
a
0
2
=1 .
b
/
y =
2
b x
2
an equation of the tangent line at (x ,y ) is
0 0
a y
(x x ) . Multiplying both sides by
0
2
b
y y
y
0
0
gives
2
y
y
2
b x
y y
b
0
2
0
(x x ) . Multiplying both sides by
a y
x
0 0
y
0
2
an equation of the tangent line at (x ,y ) is
a y
0
2
b
gives
0
2
b
2
x x
y
0
2
b
=
0
2
a
2
x
0
2
a
. Since (x ,y ) lies on
0 0
the hyperbola, we have
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
x x
0
2
2
y y
x
0
=
2
a
b
0
2
2
y
0
2
a
=1 .
b
/
1
y
+
=0
2 x 2 y
38. x + y = c y
0
x
x
an equation of the tangent line at (x ,y ) is
0 0
y
0
y y =
y
/
y =
0
y=y (x x ) . Now x=0
0
0
x
0
( x )=y + x
0
0
y+ x
0
0
0
is x + x
0
y+ x
0
y =
y . And y=0
0
x
y
(x x )
x x =
0
0
0
x
0
y
0
x=x + x
0
y , so the x intercept
0
0
0
y . The sum of the intercepts is
0
y
0
0
0
0
y
0
y , so the y intercept is
0
0
+ x+ x
0
=x +2 x
y
0
0
0
y +y =
0
x +
0
0
0
2
2
=( c ) =c .
y
0
x
, so the slope of the
y
x
y
0
1
0
. The negative reciprocal of that slope is
, which is the
tangent line at P(x ,y ) is =
0 0
x /y
y
x
2
2
2
/
39. If the circle has radius r , its equation is x +y =r /
2x+2yy =0
y =
0 0
0
0
slope of OP , so the tangent line at P is perpendicular to the radius OP .
q
p
q 1
40. y =x qy
/
y = px
p 1
/
y =
px
p 1
q 1
=
px
1
x
/
y =
1
2
1+( x )
1
y
q
qy
41. y=tan
p 1
=
px
qy
p 1 p/q
x
qx
1 x
2
d
1
( x )=
dx
1+x
1/2
=
p
=
p ( p/q ) x
q
1
1
2 x (1+x)
1 1/2
42. y= tan x =(tan x) / 1
1 1/2 d
1
y = (tan x) (tan x)=
2
dx
1
1
2 tan x
1
2
1+x
=
1
1
2
2 tan x (1+x )
1
43. y=sin (2x+1)
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
1
/
y =
2
d
(2x+1)=
dx
1 (2x+1)
2
1
2
/
1
2
=
2
4x 4x
1 (4x +4x+1)
h (x)= 1 x 44. h(x)= 1 x arcsinx
2
2=
2
/
2
1
2
H (x)=(1+x )
2
2
x x
1
2
(1 x )
2
+arcsinx
1 x
45. H(x)=(1+x )arctanx
1
1/2
( 2x) =1
xarcsinx
2
1 x
+(arctanx)(2x)=1+2xarctanx
1+x
46. y=tan
1
( x
)
2
x +1 1
/
y =
(
)
2
1+ x
x +1
x
1
2
=
2
2
(
2
2 1+x x x +1
)
x +1
2
(
2
x +1 x
1
2
2
x +1 (1+x ) x(x +1)
1
=
)
2
2
2
2 (1+x )
2
x +1
x +1 x
=
2
x +1 x
=
2
2
x +1 x
2
2
1+x 2x x +1 +x +1
x +1
2
=
2
x +1 x
1
2
2(1+x )
1
47. h(t)=cot (t)+cot (1/t)
2
d 1
t
1
1
1
1
=
2 =
+ 2 =0 .
h (t)=
2
2 dt t
2
2
2
t +1
1+t 1+(1/t)
1+t t +1
t
1+t
Note that this makes sense because h(t)=
for t>0 and h(t)=
for t<0 .
2
2
/
1
1
1
48. y=xcos x
2
1 x 1
/
y =cos x
x
+
2
1 x
1 2x
49. y=cos (e )
/
y =
1
2x 2
1 (e )
d 2x
(e )=
dx
1
x
=cos x
2
1 x
2x
2e
1 e
4x
50. y=arctan(cos )
10
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
/
y =
1
2
1+(cos )
x
sin ( sin )=
2
1+cos 2
51. f (x)=e x arctanx
x
2
1
/
=e x
2
f (x)
1+x
x
=e +(arctanx)(2x)
2
x
2
2xarctanx
1+x
This is reasonable because the graphs show that f is increasing when f
when f has a minimum.
/
is positive, and f
/
is zero
2
52. f (x)=xarcsin(1 x )
2x
=x
/
f (x)
22
1 (1 x )
2
+arcsin(1 x ) 1
2
2x
2
=arcsin(1 x )
2
2x x
4
This is reasonable because the graphs show that f is increasing when f
inflection point when f
/
1
1
2
is positive, and that f has an
changes from increasing to decreasing.
53. Let y=cos x . Then cos y=x and 0
y
dy
1
=
=
dx
sin y
/
1 cos y
=
1
2
sin y
dy
=1
dx
. (Note that sin y 0 for 0
y
).
1 x
11
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
1
54. (a) Let y=sec x . Then sec y=x and y
sec ytan y
dy
dx
0,
dy
1
=
=
dx sec ytan y
=1
2
1
,
=
2
3
2
. Differentiate with respect to x:
1
2
2
. Note that tan y=sec y 1
2
sec y sec y 1 x x 1
3
2
tan y= sec y 1 since tan y>0 when 0<y<
or <y<
.
2
2
dy
dy
1
1
2
2
2
(b) y=sec x sec y=x sec ytan y =1
=
. Now tan y=sec y 1=x 1 , so
dx
dx sec ytan y
2
tan y=
x 1 . For y
0,
, x 1 , so sec y=x=| x| and tan y 0
2
dy
1
1
2
=
=
. For y
,
, x
1, so | x|= x and tan y= x 1 dx
2
2
2
x x 1 | x| x 1
dy
1
1
1
1
=
=
=
=
.
dx sec ytan y
2
2
2
x x 1
( x) x 1 | x| x 1
(
2
2
)
2
2
55. 2x +y =3 and x=y intersect when 2x +x 3=0 2
2
extraneous since x=y is nonnegative. When x=1 , 1=y 2
2
/
x=
(2x+3)(x 1)=0
3
3
or 1 , but is
2
2
y= 1 , so there are two points of
/
2
/
/
y = 2x/y , and x=y 1=2yy y =1/(2y) . At (1,1)
1
, the slopes are m = 2(1)/1= 2 and m =1/(2 1)= , so the curves are orthogonal (since m and m
1
2
1
2
2
are negative reciprocals of each other). By symmetry, the curves are also orthogonal at (1, 1) .
intersection: (1, 1) . 2x +y =3
2
2
2
4x+2yy =0
2
2
2
2
56. x y =5 and 4x +9y =72 intersect when 4x +9(x 5)=72 13x =117
2
2
points of intersection: ( 3, 2) . x y =5
/
/
x= 3 , so there are four
2
2
/
2x 2yy =0 y =x/y , and 4x +9y =72 8x+18yy =0
3
2
/
y = 4x/9y . At (3,2) , the slopes are m = and m =
, so the curves are orthogonal. By symmetry,
1 2
2
3
the curves are also orthogonal at (3, 2) , ( 3,2) and ( 3, 2) .
57.
12
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
58. The orthogonal family represents the direction of the wind.
2
2
2
2
2
2
/
59. x +y =r is a circle with center O and ax+by=0 is a line through O . x +y =r 2x+2yy =0
/
y = x/y , so the slope of the tangent line at P (x ,y ) is x /y . The slope of the line OP is y /x ,
0 0 0
0 0
0
0 0
which is the negative reciprocal of x /y . Hence, the curves are orthogonal, and the families
0 0
of curves are orthogonal trajectories of each other.
2
2
2
2
60. The circles x +y =ax and x +y =by intersect at the origin where the tangents are vertical and
2
2
0
0
2
2
0
0
horizontal. If (x ,y ) is the other point of intersection, then x +y =ax ( 1 ) and x +y =by ( 2 ). Now
0 0
2
2
x +y =ax
/
2x+2yy =a
/
y =
a 2x
2 2
and x +y =by 2y
are orthogonal at
a 2x
b 2y
2
2
0
0
=
(x ,y )
2ax 4x =4y 2by 0
0
0 0
0
0
2y
2x
0
2
61. y=cx 0
/
/
2x+2yy =by 2
2
0
0
/
y =
0
2x
. Thus, the curves
b 2y
ax +by =2(x +y ) , which is true by ( 1 ) and ( 2 ).
0
0
0
/
2
2
y =2cx and x +2y =k
/
2x+4yy =0
/
2yy = x /
y =
x
=
2(y)
x
2
2(cx )
=
1
, so the
2cx
curves are orthogonal.
13
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
3
62. y=ax /
2
2
2
/
/
y =3ax and x +3y =b
/
3yy = x 2x+6yy =0
y =
x
=
3(y)
x
3
3(ax )
=
1
2
, so
3ax
the curves are orthogonal.
2
2
63. To find the points at which the ellipse x xy+y =3 crosses the x axis, let y=0 and solve for x .
2
2
x x(0)+0 =3
y=0
3 . So the graph of the ellipse crosses the x axis at the points (
x=
/
/
/
Using implicit differentiation to find y , we get 2x xy y+2yy =0
0 2 3
/
So y at ( 3,0) is
2(0)
/
=2 and y at (
3
3,0) is
0+2 3
2(0)+ 3
/
y (2y x)=y 2x
3,0) .
y
2x
/
y =
.
2y x
=2 . Thus, the tangent lines at these
points are parallel.
y 2x
as in Exercise 49. The slope of the tangent
2y x
1 2( 1) 3
1
line at ( 1,1) is m=
= =1 , so the slope of the normal line is = 1 , and its equation is
2(1) ( 1) 3
m
/
64. (a) We use implicit differentiation to find y =
y 1= 1(x+1)
2
2
y= x . Substituting x for y in the equation of the ellipse, we get x x( x)+( x) =3
2
3x =3 x= 1 . So the normal line must intersect the ellipse again at x=1 , and since the equation of
the line is y= x , the other point of intersection must be (1, 1) .
(b)
2 2
65. x y +xy=2
2
/
2
/
x 2yy +y 2x+x y +y 1=0
/
2
2
y (2x y+x)= 2xy y
14
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
2
2
2xy +y
/
y =
. So 2
2x y+x
2xy +y
2
2
2
= 1 2xy +y=2x y+x
y(2xy+1)=x(2xy+1)
2x y+x
y(2xy+1) x(2xy+1)=0 (2xy+1)(y x)=0
2 2
4
must have x=y . Then x y +xy=2
2
1
1
or y=x . But xy=
2
2
xy=
x +x =2
4
2
2
2 2
x y +xy=
2
1 1
2 , so we
4 2
2
x +x 2=0 (x +2)(x 1)=0 . So x = 2 , which is
2
impossible, or x =1 x= 1 . Since x=y , the points on the curve where the tangent line has a slope of
1 are ( 1, 1) and (1,1) .
x
2
2
. Let (a,b) be a point on x +4y =36 whose tangent line
4y
a
a
passes through (12,3) . The tangent line is then y 3=
(x 12) , so b 3=
(a 12) . Multiplying
4b
4b
2
2
66. x +4y =36
/
2x+8yy =0
2
/
y =
2
2
2
2
2
both sides by 4b gives 4b 12b= a +12a , so 4b +a =12(a+b) . But 4b +a =36 , so 36=12(a+b)
2
a+b=3
2
2
2
2
2
b=3 a . Substituting 3 a for b into a +4b =36 gives a +4(3 a) =36 a +36 24a+4a =36
24
24
24
9
2
5a 24a=0 a(5a 24)=0 , so a=0 or a=
. If a=0 , b=3 0=3 , and if a=
, b=3
=
. So
5
5
5
5
24 9
the two points on the ellipse are (0,3) and
,
.
5
5
a
Using y 3=
(x 12) with (a,b)=(0,3) gives us the tangent line y 3=0 or y=3 . With
4b
24 9
24/5
2
2
(a,b)=
,
, we have y 3=
(x 12) y 3= (x 12) y= x 5 . A graph of the
5
5
4( 9/5)
3
3
ellipse and the tangent lines confirms our results.
1
67. (a) If y= f (x) , then f (y)=x . Differentiating implicitly with respect to x and remembering that y
dy
dy
1
1
/
1 /
is a function of x, we get f (y) =1, so
= /
( f ) (x)= / 1
.
dx
dx
f (y)
f ( f (x))
2
3
1
1 /
/ 1
/
(b) f (4)=5 f (5)=4. By part (a), ( f ) (5)=1/ f ( f (5))=1/ f (4)=1/
= .
3
2
68. (a) f (x)=2x+cos x
one to one.
/
f (x)=2 sin x>0 for all x . Thus, f is increasing for all x and is therefore
15
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation
1
(b) Since f is one to one, f (1)=k
f (k)=1 . By inspection, we see that f (0)=2(0)+cos 0=1 , so
1
k= f (1)=0 .
1 /
/
1
/
(c) ( f ) (1)=1/ f ( f (1))=1/ f (0)=1/(2 sin 0)=
1
2
x
. Now let h be the height of the lamp, and let (a,b) be the
4y
point of tangency of the line passing through the points (3,h) and ( 5,0) . This line has slope
h 0
1
a
/
= h . But the slope of the tangent line through the point (a,b) can be expressed as y =
,
3 ( 5) 8
4b
b 0
b
a
b
2
2
2
2
2
2
or as
=
, so =
4b = a 5a a +4b = 5a . But a +4b =5 , so 5= 5a a= 1 .
a ( 5) a+5
4b a+5
2
2
69. x +4y =5
2
/
2x+4(2yy )=0
/
y =
2
Then 4b = a 5a= 1 5( 1)=4 b=1 , since the point is on the top half of the ellipse. So
h
b
1
1
=
=
= h=2 . So the lamp is located 2 units above the x axis.
8 a+5 1+5 4
16
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
/
1. a= f , b= f
//
, c= f
. We can see this because where a has a horizontal tangent, b=0 , and where b
has a horizontal tangent, c=0 . We can immediately see that c can be neither f nor f
points where c has a horizontal tangent, neither a nor b is equal to 0 .
/
, since at the
/
2. Where d has horizontal tangents, only c is 0 , so d =c . c has negative tangents for x<0 and b is the
/
only graph that is negative for x<0 , so c =b . b has positive tangents on R (except at x=0 ), and the
/
only graph that is positive on the same domain is a , so b =a . We conclude that d= f , c= f
and a= f
///
/
, b= f
//
.
3. We can immediately see that a is the graph of the acceleration function, since at the points where a
has a horizontal tangent, neither c nor b is equal to 0 . Next, we note that a=0 at the point where b has
/
a horizontal tangent, so b must be the graph of the velocity function, and hence, b =a . We conclude
that c is the graph of the position function.
4. a must be the jerk since none of the graphs are 0 at its high and low points. a is 0 where b has a
/
/
maximum, so b =a . b is 0 where c has a maximum, so c =b . We conclude that d is the position
function, c is the velocity, b is the acceleration, and a is the jerk.
5
2
/
4
5. f (x)=x +6x 7x
f (x)=5x +12x 7
8
f (t)=8t 42t +8t 6
4
/
6. f (t)=t 7t +2t /
7
5
y = 2sin 2 8. y= sin y = cos +sin 3
//
6
4
f (t)=56t 210t +24t
2
y = 4cos 2
/
/
3
//
7. y=cos 2 6
//
f (x)=20x +12
//
y = ( sin )+cos 1+cos =2cos sin 5
5
//
4
4
9. F(t)=(1 7t) F (t)=6(1 7t) ( 7)= 42(1 7t) F (t)= 42 5(1 7t) ( 7)=1470(1 7t)
2x+1
/
2
(x 1)(2) (2x+1)(1) 2x 2 2x 1
3
10. g(x)=
g (x)=
=
=
or 3(x 1)
2
2
2
x 1
(x 1)
(x 1)
(x 1)
//
3
3
6
g (x)= 3( 2)(x 1) =6(x 1) or
3
(x 1)
11. h(u)=
//
1 4u
1+3u
/
h (u)=
3
(1+3u)( 4) (1 4u)(3)
2
(1+3u)
3
=
4 12u 3+12u
2
(1+3u)
=
7
2
2
or 7(1+3u) (1+3u)
h (u)= 7( 2)(1+3u) (3)=42(1+3u) or
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
42
3
(1+3u)
12.
b
1/2
1/2
=as +bs s
1 1/2
1 3/2
1 1/2 1 3/2
/
= as bs H (s) =a 2 s +b 2 s
2
2
1
1 3/2
1
3 5/2
1 3/2 3 //
b s
=
as + bs
H (s) = 2 a 2 s
2
2
4
4
=a s +
H(s)
2
h (x)=
1 2 (x +1)
2
2
x +1 1 x
//
h (x)=
(
cx
/
14. y=xe 2
x +1
2/3
(2x)=
1
3
1/2
(2x)
4/3
3
1/2
2
(x +1)
=
2
2
[(x +1) x ]
2
1
cx
//
1/3
2
2
2
3
1/3
3
2
3/2
(x +1)
cx
cx
cx
cx
y =e (c)+(cx+1)e c=ce (1+cx+1)=ce (cx+2)
1/3
(3x )=2x (x +1)
(3x )+(x +1)
1
=
(x +1)
cx
(x +1)
2
x +1
2
2 3 (x +1)
3
/
y =
x
1/2
y =x e c+e 1=e (cx+1)
3
//
)
2
cx
15. y=(x +1) y =2x
1 2 (x +1)
2
/
13. h(x)= x +1 5/2
3
1/3
(4x)=4x(x +1)
4 3
4/3
2x (x +1)
16.
=
y
y
/
4x
x+1
x+1 4 4x
=
1
(x+1)
2
1/2
2
( x+1 )
3/2
y
//
(x+1) 2 (2x+4)
=
=
4 x+1 2x/ x+1 4(x+1) 2x
2x+4
=
=
3/2
3/2
x+1
(x+1)
(x+1)
3
1/2
(x+1)
2
3/2 2
[(x+1) ]
2x+2 3x 6
x 4
=
=
5/2
5/2
(x+1)
(x+1)
1/2
=
(x+1) [2(x+1) 3(x+2)]
3
(x+1)
17. H(t)=tan 3t 2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
/
2
H (t)=3sec 3t
//
d
2
(sec 3t)=6sec 3t (3sec 3t tan 3t)=18sec 3t tan 3t
dt
/
2
H (t)=2 3sec 3t
2
18. g(s)=s cos s
g (s)=2s cos s s sin s
//
2
2
g (s)=2cos s 2s sin s 2s sin s s cos s=(2 s )cos s 4s sin s
3 5t
/
19. g(t)=t e //
3 5t
5t
2
2 5t
g (t)=t e 5+e 3t =t e (5t+3)
5t
2
5t
2 5t
g (t) =(2t)e (5t+3)+t (5e )(5t+3)+t e (5)
=te
5t
5t
2
2(5t+3)+5t(5t+3)+5t =te (25t +30t+6)
20. h(x)=tan
1
1
/
2
h (x)=
(x )
22
2x=
1+(x )
2
21. (a) f (x)=2cos x+sin x
2x
1+x
4
//
h (x)=
4
3
(1+x )(2) (2x)(4x )
42
=
(1+x )
2 6x
4
42
(1+x )
/
f (x)=2( sin x)+2sin x(cos x)=sin 2x 2sin x
//
f (x)=2cos 2x 2cos x=2(cos 2x cos x)
(b)
/
We can see that our answers are plausible, since f has horizontal tangents where f (x)=0 , and f
/
//
has horizontal tangents where f (x)=0 .
x
3
/
22. (a) f (x)=e x x
2
f (x)=e 3x //
x
f (x)=e 6x
(b)
The graphs seem reasonable since f has horizontal tangents where f
is increasing, and f
/
/
is zero, f
/
is positive where f
is negative where f is decreasing; and the same relationships exist between f
/
//
and f .
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
y
///
=
3
(2x+3)
2
24. y=
x
2x 1
1
(2x+3)
2
/
1/2
23. y= 2x+3 =(2x+3) y =
1/2
1/2
2=(2x+3)
1
(2x+3)
2
//
y =
3/2
3/2
2= (2x+3)
5/2
5/2
2=3(2x+3)
(2x 1)(1) x(2)
/
y =
2
(2x 1)
3
//
1
=
2
or 1(2x 1) 2
(2x 1)
3
y = 1( 2)(2x 1) (2)=4(2x 1) y
///
4
///
f (t)=t( sin t)+cos t 1
///
//
f (t)=t( cos t) sin t 1 sin t
(t)=tsin t cos t 1 cos t cos t=tsin t 3cos t , so f
/
26. g(x)= 5 2x g
4
/
25. f (t)=tcos t f
4
=4( 3)(2x 1) (2)= 24(2x 1) or 24/(2x 1)
(x)=
g (x)=
3
(5 2x)
2
1
(5 2x)
2
/
///
( 2)= (5 2x)
( 2)= 3(5 2x)
2
f ( )= csc 27. f ( )=cot 1/2
1/2
5/2
5/2
///
///
, so g
(0)=0 3= 3 .
//
g (x)=
1
(5 2x)
2
3/2
3/2
( 2)= (5 2x)
5/2
= 3 .
(2)= 3(1)
//
2
f ( )= 2csc ( csc cot )=2csc cot 2
2
2
2
2
2
( )=2( 2csc cot )cot +2csc ( csc )= 2csc (2cot +csc )
///
2
2
2
f
= 2(2) [2( 3 ) +(2) ]= 80
6
f
/
28. g(x)=sec x
g (x)=sec x tan x
//
2
3
2
3
2
3
g (x)=sec x sec x+tan x(sec x tan x)=sec x+sec x tan x=sec x+sec x(sec x 1)=2sec x sec x
///
2
2
///
g (x)=6sec x(sec x tan x) sec x tan x=sec x tan x(6sec x 1) g
= 2 (1)(6 2 1)=11 2
4
2
2
/
29. 9x +y =9
//
y = 9
18x+2yy =0
y 1 x y
/
= 9
2
y
/
2yy = 18x
y x( 9x/y)
2
/
y = 9x/y
2
= 9
y
2
2
2
y +9x
y
//
3
= 9
9
y
3
[ since x and y must satisfy the
3
original equation, 9x +y =9] . Thus, y = 81/y .
30.
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
y
+
=0
2 x 2 y
x + y =1
y
1
x
//
/
y (2 y )
=
/
1
1
y
(2 x )
y
/
y =
x
1
y
x
=
y
x
2x
1
x
y
y
1+
x
=
2x
x
x+ y
1
=
=
since x and y must satisfy the original equation, x + y =1 .
2x x
2x x
3
3
31. x +y =1
2
2
/
2
x
/
3x +3y y =0
y =
2
y
2
2
2
y (2x) x 2yy
//
y =
2
2
2xy 2x y
/
2
4
y
=
22
x
=
4
(y )
4
2xy +2x y
6
y
3
=
6
y
3
3
2xy(y +x )
2x
=
y
y
,
5
3
since x and y must satisfy the original equation, x +y =1 .
4
4
4
3
32. x +y =a 3
/
3
2
3
2
y =
/
3
/
y
= 3x y n
34. f (x)=
f
///
/
f
///
//
f (x)=n(n 1)x
4
3
/
2x
3 2x
2x
(x)=2 2e =2 e 1/2
f (x)=
1
2
1
2
3/2
= 3x f
( n)
a
y
4
7
4 2
=
3a x
y
7
(x)=n(n 1)(n 2) 2 1x
3
//
n n
=n!
2
f (x)=( 1)( 2)(5x 1) 5 (n)
n
(n+1)
n
f (x)=( 1) n!5 (5x 1)
//
f (x)=
x
4 3
2
2x
2 2x
f (x)=2 2e =2 e /
n 2
4
y y
2
/
f (x)=2e 2
4
y +x
2
= 3x f (x)= 1(5x 1) 5
(x)=( 1)( 2)( 3)(5x 1) 5 36. f (x)= x =x
//
n 1
1
1
=(5x 1) 5x 1
2x
3
y
f (x)=nx
35. f (x)=e 3
6
(y )
33. f (x)=x x
2 2
32
3 3
y = x /y y x
/
y 3x x 3y y
//
3
4y y = 4x 4x +4y y =0
f
(n)
1 x
2
///
n 2x
f (x)=2 e
1/2
(x)=
1
2
1
2
3
2
x
5/2
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
f
f
f
( 4)
( 5)
( n)
(x)=
(x)=
1
2
1
2
1
(x)=
2
1
2
1
2
1
2
3
37. f (x)=1/(3x )=
f
f
///
(x)=
( n)
3
2
3
2
3
2
1 3
x 3
5
2
7/2
=
1 3 5
4
7
2
x
/
7/2
9/2
=
1 3 5 7
5
x
9/2
2
1
n+1
2
f (x)=
x
2
5
2
1
6
( 3)( 4)( 5)x 3
x
x
( 2n 1 ) /2
n 1
=( 1)
1 3 5 (2n 3)
x
n
( 2n 1 ) /2
2
1
4
( 3)x 3
//
f (x)=
1
5
( 3)( 4)x 3
n
n
1
( n+3 ) ( 1) 3 4 5 (n+2) 2 ( 1) (n+2)!
(x)= ( 3)( 4) [ (n+2)]x
=
=
n+3
n+3
3
2
3x
6x
2
D sin x= sin x
38. Dsin x=cos x
3
4
D sin x= cos x
D sin x=sin x . The derivatives of sin x occur in
74
2
2
2
a cycle of four. Since 74=4(18)+2 , we have D sin x=D sin x= sin x .
/
//
3
3
39. Let f (x)=cos x . Then Df (2x)=2 f (2x) , D f (2x)=2 f (2x) , D f (2x)=2 f
(n)
///
(2x) , ... ,
n (n)
D f (2x)=2 f (2x) . Since the derivatives of cos x occur in a cycle of four, and since 103=4(25)+3 ,
we have f
(103)
x
40. f (x)=xe f
///
(3)
103
(x)= f (x)=sin x and D
x
/
x
n
x
x
x
f (x)=( 1) (x n)e
1000
So D
x
x
f (x)=x( e )+e =(1 x)e x
(x)=(x 2)( e )+e =(3 x)e (n)
103 (103)
cos 2x=2
(4)
f
103
(2x)=2
x
//
sin 2x .
x
x
f (x)=(1 x)( e )+e ( 1)=(x 2)e x
x
x
f (x)=(3 x)( e )+e ( 1)=(x 4)e .
x
xe =(x 1000)e .
41. By measuring the slope of the graph of s= f (t) at t=0 , 1 , 2 , 3 , 4 , and 5 , and using the method
/
of Example 1 in Section 3.2, we plot the graph of the velocity function v= f (t) in the first figure. The
//
/
acceleration when t=2 s is a= f (2) , the slope of the tangent line to the graph of f when t=2 . We
27
//
/
2
=9 ft / s . Similar measurements
estimate the slope of this tangent line to be a(2)= f (2)=v (2)
3
enable us to graph the acceleration function in the second figure.
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
.
42. (a) Since we estimate the velocity to be a maximum at t=10 , the acceleration is 0 at t=10 .
2
(b) Drawing a tangent line at t=10 on the graph of a , a appears to decrease by 10 ft / s over a
/
2
3
period of 20 s. So at t=10 s, the jerk is approximately 10/20= 0.5 (ft/s ) s or ft / s .
3
2
/
43. (a) s=2t 15t +36t+2
2
/
v(t)=s (t)=6t 30t+36
(b) a(1)=12 1 30= 18 m / s
a(t)=v (t)=12t 30
2
2
2
2
(c) v(t)=6(t 5t+6)=6(t 2)(t 3)=0 when t=2 or 3 and a(2)=24 30= 6 m / s , a(3)=36 30=6 m / s .
3
2
/
44. (a) s=2t 3t 12t
2
/
v(t)=s (t)=6t 6t 12
a(t)=v (t)=12t 6
2
(b) a(1)=12 1 6=6 m / s
2
(c) v(t)=6(t t 2)=6(t+1)(t 2)=0 when t= 1 or 2 . Since t 0 , t 1 and
2
a(2)=24 6=18 m / s .
45. (a) s=sin
/
v(t)=s (t)=cos
/
a(t)=v (t)=
(b) a(1)=
6
6
6
t 6
sin
2
36
t +cos
sin
(c) v(t)=0 for 0 t 2
6
6
6
t
sin
t , 0 t 2 .
6
6
6
6
cos
1 +cos
cos
t 6
6
=
6
t =sin
t 1
=
6
=
6
cos
t
2
36
6
2
36
t sin
sin
3
1
+
2
2
6
t +cos
6
=
t
6
t
2
72
(1+ 3 ) 0.3745 m / s
2
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
sin
6
1=
cos
tan
6
6
t
t
t =1
1
6
2
3
a( )=
2
36
sin
3
t=tan 1
3
6 2
2
6 +cos
4
=
3
=1.5 s. Thus,
2
3
6 2
/
46. (a) s=2t 7t +4t+1
2
2
2
2
+
2
2
36
=
2
2
36
2 0.3877 m / s .
1
3
14= 10 m / s ,
/
a(t)=v (t)=12t 14
2
)
2
=
v(t)=s (t)=6t 14t+4
(b) a(1)=12 14= 2 m / s
(
t=
(c) v(t)=2 3t 7t+2 =2(3t 1)(t 2)=0 when t=
1
or 2 and a
3
1
3
=12
2
2
a(2)=12(2) 14=10 m / s .
4
3
/
3
2
/
2
47. (a) s(t)=t 4t +2 v(t)=s (t)=4t 12t a(t)=v (t)=12t 24t=12t(t 2)=0 when t=0 or 2 .
(b) s(0)=2 m, v(0)=0 m / s, s(2)= 14 m, v(2)= 16 m / s
3
2
/
2
/
48. (a) s(t)=2t 9t v(t)=s (t)=6t 18t (b) s(1.5)= 13.5 m, v(1.5)= 13.5 m / s
3
2
49. (a) s= f (t)=t 12t +36t , t 0
a(t)=v ( t ) =12t 18=0 when t=1.5 .
/
2
v(t)= f (t)=3t 24t+36 .
/
2
a(t)=v (t)=6t 24 . a(3)=6(3) 24= 6 ( m/s ) / s or m / s .
(b)
(c) The particle is speeding up when v and a have the same sign. This occurs when 2<t<4 and when
t>6 . It is slowing down when v and a have opposite signs; that is, when 0 t<2 and when 4<t<6 .
50. (a) x(t)=
t
1+t
/
a(t)=v (t)=
2
/
v(t)=x (t)=
23
22
(1+t )
2
2t (t 3)
2
(1+t )(1) t(2t)
. a(t)=0
2
2t (t 3)=0
=
1 t
2
22
.
(1+t )
t=0 or 3
(1+t )
(b)
8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
(c) v and a have the same sign and the particle is speeding up when 1<t< 3 . The particle is slowing
down and v and a have opposite signs when 0<t<1 and when t> 3 .
/
/
v(t)=y (t)=A cos t 51. (a) y(t)=Asin t 2
2
2
sin t
a(t)=v (t)= A
2
(b) a(t)= A sin t= (Asin t)= y(t) , so a(t) is proportional to y(t) .
(c) speed = v(t) =A cos t is a maximum when cos t=
1 . But when cos t=
1 , we have
2
2
sin t= A
sin t=0 , and a(t)= A
(0)=0 .
dv dv ds dv
dv
=
=
v(t)=v(t)
. The derivative dv/dt is the rate of change
dt ds dt ds
ds
of the velocity with respect to time (in other words, the acceleration) whereas the derivative dv/ds is
the rate of change of the velocity with respect to the displacement.
52. By the Chain Rule, a(t)=
2
/
//
//
53. Let P(x)=ax +bx+c . Then P (x)=2ax+b and P (x)=2a . P (2)=2
/
P (2)=3
P(2)=5
2(1)(2)+b=3
4+b=3
2
1(2) +( 1)(2)+c=5
3
2a=2
a=1 .
b= 1 .
2
c=3 . So P(x)=x x+3 .
2+c=5
2
/
2
//
54. Let Q(x)=ax +bx +cx+d . Then Q (x)=3ax +2bx+c , Q (x)=6ax+2b and Q
/
//
Q ( 1 ) =a+b+c+d=1 , Q ( 1 ) =3a+2b+c=3 , Q (1)=6a+2b=6 and Q
///
///
(x)=6a . Thus,
(1)=6a=12 . Solving these four
3
2
equations in four unknowns a , b , c and d we get a=2 , b= 3 , c=3 and d= 1 , so Q(x)=2x 3x +3x 1
.
/
//
//
/
55. y=Asin x+Bcos x y =Acos x Bsin x y = Asin x Bcos x . Substituting into y +y 2y=sin x
gives us ( 3A B)sin x+(A 3B)cos x=1sin x , so we must have 3A B=1 and A 3B=0 . Solving for A
1
3
and B , we add the first equation to three times the second to get B=
and A=
.
10
10
2
56. y=Ax +Bx+C //
/
/
y =2Ax+B
//
y =2A . We substitute these expressions into the equation
2
y +y 2y=x to get
2
2
(2A)+(2Ax+B) 2(Ax +Bx+C) = x
2
2A+2Ax+B 2Ax 2Bx 2C
= x2
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
2
2
( 2A)x +(2A 2B)x+(2A+B 2C) = (1)x +(0)x+(0)
2
The coefficients of x on each side must be equal, so 2A=1
1
and 2A+B 2C=0
2
A=B=
rx
/
rx
x
/
//
1
1
2C=0
2
2 rx
//
1
. Similarly, 2A 2B=0
2
A=
3
.
4
C=
/
2 rx
rx
rx
rx
2
rx
57. y=e y =re y =r e , so y +5y 6y=r e +5re 6e =e (r +5r 6)=e (r+6)(r 1)=0
(r+6)(r 1)=0 r=1 or 6.
x
2 x
//
/
//
58. y=e y = e y = e . Thus, y+y =y 1
5
x
=
, since e 0.
2
2
59. f (x)=xg(x )
//
/
/
/
2
2
x
x
2 x
e + e = e 2
2
/
2
3
//
x
2
e ( 1)=0
f (x)=x g (x ) 2x+g(x ) 1=g(x )+2x g (x )
2
2
//
2
/
2
/
2
2
f (x)=g (x ) 2x+2x g (x ) 2x+g (x ) 4x=6xg (x )+4x g (x )
g(x)
60. f (x)=
x
2
//
f (x)=
/
/
f (x)=
xg (x) g(x)
2
x
/
//
/
/
x [g (x)+xg (x) g (x)] 2x[xg (x) g(x)]
x
4
2
=
//
/
x g (x) 2xg (x)+2g(x)
x
3
61.
/
g ( x)
f (x) =g( x ) f (x)=g ( x ) 1 x 1/2=
2
2 x
1 1/2 /
1 1/2
//
2 x g ( x )
x g ( x ) 2
x
//
x
2
2
f (x) =
=
2
(2 x )
/
//
=
/
1/2
//
/
[ x g ( x ) g ( x )]
4x
/
x g ( x ) g ( x )
4x x
62.
10
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
5
3
/
f (x)=3x 10x +5
4
2
//
f (x)=15x 30x 3
2
f (x)=60x 60x=60x(x 1)=60x(x+1)(x 1)
//
So f (x)>0 when 1<x<0 or x>1 , and on these intervals the graph of f lies above its tangent lines;
//
and f (x)<0 when x< 1 or 0<x<1 , and on these intervals the graph of f lies below its tangent lines.
63. (a)
f (x)
1
=
2
(2x+1)
/
f (x)=
2
x +x
2
//
f (x) =
2
(x +x)
2
2
(x +x) ( 2)+(2x+1)(2)(x +x)(2x+1)
2
4
2
=
2(3x +3x+1)
2
(x +x)
2
f
///
(x) =
3
3
(x +x)
2
2
2
(x +x) (2)(6x+3) 2(3x +3x+1)(3)(x +x) (2x+1)
2
6
(x +x)
3
=
2
6(4x +6x +4x+1)
2
4
(x +x)
2
(4)
f (x) =
4
2
3
2
2
3
(x +x) ( 6)(12x +12x+4)+6(4x +6x +4x+1)(4)(x +x) (2x+1)
2
8
(x +x)
4
=
3
2
24(5x +10x +10x +5x+1)
2
5
(x +x)
(5)
f (x) =?
(b) f (x)=
f
///
1
1
1
= x(x+1) x x+1
4
4
(x)=( 3)(2)x +(3)(2)(x+1) 64. (a) For f (x)=
7x+17
2
2x 7x 4
///
(x)=
144
4
30
4
(2x+1)
2
f
( n)
, a CAS gives us f
n
(x)=( 1) n! x
///
3
//
3
f (x)=2x 2(x+1) ( n+1 )
4
(x)=
(x+1)
3
( n+1 )
2
6(56x +544x 2184x +6184x 6139)
2
4
(2x 7x 4)
(b) Using a CAS we get f (x)=
f
2
/
f (x)= x +(x+1) 7x+17
2
2x 7x 4
=
3
5
+
. Now we differentiate three times to obtain
2x+1 x 4
.
(x 4)
65.
2 x
/
2 x
x
x
2
For f (x)=x e , f (x)=x e +e (2x) =e (x +2x). Similarly, we have
11
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
//
x
2
x
2
(4)
x
2
(5)
x
2
f (x) =e (x +4x+2)
///
f
(x) =e (x +6x+6)
f (x) =e (x +8x+12)
f (x) =e (x +10x+20)
It appears that the coefficient of x in the quadratic term increases by 2 with each differentiation. The
pattern for the constant terms seems to be 0=1 0, 2=2 1, 6=3 2, 12=4 3, 20=5 4. So a reasonable
(n)
guess is that f (x)=e
x
2
x +2nx+n(n 1) .
(n)
Proof: Let S be the statement that f (x)=e
x
2
x +2nx+n(n 1) . 1. S is true because
n
/
x
1
2
f (x)=e (x +2x).
(k)
2. Assume that S is true; that is, f (x)=e
x
k
f
d
(x) = dx
This shows that S
k+1
(k)
(k+1)
=e
x
2
x +2kx+k(k 1) . Then
x
2
f (x) =e (2x+2k)+ x +2kx+k(k 1) e
2
2
x +(2k+2)x+(k +k) =e
x
x
2
x +2(k+1)x+(k+1)k
is true.
(n)
3. Therefore, by mathematical induction, S is true for all n; that is, f (x)=e
n
x
2
x +2nx+n(n 1) for
every positive integer n.
/
//
F =( f
//
/
/
/
/
//
//
g+ f g )+( f g + fg )= f
///
///
//
/
//
/
/
/
/
g+2 f g + fg
//
/
/
/
F = f g+ fg 66. (a) Use the Product Rule repeatedly: F= fg
//
//
.
///
///
//
/
/
//
(b) F = f g+ f g +2( f g + f g )+ f g + fg = f g+3 f g +3 f g + fg
/// /
/// /
// //
// //
/ ///
/ ///
( 4) ( 4)
( 4)
F = f g+ f g +3( f g + f g )+3( f g + f g )+ f g + fg
/// /
// //
/ ///
( 4)
( 4)
= f g+4 f g +6 f g +4 f g + fg
///
(c) By analogy with the Binomial Theorem, we make the guess:
F
( n)
=f
( n)
g+nf
( n 1 )
/
/ ( n 1 )
( n 2 ) //
( n k ) ( k )
( n)
g + n f
g + + n f
g + +nf g
+ fg
2
k
n!
n(n 1)(n 2) (n k+1)
where n =
=
.
k!(n k)!
k!
k
67. The Chain Rule says that
12
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives
dy dy du
=
, so
dx du dx
2
d y
2
=
dx
d
dx
dy
dx
d
du
=
d
dx
=
dy
du
dy du
du dx
d y
2
2
=
dx
3
2
=
du
2
2
du
du
dx
d y
du
2
du
dx
du
dx
d y
3
=
du
3
2
2
2
3
dy d u
+
2
du
dx
2
2
d
+
dx
2
d
dx
+
du
dx
dy d u
2
du
dx
du
dx
2
2
2
2
d y
du
2
du d u d y
+2
+
2
2
dx
dx du
2
2
+
d
du
d
dx
//
///
2
dy
du
dy
du
d u
2
+
dx
2
du
dx
2
d
dx
d u
2
dx
dy
du
d u
2
dx
3
d u dy
+
3 du
dx
3
du d u d y dy d u
+3
+
2
2 du
3
dx
dx du
dx
69. We will show that for each positive integer n , the n th derivative f
/
[Product Rule]
2
2
du
dx
du
dx
dy d u
+
2
du
dx
2
du
dx
d y
d y
d
du
=
du
2
d
dx
2
2
du
dx
d y
2
d d y d
=
=
3
2 dx
dx
dx
dx
du dy d
+
dx du dx
du dy d u d y
+
=
2
2
dx du
dx
du
2
d y
dy
du
2
du
dx
68. From Exercise 65,
d
dx
=
( p 1)
(n)
( p)
exists and equals one of f ,
/
//
///
( p 1)
f ,f ,f
. Since f = f , the first p derivatives of f are f , f , f
, ... , f
, ... , f
,
and f. In particular, our statement is true for n=1 . Suppose that k is an integer, k 1 , for which f is k
times differentiable with f
S={ f , f
so f
///
(k+1)
/
,f
//
, ... , f
(k)
in the set
( p 1)
} . Since f is p times differentiable, every member of S is differentiable,
exists and equals the derivative of some member of S . Thus, f
( p)
(k+1)
is in the set { f
/
,f
//
,
( p)
, ... , f } , which equals S since f = f . We have shown that the statement is true for n=1 and
f
that its truth for n=k implies its truth for n=k+1 . By mathematical induction, the statement is true for
all positive integers n.
13
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions
1. The differentiation formula for logarithmic functions,
d
1
(log x)=
, is simplest when a=e
a
dx
xln a
because ln e=1.
2
/
1
d 2
2x
(x
+10)=
2
2
dx
x +10
x +10
/
1
cos 2. f (x)=ln (x +10)
f (x)=
3. f ( )=ln (cos )
f ( )=
/
4. f (x)=cos (ln x)
f (x)= sin (ln x)
f (x)=
2
10
x
x 1
=log
1/5
5
8. f (x)=ln
5
10
f (x)=
1
(ln x)
5
4/5
1
1
1
or xln 10 (x 1)ln 10
x(x 1)ln 10
d
(ln x)=
dx
1
4/5
5(ln x)
1
=
x
1
5
5x
4
(ln x)
1 1 1
/
f (x)= =
5 x 5x
1
x
/
9. f (x)= x ln x
/
f (x)=
10
1
ln x
5
1/5
x =ln x =
1 sin (ln x)
=
x
x
x log (x 1)
/
7. f (x)= ln x =(ln x) = tan 1
d
3
3
(1 3x)=
or
(1 3x)ln 2 dx
(1 3x)ln 2
(3x 1)ln 2
/
5. f (x)=log (1 3x)
6. f (x)=log
d
sin (cos )=
d
cos f (x)= x
1+ln t
1 ln t
(1 ln t)(1/t) (1+ln t)( 1/t)
+(ln x)
1
2 x
=
1
ln x 2+ln x
+
=
x 2 x
2 x
10. f (t)=
/
f (t)=
2
=
(1/t) (1 ln t)+(1+ln t)
2
(1 ln t)
3
11. F(t)=ln
(2t+1)
4
(1 ln t)
3
2
=
2
t(1 ln t)
4
=ln (2t+1) ln (3t 1) =3ln (2t+1) 4ln (3t 1)
(3t 1)
1
1
6
12
6(t+3)
/
F (t)=3
2 4
3=
, or combined,
.
2t+1
3t 1
2t+1 3t 1
(2t+1)(3t 1)
12.
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions
(
)
2
1
/
h(x)=ln x+ x 1 h (x)=
x
1+
2
=
2
x+ x 1
2
1
2
x 1
x+ x 1
x 1 +x
2
1
=
2
x 1
x 1
a x
=ln (a x) ln (a+x)
a+x
1
1
(a+x) (a x) 2a
/
g (x)=
( 1)
=
=
2 2
a x
a+x
(a x)(a+x)
a x
13. g(x)=ln
y
1
/
14. F(y)=yln (1+e )
F (y)=y
y
1+e
15. f (u)=
f (u) =
=
e +ln (1+e ) 1=
y
1+e
y
+ln (1+e )
y
ln u
1+ln (2u)
1+ln (2u) /
y
ye
y
1
1
ln u
2
u
2u
1+ln (2u)
1+(ln 2+ln u) ln u
2
u 1+ln (2u)
4
2
2
=
1
1+ln (2u) ln u
u
1+ln (2u)
2
1+ln 2
=
u 1+ln (2u)
4
2
2
2
17. y=ln |2 x 5x |
1
/
y =
2
2
x
3u+2 1
=
ln (3u+2) ln (3u 2) 3u 2 2
x
x
10x+1
or
2
5x +x 2
2 x 5x
2 x 5x
18. G(u)=ln
y =4
10x 1
( 1 10x)=
1
1
4
+2
cos x= +2cot x
x
sin x
x
/
16. y=ln (x sin x)=ln x +ln (sin x) =4ln x+2ln sin x
/
G (u)=
1
2
x
19. y=ln (e +xe )=ln (e (1+x))=ln (e )+ln (1+x)= x+ln (1+x)
x 2
20. y=[ln (1+e )] /
x
y =2[ln (1+e )]
1
1+e
21. y=xln x
/
y =x(1/x)+(ln x) 1=1+ln x
x
x
e=
x
3
3
3u+2 3u 2
/
y = 1+
=
6
2
9u 4
1
1 x+1
x
=
=
1+x
1+x
1+x
x
2e ln (1+e )
1+e
x
//
y =1/x
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions
22. y=
ln x
2
2
x (1/x) (ln x)(2x)
/
y =
2 2
(x )
x
//
y =
3
2
x ( 2/x) (1 2ln x)(3x )
32
x
10
x
/
1
1
=
xln 10 ln 10
=
1
x
6ln x 5
//
3
y =
4
1
ln 10
1
=
2
2
x ln 10
x
2
sec xtan x+sec x
y =
=sec x
sec x+tan x
1
//
y =sec xtan x
x
1 ln (x 1)
25. f (x)=
1
x 1
[1 ln (x 1)] 1 x
f (x) =
=
1 2ln x
x
/
24. y=ln (sec x+tan x)
/
6
=
x
x ( 2 3+6ln x)
=
x
y =
4
2
(x )
23. y=log
x(1 2ln x)
=
2
=
(x 1)[1 ln (x 1)]+x
x 1
=
2
[1 ln (x 1)]
2x 1 (x 1)ln (x 1)
x 1 (x 1)ln (x 1)+x
[1 ln (x 1)]
(x 1)
2
2
(x 1)[1 ln (x 1)]
Dom( f ) ={ x| x 1>0 and 1 ln (x 1) 0}={ x| x>1 and ln (x 1) 1}
1
={ x| x>1 and x 1 e }={ x| x>1 and x 1+e}=(1, 1+e) (1+e, 26. f (x) =
1
1+ln x
/
f (x) = 1/x
2
(1+ln x)
{ x| x>0 and x 1/e} = (0, 1/e) (1/e, ).
2
2
27. f (x) = x ln (1 x )
/
1
=
2
f (x) = 2xln (1 x )+
2
)
. Dom( f ) = { x| x>0 and ln x 1} =
x(1+ln x)
2
x ( 2x)
2
2
= 2xln (1 x )
1 x
2x
3
2
.
1 x
2
Dom( f ) = { x|1 x >0} = { x | | x|<1} = ( 1, 1).
1
1 1
/
28. f (x) = ln ln ln x f (x) =
.
ln ln x ln x x
Dom( f ) = { x|ln ln x>0} = { x|ln x>1} = { x| x>e} = (e, ).
29. f (x) =
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions
x
ln x
ln x x(1/x)
/
f (x) =
2
2
(ln x)
2
1
x
1
ln x
1
x
/
2
=0
1
2
f (x) =
1 1
/
= 2xln x+x
/
f (e) =
f (1) = 2ln 1+1 = 1
1
, so an equation of the tangent line at
e
1
1
(x e), or y = x 1, or x ey = e.
e
e
(e, 0) is y 0 =
/
3
32. y=ln (x 7)
y =
1
3
x 7
y 0=12(x 2) or y=12x 24.
2
/
3x y (2)=
12
=12, so an equation of a tangent line at (2, 0) is
8 7
/
33. f (x)=sin x+ln x
/
f (e) =
f (x) = 2xln x+x
31. y = f (x) = ln ln x
when f
/
(ln x)
/
30. f (x) = x ln x
34. y=
ln x 1
=
f (x)=cos x+1/x. This is reasonable, because the graph shows that f increases
/
is positive, and f (x)=0 when f has a horizontal tangent.
ln x
x
/
y =
x(1/x) ln x
2
=
1 ln x
2
/
. y (1)=
1 0
2
/
=1 and y (e)=
x
x
1
lines are y 0=1(x 1) or y=x 1 and y 1/e=0(x e) or y=1/e.
5
4
6
5
35. y=(2x+1) (x 3) 4
y =y
3
10
24x
+ 4
2x+1
x 3
5
4
6
=(2x+1) (x 3)
2
=0
equations of tangent
e
6
ln y=ln ((2x+1) (x 3) )
1 /
1
4
ln y=5ln (2x+1)+6ln (x 3)
y =5
2+6
y
2x+1
/
1 1
1
4
x 3
3
4x 3
10
24x
+ 4
2x+1
x 3
.
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions
2
x
2
2
10
x
36. y= x e (x +1) 1 / 1 1
y = +2x+10
y
2 x
2
37. y=
1
2x
2
2
2
2
/
10
2
10
1
2
2
ln x+x +10ln (x +1)
2
1
20x
+2x+ 2
2x
x +1
x
ln y=
y = x e (x +1)
x +1
4
sin xtan x
2
ln y=ln x +ln e +ln (x +1) 2
4
2
2
ln y=ln (sin xtan x) ln (x +1) (x +1)
2
4
2
2
2
ln y=ln (sin x) +ln (tan x) ln (x +1) ln y=2ln |sin x|+4ln |tan x| 2ln (x +1)
1 /
1
1
1
2
y =2
cos x+4
sec x 2 2 2x
y
sin x
tan x
x +1
/
y =
2
2
4
4sec x
2cot x+
tan x
sin xtan x
2
2
(x +1)
2
38. y=
x +1
4
2
x 1
2
x +1
/ 4
y =
2
x 1
x
/
x
2
sin x
+ln xcos x
x
42. y=(sin x) x
43. y=(ln x) 1
=
2
2
x 1
1 / 1
y = y
4
2
4
2x
x +1
2
4
x 1
x 1
/
/
y
1
=
y x
1
x
+(ln x) 1
2
1
1
4
2x
2
x +1
1 x
2
x 1
x +1
4
2
4
x 1
y =y(1+ln x) /
1/x
y =x
x
2x
2
x
=
1
/
y /y=x(1/x)+(ln x) 1 ln y=xln x
ln y=ln x
x
x
x +1
sin x
2
x +1
1
1
2
2
ln (x +1)
ln (x 1)
4
4
1
ln y= ln x
x
sin x
y =y
1
2
x
1/x
41. y=x
ln y=
ln y=ln x 39. y=x 40. y=x
4x
/
x
y =x (1+ln x)
1 ln x
2
x
/
ln y=sin xln x
/
sin x
y =x
y
1
=(sin x) +(ln x)(cos x)
y
x
sin x
+ln xcos x
x
/
ln y=xln (sin x)
x
ln y=ln (ln x) y
1
=x
cos x+ ln (sin x) 1
y
sin x
/
x
y =(sin x) xcot x+ln (sin x)
/
ln y=xln ln x
y
1 1
=x
+(ln ln x) 1
y
ln x x
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions
x
+ln ln x xln x
/
y =y
44. y=x
ln x
e
2
d 2 2
(x +y ) dx
1
/
y =
2
2
x +y
/
/
2
x y +y y 2yy =2x y
x
48. x =y ///
3
2
/
2x+2yy
ln x+
1
x
y =
2
2
2
2
1
/
4
2
/
2
/
/
x y +y y =2x+2yy x +y 2y
//
f (x)=1/(x 1)=(x 1) /
x +y
2x
1
1
/
/
+(ln x) y =x y +ln y
x
y
y
/
y ln x
2
x /
y
y =ln y
y
x
/
y =
ln y y/x
ln x x/y
f (x)= (x 1) f (x)= 2 3(x 1) n
n 1
n 1
(n)
y =
/
(4)
(x)=2(x 1) /
(x +y 2y) y =2x yln x=xln y
49. f (x)=ln (x 1)
f
y =x e
y
1 1
=cos x
+(ln ln x)( sin x)
y
ln x x
ln y=cos xln (ln x)
47. y=ln (x +y )
2
x
e x
2ln x
x
ln x
cos x
sin xln ln x
xln x
cos x
/
/
y =x
/
y =(ln x)
2
/
/
ln y=e ln x
46. y=(ln x)
1
x
y
x 1
x
=e +(ln x) e y
x
x
cos x
2
/
y
=2ln x
y
ln y=ln xln x=(ln x) 45. y=x 1
+ln ln x
ln x
x
2
x
/
/
y =(ln x)
f (x)=( 1) 2 3 4 ... (n 1)(x 1) =( 1)
(n 1)!
n
(x 1)
8
9
8
/
8
7
7
7
50. y=x ln x , so D y=D y =D (8x ln x+x ). But the eighth derivative of x is 0 , so we now have
8
7
7
6
6
7
6
D (8x ln x) = D (8 7x ln x+8x )=D (8 7x ln x)
= D6(8 7 6x5ln x)=...=D(8! x0ln x)=8!/x.
1
/
, so f (0)=1. Thus,
1+x
ln (1+x)
f (x)
f (x) f (0)
/
lim
=lim
=lim
= f (0)=1.
x
x 0
x 0
x 0 x
x 0
/
51. If f (x)=ln (1+x), then f (x)=
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions
52. Let m=n/x. Then n=xm, and as n , m . Therefore,
x n
1 mx
1 m x x
lim 1+
= lim
1+
= lim
1+
=e by Equation 6.
n
m
m
n m m 7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
1 0 0
(e e )=0
2
1 0 0 1
(b) cosh 0= (e +e )= (1+1)=1
2
2
1. (a) sinh 0=
2. (a) tanh 0=
0
0
0
0
(e e )/2
=0
(e +e )/2
(b) tanh 1=
1
1
1
1
e e
2
=
(b) sinh 2=
2
0.76159
e +1
e +e
3. (a) sinh (ln 2)=
e 1
e
ln 2
ln 2
e
2
=
e
ln 2
(e
2
ln 2 1
)
1
=
2 2
=
2
2
1
2 3
=
2
4
1 2 2
(e e ) 3.62686
2
1 3 3
(e +e ) 10.06766
2
1
ln 3 ln 3 3+
e +e
3 5
(b) cosh (ln 3)=
=
=
2
2
3
4. (a) cosh 3=
5. (a) sech 0=
1
1
1
= =1
cosh 0 1
(b) cosh 1=0 because cosh 0=1 .
6. (a) sinh 1=
1 1 1
(e e ) 1.17520
2
1
(
2
)
(b) Using Equation 3, we have sinh 1=ln 1+ 1 +1 =ln (1+ 2 ) 0.88137 .
7. sinh ( x)=
1 x e e
2
8. cosh ( x)=
1 x e +e
2
9. cosh x+sinh x=
( x)
( x)
=
1 x x
1 x x
(e e )=
(e e )= sinh x
2
2
=
1 x x 1 x x
(e +e )= (e +e )=cosh x
2
2
1 x x 1 x x 1
x
x
(e +e )+ (e e )= (2e )=e
2
2
2
1
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
10. cosh x sinh x=
1 x x 1 x x 1
x
(e +e )
(e e )= (2e )=e
2
2
2
1 x x
(e e )
2
11. sinh xcosh y+cosh xsinh y=
x
1 y y
(e +e ) +
2
1 x x
(e +e )
2
1 y y
(e e )
2
1 x x
(e e )
2
1 y y
(e e )
2
1
x+y x y x+y x y
x+y x y x+y x y
(e +e e
e )+(e e +e
e )
4
1
x+y
x y 1
x+y (x+y)
= (2e 2e )=
e e
=sinh (x+y)
4
2
=
1 x x
(e +e )
2
12. cosh xcosh y+sinh xsinh y=
1 y y
(e +e ) +
2
1
x+y x y x+y x y
x+y x y x+y x y
(e +e +e +e )+(e e e +e )
4
1
x+y
x y 1
x+y (x+y)
= (2e +2e )=
e +e
=cosh (x+y)
4
2
=
2
2
2
13. Divide both sides of the identity cosh x sinh x=1 by sinh x :
2
cosh x
2
sinh x
14.
2
sinh x
2
sinh x
=
1
2
2
2
coth x 1=csch x.
sinh x
sinh xcosh y cosh xsinh y
+
sinh (x+y) sinh xcosh y+cosh xsinh y
cosh xcosh y cosh xcosh y
tanh (x+y) =
=
=
cosh (x+y) cosh xcosh y+sinh xsinh y cosh xcosh y sinh xsinh y
+
cosh xcosh y cosh xcosh y
tanh x+tanh y
=
1+tanh xtanh y
15. Putting y=x in the result from Exercise 11, we have
sinh 2x=sinh (x+x)=sinh xcosh x+cosh xsinh x=2sinh xcosh x.
16. Putting y=x in the result from Exercise 12, we have
2
2
cosh 2x=cosh (x+x)=cosh xcosh x+sinh xsinh x=cosh x+sinh x .
17.
ln x
ln x
sinh (ln x) (e e
= ln x tanh (ln x) =
cosh (ln x)
(e +e
)/2
ln x
)/2
=
x (e
ln x 1
x+(e
)
ln x 1
)
=
x x
1
x+x
1
2
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
2
2
x 1/x (x 1)/x x 1
=
= 2
= 2
x+1/x
(x +1)/x x +1
18.
1 x x 1 x x
(e +e )+ (e e )
1+tanh x
1+(sinh x)/cosh x cosh x+sinh x 2
2
=
=
=
1 tanh x
1 (sinh x)/cosh x cosh x sinh x 1 x x 1 x x
(e +e )
(e e )
2
2
x
=
x
x
x
e +e +e e
x
x
x
x
e +e e +e
=
2e
x
2x
x
=e
2e
x
cosh x+sinh x e
2x
Or: Using the results of Exercises 9 and 10,
=
=e
x
cosh x sinh x
e
n
xn
nx
19. By Exercise 9, (cosh x+sinh x) =(e ) =e =cosh nx+sinh nx .
4
9
25
5
2
2
. cosh x=sinh x+1=
+1=
cosh x= (since cosh x>0 ).
3
16
16
4
4
3/4 3
5
coth x=1/cosh x= , tanh x=sinh x/cosh x=
= , and cothx=1/tanh x= .
5
5/4 5
3
20. sinh x=
3
4
csch x=1/sinh x=
4
5
4 2 9
3
2
2
>0 , so x>0 . cothx=1/tanh x= , sech x=1 tanh x=1
=
sech x=
5
4
5
25
5
5
4 5 4
3
(since sech x>0 ), cosh x=1/sech x= , sinh x=tanh xcosh x= = , and csch x=1/sinh x= .
3
5 3 3
4
21. tanh x=
22.
23. (a)
3
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
x
x
lim tanh x=lim
x
x
x
e e
x
x
x
e
x
x
x
x
e
e e
(b) lim tanh x= lim
=lim
x
e +e
x
x
2x
e
e +e
e
1 e
2x
1+e
1 0
=1
1+0
2x
x
x
=
= lim
x
e 1
2x
=
e +1
0 1
= 1
0+1
x
x
e e
=
(c) lim sinh x=lim
2
x x x
x
e e
= (d) lim sinh x= lim
2
x x 2
(e) lim sech x=lim
=0
x
x x e +e x
x
x
(f) lim cothx=lim
x
e +e
x
x
x
2x
=lim
1+e
2x
x 1 e
e e
e
cosh x
(g) lim cothx=lim
= , since sinh x
+
+ sinh x
x
x
e
x
0
x
x
0
x
cosh x
= sinh x
0
2
(i) lim csch x= lim
x
24. (a)
1+0
=1
1 0
0 through positive values and cosh x
1.
0
(h) lim cothx=lim
=
x
x
(c)
d
d
csch x=
dx
dx
(d)
d
d
sech x=
dx
dx
0 through negative values and cosh x
1.
=0
e e
1 x x
1 x x
(e +e ) = (e e )=sinh x
2
2
d
d
cosh x=
dx
dx
d
d
tanh x=
(b)
dx
dx
x
, since sinh x
sinh x
cosh x
=
cosh xcosh x sinh xsinh x
2
2
=
2
cosh x sinh x
2
cosh x
cosh x
1
cosh x
1
cosh x
=
=
= csch xcoth x
2
sinh x
sinh x sinh x
sinh x
1
sinh x
1
sinh x
=
=
= sech xtanh x
2
cosh x
cosh x cosh x
cosh x
1
=
2
2
=sech x
cosh x
(e)
d
d
coth x =
dx
dx
cosh x
sinh x
=
sinh xsinh x cosh xcosh x
2
sinh x
2
=
2
sinh x cosh x
2
sinh x
=
1
2
sinh x
2
= csch x
4
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
1
2
2
25. Let y=sinh x . Then sinh y=x and, by Example 1(a), cosh y sinh y=1
2
y
2
[ with cosh y>0 ]
(
2
cosh y= 1+sinh y = 1+x . So by Exercise 9, e =sinh y+cosh y=x+ 1+x 1
2
2
y=ln x+ 1+x
).
2
26. Let y=cosh x . Then cosh y=x and y
0 , so sinh y= cosh y 1 = x 1 . So, by Exercise 9,
y
(
2
e =cosh y+sinh y=x+ x 1 )
2
y=ln x+ x 1 .
1 y y
Another method: Write x=cosh y= (e +e ) and solve a quadratic, as in Example 3.
2
y
y
2y
y
sinh y (e e )/2 e
e 1
27. (a) Let y=tanh x . Then x=tanh y=
= y y = 2y y
cosh y
(e +e )/2 e e +1
1
2y
2y
2y
2y
2y
xe +x=e 1 1+x=e xe 1+x=e (1 x)
1+x
1
1+x
2y 1+x
e =
2y=ln
y= ln
1 x
1 x
2
1 x
.
1
(b) Let y=tanh x . Then x=tanh y , so from Exercise 18 we have
1+x
1
1+x
2y 1+tanh y 1+x
e =
=
2y=ln
y= ln
.
1 tanh y 1 x
1 x
2
1 x
28. (a)
(i)
1
y=csch x
csch y=x ( x 0 )
1
(ii) We sketch the graph of csch
y=x.
by reflecting the graph of csch (see Exercise 22) about the line
1
(iii) Let y=csch x . Then x=csch y=
2
y
y
e e
2
y
y
xe xe =2
1
csch x=ln
1
+
x
y
x(e ) 2e x=0
1+ x +1
y 1
But e >0 , so for x>0 , e =
and for x<0 , e =
x
y
y
y2
y
e=
1
2
x +1
.
x
2
x +1
. Thus,
x
2
x +1
| x|
.
5
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
(b)
(i)
1
y=sech x sech y=x and y>0.
1
(ii) We sketch the graph of coth by reflecting the graph of coth (see Exercise 22) about the line
y=x.
1
2
(iii) Let y=coth x , so x=sech y=
y
y
y
y
xe +xe =2
y2
y
x(e ) 2e +x=0
y
e=
1
e +e
But y>0
y
e >1 . This rules out the minus sign because
1
2
1 x
.
x
2
1 x
>1 1
x
2
1 x >x
2
2
2
2
1 x> 1 x 1 2x+x >1 x 1+ 1 x
x>1 , but x=sech y 1 . Thus, e =
x
y
2
x >x
2
1+ 1 x
x
1
sech x=ln
.
(c)
(i)
1
y=coth x
coth y=x
1
(ii) We sketch the graph of coth by reflecting the graph of coth (see Exercise 22) about the line
y=x.
1
Let y=coth x . Then x=coth y=
(iii)
2y=ln
x+1
x 1
y
y
e +e
y
y
e e
1
x+1
1
coth x= ln
2
x 1
y
y
y
y
xe xe =e +e y
y
(x 1)e =(x+1)e 2y
e =
x+1
x 1
6
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
1
29. (a) Let y=cosh x . Then cosh y=x and y
0
dy
1
=
=
dx sinh y
1
=
2
1
sinh y
dy
=1
dx
(since sinh y
0 for y
0 ). Or: Use Formula 4.
2
x 1
cosh y 1
1
2
(b) Let y=tanh x . Then tanh y=x
sech y
dy
=1
dx
dy
=
dx
1
2
1
=
sech y
2
1 tanh y
=
1
2
.
1 x
Or: Use Formula 5.
1
csch ycoth y
(c) Let y=csch x. Then csch y=x
2
coth y=
dy
=1
dx
dy
1
=
. By Exercise 13,
dx
csch ycoth y
2
csch y+1 =
2
x +1 . If x>0 , then coth y>0 , so coth y= x +1 . If x<0 , then coth y<0 , so
dy
1
1
2
coth y= x +1 . In either case we have
=
=
.
dx
csch ycoth y
2
| x| x +1
dy
1
(d) Let y=sech x. Then sech y=x sech ytanh y
=1
dx
dy
1
1
1
=
=
=
. (Note that y>0 and so tanh y>0 .)
dx
sech ytanh y
2
2
sech y 1 sech y
x 1 x
dy
1
1
1
2 dy
1
(e) Let y=coth x. Then coth y=x csch y
=1
=
=
=
by Exercise
2
2
2
dx
dx
csch y 1 coth y 1 x
13.
/
2
30. f (x)=tanh 4x
f (x)=4sech 4x
31. f (x)=xcosh x
f (x)=x(cosh x) +(cosh x)(x) =xsinh x+cosh x
2
32. g(x)=sinh x
2
33. h(x)=sinh (x )
/
/
/
/
g (x)=2sinh xcosh x
/
2
2
h (x)=cosh (x ) 2x=2xcosh (x )
34. F(x)=sinh xtanh x
/
2
F (x)=sinh xsech x+tanh xcosh x
1 cosh x
1+cosh x
(1+cosh x)( sinh x) (1 cosh x)(sinh x)
35. G(x)=
/
G (x)
=
2
(1+cosh x)
7
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
=
sinh x sinh xcosh x sinh x+sinh xcosh x
2
(1+cosh x)
t
/
36. f (t)=e sech t 1+t h (t)= csch
t
2
1/2
(2t)=
tcsch
1+t
2 t
t
t
2 t
/
/
1
cosh 3x
sinh 3x 3=3e
/
sinh 3x
1
1
2
y =x cosh 3x
2
1
/
x
y =
2
1 ( x )
1
1
x
2
1 x
2
1
2
1 x =xtanh x+
44. y=xtanh x+ln
/
+
1 x
1
1
2
1
+sinh (2x)
2
1+4x
1/2
=
1
2 x (1 x)
1
2
ln (1 x )
2
1
1
2
( 2x)=tanh x
1 x
2
45. y=xsinh (x/3) 9+x x
/
1
1/3
2x
y =sinh
+x
=sinh
3
2
2
1+(x/3)
2 9+x
46. y=sech
1
x
2+sinh (2x) 2x=2x
1+(2x)
y =tanh x+
2
y =cosh (cosh x) sinh x
y =e
1
2
1
cosh t=coth t
sinh t
/
42. y=x sinh (2x)
43. y=tanh
1+t
H (t)=sech (e ) e =e sech (e )
40. y=sinh (cosh x)
2
1
2
1+t (1+t )
2
2
2
/
t
t
f (t)=
39. H(t)=tanh (e )
cosh 3x
t
/
38. f (t)=ln (sinh t)
41. y=e
2
(1+cosh x)
f (t)=e ( sech ttanh t)+(sech t)e =e sech t (1 tanh t)
2
37. h(t)=coth
2sinh x
=
1
x
3
+
x
2
9+x
x
=sinh
2
9+x
1
x
3
2
1 x 8
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
2x
1
/
y =
2
1 x
1
47. y=coth
2
2
1 (1 x ) 2
2
1
y =
2
(1 x )| x|
1 x
/
x +1 x
=
2x
2
1 (x +1) 2
1
=
2
x +1
x
2
x +1
48.
For y=acosh (x/a) with a>0 , we have the y intercept equal to a. As a increases, the graph flattens.
49. (a) y=20cosh (x/20) 15
/
x=7, we have y (7)=sinh
(b) If 1
=sinh (x/20) . Since the right pole is positioned at
20
/
y =20sinh (x/20)
7
0.3572 .
20
is the angle between the tangent line and the x axis, then tan = slope of the line =sinh
, so =tan
1
7
20
sinh
7
20
0.343 rad 19.66 . Thus, the angle between the line and the pole is
=90 70.34 .
50. We differentiate the function twice, then substitute into the differential equation: y=
dy T
=
sinh
dx g
gx
T
g
gx
=sinh
T
T
2
d y
2
=cosh
dx
2
We evaluate the two sides separately: LHS=
d y
2
dx
RHS=
g
T
1+
dy
dx
2
=
g
T
51. (a) y=Asinh mx+Bcosh mx
//
2
2
1+sinh
2
=
gx
T
T
gx
cosh
g
T
g g
gx
=
cosh
.
T
T
T
g
gx
cosh
,
T
T
gx g gx
=
by the identity proved in Example 1(a).
T
T T
/
y =mAcosh mx+mBsinh mx
2
2
y =m Asinh mx+m Bcosh mx=m (Asinh mx+Bcosh mx)=m y
//
(b) From part (a), a solution of y =9y is y(x)=Asinh 3x+Bcosh 3x. So 4=y(0)=Asinh 0+Bcosh 0=B,
9
Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions
/
so B= 4. Now y (x)=3Acosh 3x 12sinh 3x
sinh x
52. lim
x
e
x
x
=lim
x
x
e e
2e
x
2x
1 e
=lim
2
x =
/
6=y (0)=3A
A=2, so y=2sinh 3x 4cosh 3x .
1 0 1
=
2
2
/
53. The tangent to y=cosh x has slope 1 when y =sinh x=1
1
x=sinh 1=ln (1+ 2 ) , by Equation 3.
2
Since sinh x=1 and y=cosh x= 1+sinh x , we have cosh x= 2. The point is (ln (1+ 2 ), 2 ).
54.
1 ln (sec +tan ) ln (sec +tan )
e
+e
2
1
1
sec tan sec +tan +
=
sec +tan +
sec +tan 2
(sec +tan )(sec tan )
sec tan 1
sec +tan +
=
(sec +tan +sec tan )=sec 2
2
2
sec tan cosh x =cosh [ln (sec +tan )]=
1
2
1
=
2
=
x
x
55. If ae +be = cosh (x+ ) [ or sinh (x+ ) ], then
x
x x+
x x x x
x
ae +be = (e e
)= (e e e e )=
e
e
e
e . Comparing coefficients
2
2
2
2
x
x
of e and e , we have a= e (1) and b=
e (2) . We need to find and . Dividing
2
2
equation (1) by
a
a
1
a
2
equation (2) gives us = e (*)2 =ln (
) = ln . Solving equations (1) and
b
b
2
b
2a
2a
2
and e =
, so
(2) for e gives us e =
=
= 4ab =2 ab .
2b
2b
a
a
(*) If >0 , we use the + sign and obtain a cosh function, whereas if <0 , we use the sign and
b
b
obtain a sinh function.
1
a
x
x
In summary, if a and b have the same sign, we have ae +be =2 ab cosh x+ ln
, whereas, if
2
b
1
a
x
x
a and b have the opposite sign, then ae +be =2 ab sinh x+ ln .
2
b
10