Calculus one and several variables 10E Salas solutions manual ch17
Calculus one and several variables 10E Salas solutions manual ch03
-
date post
27-Oct-2014 -
Category
Documents
-
view
566 -
download
12
description
Transcript of Calculus one and several variables 10E Salas solutions manual ch03
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.1 65
CHAPTER 3
SECTION 3.1
1. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
[2 − 3(x + h)] − [2 − 3x]h
= limh→0
−3hh
= limh→0
−3 = −3
2. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
k − k
h= lim
h→00 = 0
3. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
[5(x + h) − (x + h)2] − (5x− x2)h
= limh→0
5h− 2xh− h2
h= lim
h→0(5 − 2x− h) = 5 − 2x
4. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
[2(x + h)3 + 1] − [2x3 + 1]h
= limh→0
2(x3 + 3x2h + 3xh2 + h3) − 2x3
h= lim
h→0
6x2h + 6xh2 + 2h3
h
= limh→0
(6x2 + 6xh + 2h2) = 6x2
5. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)4 − x4
h
= limh→0
(x4 + 4x3h + 6x2h2 + 4xh3 + h4) − x4
h
= limh→0
(4x3 + 6x2h + 4xh2 + h3) = 4x3
6. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
1x + h + 3
− 1x + 3
h
limh→0
(x + 3) − (x + h + 3)h(x + h + 3)(x + 3)
= limh→0
−h
h(x + h + 3)(x + 3)
limh→0
−1(x + h + 3)(x + 3)
=−1
(x + 3)2
7. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√x + h− 1 −
√x− 1
h
= limh→0
(x + h− 1) − (x− 1)h(√x + h− 1 +
√x− 1 )
= limh→0
1√x + h− 1 +
√x− 1
=1
2√x− 1
8. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
[(x + h)3 − 4(x + h)] − [x3 − 4x]h
= limh→0
3x2h + 3xh2 + h3 − 4hh
= limh→0
(3x2 + 3xh + h2 − 4) = 3x2 − 4
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
66 SECTION 3.1
9. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
1(x + h)2
− 1x2
h
= limh→0
x2 − (x2 + 2hx + h2)hx2(x + h)2
= limh→0
−2x− h
x2(x + h)2= − 2
x3
10. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
1√x + h
− 1√x
h
limh→0
√x−
√x + h
h√x√x + h
= limh→0
(√x−
√x + h
) (√x +
√x + h
)h√x√x + h
(√x +
√x + h
) = limh→0
x− (x + h)h√x√x + h
(√x +
√x + h
)limh→0
−h
h√x√x + h
(√x +
√x + h
) =−1
2x√x
11. f(x) = x2 − 4x; c = 3:
difference quotient:
f(3 + h) − f(3)h
=(3 + h)2 − 4(3 + h) − (−3)
h
=9 + 6h + h2 − 12 − 4h + 3
h=
2h + h2
h= 2 + h
Therefore, f ′(3) = limh→0
f(3 + h) − f(3)h
= limh→0
(2 + h) = 2
12. f(x) = 7x− x2; c = 2:
difference quotient:
f(2 + h) − f(2)h
=7(2 + h) − (2 + h)2 − (10)
h
=14 + 7h− 4 − 4h− h2 − 10
h=
3h− h2
h= 3 − h
Therefore, f ′(2) = limh→0
f(2 + h) − f(2)h
= limh→0
(3 − h) = 3
13. f(x) = 2x3 + 1; c = 1:
difference quotient:
f(−1 + h) − f(−1)h
=2(−1 + h)3 + 1 − (−1)
h
=2
[−1 + 3h− 3h2 + h3
]+ 2
h=
6h− 6h2 + 2h3
h= 6 − 6h + 22
Therefore, f ′(−1) = limh→0
f(−1 + h) − f(−1)h
= limh→0
(6 − 6h + 2h2) = 6
14. f(x) = 5 − x4; c = −1:
difference quotient:
f(1 + h) − f(1)h
=5 − (1 + h)4 − (4)
h
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.1 67
=5 − 1 − 4h− 6h2 − 4h3 − h4 − 4
h
=−4h− 6h2 − 4h3 − h4
h= −4 − 6h− 4h2 − h3
Therefore, f ′(3) = limh→0
f(1 + h) − f(1)h
= limh→0
(−4 − 6h− 4h2 − h3
)= −4
15. f(x) =8
x + 4; c = −2:
difference quotient:
f(−2 + h) − f(−2)h
=
8(−2 + h) + 4
− 4
h=
8h + 2
− 4
h
=8 − 4h− 8h(h + 2)
=−4
h + 2
Therefore, f ′(−2) = limh→0
f(−2 + h) − f(−2)h
= limh→0
−4h + 2
= −2
16. f(x) =√
6 − x; c = 2:
difference quotient:
f(2 + h) − f(2)h
=
√6 − (2 + h) − (2)
h
=√
4 − h− 2h
=√
4 − h− 2h
·√
4 − h + 2√4 − h + 2
=−1√
4 − h + 2
Therefore, f ′(2) = limh→0
f(2 + h) − f(2)h
= limh→0
−1√4 − h + 2
= − 14
17. f(4) = 4. Slope of tangent at (4, 4) is f ′(4) = −3. Tangent y − 4 = −3(x− 4).
18. f(4) = 2. Slope of tangent at (4, 2) is f ′(4) =1
2√
4=
14. Tangent y − 2 =
14(x− 4).
19. f(−2) =14. Slope of tangent at (−2,
14) is
14. Tangent: y − 1
4=
14(x + 2).
20. f(2) = −3. Slope of tangent at (2,−3) is f ′(2) = −3(2)2 = −12. Tangent: y + 3 = −12(x− 2) ;
21. (a) f is not continuous at c = −1 and c = 1; f has a removable discontinuity at c = −1 and a jump
discontinuity at c = 1.
(b) f is continuous but not differentiable at c = 0 and c = 3.
22. (a) f is not continuous at c = 2; f has a jump discontinuity at 2
(b) f is continuous but not differentiable at c = −2 and c = 3.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
68 SECTION 3.1
23. at x = −1 24. at x = 52 25. at x = 0
26. at x = −2, 2 27. at x = 1 28. at x = 2
29. f ′(1) = 4
limh→0−
f(1 + h) − f(1)h
= limh→0−
4(1 + h) − 4h
= 4
limh→0+
f(1 + h) − f(1)h
= limh→0+
2(1 + h)2 + 2 − 4h
= 4
30. f ′(1) = 6
limh→0−
f(1 + h) − f(1)h
= limh→0−
3(1 + h)2 − 3h
= 6
limh→0+
f(1 + h) − f(1)h
= limh→0+
[2(1 + h)3 + 1] − 3h
= 6
31. f ′(−1) does not exist
limh→0−
f(−1 + h) − f(−1)h
= limh→0−
h− 0h
= 1
limh→0+
f(−1 + h) − f(−1)h
= limh→0+
h2 − 0h
= 0
32. f ′(3) does not exist because f is not continuous at x = 3.
33. 34. 35.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.1 69
36. 37. 38.
39. Since f(1) = 1 and limx→1+
f(x) = 2, f is not continuous at 1. Therefore, by (3.1.3), f is not differentiable
at 1.
40. (a) f ′(x) =
{2(x + 1), x < 0
2(x− 1), x > 0.
(b) limh→0−
f(0 + h) − f(0)h
= limh→0−
(h + 1)2 − 1h
= limh→0−
(h + 2) = 2,
limh→0+
f(0 + h) − f(0)h
= limh→0+
(h− 1)2 − 1h
= limh→0+
(h− 2) = −2.
41. Continuity at x = 1 : limx→1−
f(x) = 1 = limx→1+
f(x) = A + B. Thus A + B = 1.
Differentiability at x = 1:
limh→0−
f(1 + h) − f(1)h
= limh→0−
(1 + h)3 − 1h
= 3 = limh→0+
f(1 + h) − f(1)h
= A
Therefore, A = 3, =⇒ B = −2.
42. Continuity at x = 2 =⇒ 4B + 2A = 2; differentiability at x = 2 =⇒ 4B + A = 4;
A = −2, B = 32
43. f(x) = c, c any constant. 44. f(x) =
{1, x �= 0
0, x = 0.
45. f(x) = |x + 1|; or f(x) =
{0, x �= −1
1, x = −1.
46. f(x) = |x2 − 1|
47. f(x) = 2x + 5 48. f(x) = − |x|
49. (a) limx→2+
f(x) = limx→2−
f(x) = f(2) = 2 Thus, f is continuous at x = 2.
(b) limh→0−
f(2 + h) − f(2)h
= limh→0−
(2 + h)2 − (2 + h) − 2h
= 3
limh→0+
f(2 + h) − f(2)h
= limh→0+
2(2 + h) − 2 − 2h
= 2 f is not differentiable at 2.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
70 SECTION 3.1
50.
f ′(x) = limh→0
(x + h)√x + h− x
√x
h
= limh→0
x(√
x + h−√x)
+ h√x + h
h
= limh→0
xh
h(√
x + h +√x) + lim
h→0
√x + h
= 12
√x +
√x = 3
2
√x
51. (a) f is not continuous at 0 since limx→0−
f(x) = 1 and limx→0+
f(x) = 0. Therefore f is not differentiable
at 0.
(b)
1x
1
y
52. (a) limh→0
f(h) − 0h
=
⎧⎪⎪⎨⎪⎪⎩
limh→0
h
h= 1, h rational
limh→0
0h
= 0, h irrational.
Therefore, limh→0
f(0 + h) − f(0)h
does not exist.
(b) limh→0
g(h) − 0h
=
⎧⎪⎪⎨⎪⎪⎩
limh→0
h2
h= 0, h rational
limh→0
0h
= 0, h irrational
Therefore, g is differentiable at 0 and g′(0) = 0.
53. (a) If the tangent line is horizontal, then the normal line is vertical: x = c.
(b) If the tangent line has slope f ′(c) �= 0, then the normal line has slope − 1f ′(c)
:
y − f(c) = − 1f ′(c)
(x− c).
(c) If the tangent line is vertical, then the normal line is horizontal: y = f(c).
54. The center of the circle.
55. The normal line has slope −4 : y − 2 = −4(x− 4).
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.1 71
56.
x
y
2 4
2
4
57. f(x) = x2, f ′(x) = 2x
The tangent line at (3, 9) has equation y − 9 = 6(x− 3) and x-intercept x =32.
The normal line at (3, 9) has equation y − 9 = − 16 (x− 3) and x-intercept x = 57.
s = 57 − 32
=1112
.
58. (a) The slope of the normal at P is y/x.
(b) The slope of the tangent at P is −x/y.
(c) √1 − (x + h)2 −
√1 − x2
h=
√1 − (x + h)2 −
√1 − x2
h·√
1 − (x + h)2 +√
1 − x2√1 − (x + h)2 +
√1 − x2
=1 − (x + h)2 − (1 − x2)
h[√
1 − (x + h)2 +√
1 − x2]
=−2xh− h2
h[√
1 − (x + h)2 +√
1 − x2] =
−2x− h[√1 − (x + h)2 +
√1 − x2
]
limh→0
f(x + h) − f(x)h
= limh→0
−2x− h√1 − (x + h)2 +
√1 − x2
= − 2x2√
1 − x2= −x
y
where y =√
1 − x2.
59. (a) Since | sin(1/x)| ≤ 1 it follows that
−x ≤ f(x) ≤ x and − x2 ≤ g(x) ≤ x2
Thus limx→0
f(x) = f(0) = 0 and limx→0
g(x) = g(0) = 0, which implies that f and g
are continuous at 0.
(b) limh→0
h sin(1/h) − 0h
= limh→0
sin(1/h) does not exist.
(c) limh→0
h2 sin(1/h) − 0h
= limh→0
h sin(1/h) = 0. Thus g is differentiable at 0 and g′(0) = 0.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
72 SECTION 3.1
60. f ′(2) = limh→0
f(2 + h) − f(2)h
= limh→0
[(2 + h)3 + 1] − (23 + 1)h
= limh→0
12h + 6h2 + h3
h= 12
f ′(2) = limx→2
f(x) − f(2)x− 2
= limx→2
(x3 + 1) − 9x− 2
= limx→2
(x− 2)(x2 + 2x + 4)x− 2
= limx→2
(x2 + 2x + 4) = 12
61. f ′(1) = limh→0
f(1 + h) − f(1)h
= limh→0
[(1 + h)2 − 3(1 + h)] − (−2)]h
= limh→0
−h + h2
h= −1
f ′(1) = limx→1
f(x) − f(1)x− 1
= limx→1
(x2 − 3x) − (−2)x− 1
= limx→1
(x− 2)(x− 1)x− 1
= limx→1
(x− 2) = −1
62. f ′(3) = limh→0
f(3 + h) − f(3)h
= limh→0
√1 + (3 + h) − 2
h= lim
h→0
√4 + h− 2
h= lim
h→0
h
h√
4 + h + 2=
14
f ′(3) = limx→3
f(x) − f(3)x− 3
= limx→1
√1 + x− 2x− 3
= limx→3
x− 3(x− 3)(
√1 + x + 2)
= limx→3
1√1 + x + 2
=14
63. f ′(−1) = limh→0
f(−1 + h) − f(−1)h
= limh→0
(−1 + h)1/3 + 1h
= limh→0
(−1 + h)1/3 + 1h
· (−1 + h)2/3 − (−1 + h)1/3 + 1(−1 + h)2/3 − (−1 + h)1/3 + 1
= limh→0
h
h [−1 + h)2/3 − (−1 + h)1/3 + 1]=
13
f ′(−1) = limx→−1
f(x) − f(−1)x− (−1)
= limx→−1
x1/3 + 1x + 1
= limx→−1
x1/3 + 1x + 1
· x2/3 − x1/3 + 1
x2/3 − x1/3 + 1
= limx→−1
x + 1(x + 1)(x2/3 − x1/3 + 1)
=13
64. f ′(0) = limh→0
f(0 + h) − f(0)h
= limh→0
1h + 2
− 12
h= lim
h→0
−h
2h(h + 2)= lim
h→0
−12(h + 2)
= − 14
f ′(0) = limx→0
f(x) − f(0)x
= limx→0
1x + 2
− 12
x= lim
x→0
−x
2x(x + 2)= lim
x→0
−12(x + 2)
= − 14
65. (a) D =(2 + h)5/2 − 25/2
h− 1 ≤ h ≤ 1
(b) f ′(2) ∼= 7.071
66. (a) D =(2 + h)2/3 − 22/3
h− 1 ≤ h ≤ 1
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.1 73
(b) f ′(2) ∼= 0.529
67. (a) f ′(x) =5
2√
5x− 4; f ′(3) =
52√
11(b) f ′(x) = −2x + 16x3 − 6x5; f ′(−2) = 68
(c) f ′(x) = − 3(3 − 2x)2 + 3x)2
− 22 + 3x
; f ′(−1) = 13
68. (a) f ′(1) does not exist (b) f ′(−2) = 0 (c) f ′(3) does not exist
69. (c) f ′(x) = 10x− 21x2 (d) f ′(x) = 0 at x = 0, 1021
70. (c) f ′(x) = 3x2 + 2x− 4 (d) f ′(x) = 0 at x ∼= −1.5352, 0.8685
71. (a) f ′( 32 ) = − 11
4 ; tangent line: y − 218 = − 11
4
(x− 3
2
), normal line: y − 21
8 = 411
(x− 3
2
)(c) (1.453, 1.547)
72. (a) Set f(x) = x3. Then
f(x + h) − f(x)h
=(x + h)3 − x3
h
=x3 + 3x2h + 3xh2 + h3 − x3
h
=3x2h + 3xh2 + h3
h= 3x2 + 3xh + h2
limh→0
f(x + h) − f(x)h
= limh→0
(3x2 + 3xh + h2) = 3x2.
(b) Let S be the set of integers for which the statement is true. We have shown that 1, 2, 3 ∈ S.
Assume that k ∈ S. This tells us that if f(x) = xk, then f ′(x) = limh→0
(x + h)k − xk
h= kxk−1.
Set f(x) = xk+1. Then, by the hint,
f ′(x) = limh→0
(x + h)k+1 − xk+1
h= lim
h→0
x(x + h)k − x · xk + h(x + h)k
h
= limh→0
x
[(x + h)k − xk
h
]+ lim
h→0(x + h)k
= x · kxk−1 + xk = (k + 1)xk.
Therefore, k + 1 ∈ S. Thus S contains the set of positive integers.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
74 SECTION 3.2
SECTION 3.2
1. F ′(x) = −1 2. F ′(x) = 2 3. F ′(x) = 55x4 − 18x2
4. F ′(x) =−6x3
5. F ′(x) = 2ax + b 6. F ′(x) = x3 − x2 + x− 1
7. F ′(x) = 2x−3 8. F ′(x) =x3(2x) − (x2 + 2)3x2
x6= − x2 + 6
x4
9. G′(x) = (x2 − 1)(1) + (x− 3)(2x) = 3x2 − 6x− 1 10. F ′(x) = 1 +1x2
11. G′(x) =(1 − x)(3x2) − x3(−1)
(1 − x)2=
3x2 − 2x3
(1 − x)2
12. F ′(x) =(cx− d)a− (ax− b)c
(cx− d)2=
bc− ad
(cx− d)2
13. G′(x) =(2x + 3)(2x) − (x2 − 1)(2)
(2x + 3)2=
2(x2 + 3x + 1)(2x + 3)2
14. G′(x) =(x + 1)(28x3) − (7x4 + 11)(1)
(x + 1)2=
21x4 + 28x3 − 11(x + 1)2
15. G′(x) = (x3 − 2x)(2) + (2x + 5)(3x2 − 2) = 8x3 + 15x2 − 8x− 10
16. G′(x) =(x2 − 1)(3x2 + 3) − (x3 + 3x)2x
(x2 − 1)2=
x4 − 6x2 − 3(x2 − 1)2
17. G′(x) =(x− 2)(1/x2) − (6 − 1/x)(1)
(x− 2)2=
−2(3x2 − x + 1)x2(x− 2)2
18. G′(x) =x2(4x3) − (1 + x4)2x
x4=
2(x4 − 1)x3
19. G′(x) = (9x8 − 8x9)(
1 − 1x2
)+
(x +
1x
)(72x7 − 72x8) = −80x9 + 81x8 − 64x7 + 63x6
20. G′(x) =(−1x2
) (1 +
1x2
)+
(1 +
1x
) (−2x3
)= − 1
x2− 2
x3− 3
x4
21. f ′(x) = −(x− 2)−2; f ′(0) = − 14 , f ′(1) = −1
22. f ′(x) = 3x3 + 2x; f ′(0) = 0, f ′(1) = 5
23. f ′(x) =(1 + x2)(−2x) − (1 − x2)(2x)
(1 + x2)2=
−4x(1 + x2)2
; f ′(0) = 0, f ′(1) = −1
24. f ′(x) =(x2 + 2x + 1)(4x + 1) − (2x2 + x + 1)(2x + 2)
(x2 + 2x + 1)2=
(x + 1)(4x + 1) − 2(2x2 + x + 1)(x + 1)3
;
f ′(0) = −1, f ′(1) = 14
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.2 75
25. f ′(x) =(cx + d)a− (ax + b)c
(cx + d)2=
ad− bc
(cx + d)2, f ′(0) =
ad− bc
d2, f ′(1) =
ad− bc
(c + d)2
26. f ′(x) =(cx2 + bx + a)(2ax + b) − (ax2 + bx + c)(2cx + b)
(cx2 + bx + a)2; f ′(0) =
b(a− c)a2
, f ′(1) =2(a− c)a + b + c
27. f ′(x) = xh′(x) + h(x); f ′(0) = 0h′(0) + h(0) = 0(2) + 3 = 3
28. f ′(x) = 6xh(x) + 3x2h′(x) − 5; f ′(0) = −5
29. f ′(x) = h′(x) +h′(x)
[h(x)]2, f ′(0) = h′(0) +
h′(0)[h(0)]2
= 2 +232
=209
30. f ′(x) = h′(x) +h(x) − xh′(x)
h2(x); f ′(0) = h′(0) +
h(0)h2(0)
= 2 +13
=73
31. f ′(x) =(x + 2)(1) − x(1)
(x + 2)2=
2(x + 2)2
,
slope of tangent at (−4, 2) : f ′(−4) = 12 ; equation for tangent: y − 2 = 1
2 (x + 4)
32. f ′(x) = (x3 − 2x + 1)(4) + (4x− 5)(3x2 − 2) = 16x3 − 15x2 − 16x + 14
slope of tangent at (2, 15) : f ′(2) = 50; equation for tangent: y − 15 = 50(x− 2)
33. f ′(x) = (x2 − 3)(5 − 3x2) + (5x− x3)(2x);
slope of tangent at (1,−8) : f ′(1) = (−2)(2) + (4)(2) = 4; equation for tangent: y + 8 = 4(x− 1)
34. f ′(x) = 2x +10x2
;
slope of tangent at (−2, 9) : f ′(−2) = − 32 ; equation for tangent: y − 9 = − 3
2 (x + 2)
35. f ′(x) = (x− 2)(2x− 1) + (x2 − x− 11)(1) = 3(x− 3)(x + 1);
f ′(x) = 0 at x = −1, 3; (−1, 27), (3,−5)
36. f ′(x) = 2x +16x2
=2(x3 + 8)
x2; f ′(x) = 0 at x = −2; (−2, 12)
37. f ′(x) =(x2 + 1)(5) − 5x(2x)
(x2 + 1)2=
5(1 − x2)(x2 + 1)2
, f ′(x) = 0 at x = ±1; (−1,−5/2), (1, 5/2)
38. f ′(x) = (x + 2)(2x− 2) + (x2 − 2x− 8)(1) = 3x2 − 12 = 3(x2 − 4);
f ′(x) = 0 at x = ±2; (−2, 0), (2,−32)
39. f(x) = x4 − 8x2 + 3; f ′(x) = 4x3 − 16x = 4x(x2 − 4) = 4x(x− 2)(x + 2)
(a) f ′(x) = 0 at x = 0, ±2
(b) f ′(x) > 0 on (−2, 0) ∪ (2,∞)
(c) f ′(x) < 0 on (−∞,−2) ∪ (0, 2)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
76 SECTION 3.2
40. f(x) = 3x4 − 4x3 − 2; f ′(x) = 12x3 − 12x2 = 12x2(x− 1)
(a) f ′(x) = 0 at x = 0, 1.
(b) f ′(x) > 0 on (1,∞).
(c) f ′(x) < 0 on (−∞, 0) ∪ (0, 1)
41. f(x) = x +4x2
; f ′(x) = 1 − 8x3
=x3 − 8x3
=(x− 2)(x2 + 2x + 4)
x3
(a) f ′(x) = 0 at x = 2.
(b) f ′(x) > 0 on (−∞, 0) ∪ (2,∞)
(c) f ′(x) < 0 on (0, 2)
42. f(x) =x2 − 2x + 4
x2 + 4; f ′(x) =
(x2 + 4)(2x− 2) − (x2 − 2x + 4)(2x)(x2 + 4)2
=2(x2 − 4)(x2 + 4)2
(a) f ′(x) = 0 at x = 2,−2.
(b) f ′(x) > 0 on (−∞,−2) ∪ (2,∞)
(c) f ′(x) < 0 on (−2, 2)
43. slope of line 4; slope of tangent f ′(x) = −2x; −2x = 4 at x = −2; (−2,−10)
44. slope of line 3/5; slope of tangent f ′(x) = 3x2 − 3;
perpendicular when 3x2 − 3 = − 53 ; x = ± 2
3 ;(− 2
3 ,4627
),
(23 ,− 46
27
)45. f(x) = x3 + x2 + x + C 46. f(x) = x4 − x2 + 4x + C
47. f(x) =2x3
3− 3x2
2+
1x
+ C 48. f(x) =x5
5+
x4
2+√x + C
49. We want f to be continuous at x = 2. That is, we want
limx→2−
f(x) = f(2) = limx→2+
f(x).
This gives
(1) 8A + 2B + 2 = 4B −A.
We also want
limx→2−
f ′(x) = limx→2+
f ′(x).
This gives
(2) 12A + B = 4B.
Equations (1) and (2) together imply that A = −2 and B = −8.
50. First we need, limx→−1−
f(x) = limx→−1+
f(x), =⇒ A + B = −B −A + 4 or A + B = 2.
Next we need, limx→−1−
f ′(x) = limx→−1+
f ′(x) =⇒ −2A = 5B + A or 3A + 5B = 0.
Solving these equations gives A = 5, B = −3.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.2 77
51. slope of tangent at (5, 5) is f ′(5) = −4
tangent y − 5 = −4(x− 5) intersects
x-axis at ( 254 , 0)
normal y − 5 = 14 (x− 5) intersects
x-axis at (−15, 0)
area of triangle is12 (5)(15 + 25
4 ) = 4258
52. slope of tangent at (2, 5) is f ′(2) = −4
tangent y − 5 = −4(x− 2) intersects
x-axis at ( 134 , 0)
normal y − 5 = 14 (x− 2) intersects
x-axis at (−18, 0)
area of triangle is12 (5)(18 + 13
4 ) = 4258
53. If the point (1, 3) lies on the graph, we have f(1) = 3 and thus
(∗) A + B + C = 3.
If the line 4x + y = 8 (slope −4) is tangent to the graph at (2, 0), then
f(2) = 0 and f ′(2) = −4. Thus,
(∗∗) 4A + 2B + C = 0 and 4A + B = −4.
Solving the equations in (∗) and (∗∗), we find that A = −1, B = 0, C = 4.
54. First, f(1) = 0 and f ′(1) = 3 =⇒ A + B + C + D = 0 and 3A + 2B + C = 3
Next, f(2) = 9 and f ′(2) = 18 =⇒ 8A + 4B + 2C + D = 9 and 12A + 4B + C = 18
Solving these equations gives A = 3, B = −6, C = 6, D = −3.
55. Let f(x) = ax2 + bx + c. Then f ′(x) = 2ax + b and f ′(x) = 0 at x = −b/2a.
56. The derivative of p is the quadratic p′(x) = 3ax2 + 2bx + c. Its discriminant is
D = (2b)2 − 4(3a)(c) = 4b2 − 12ac
(a) p has two horizontal tangents iff p′ has two real roots iff D > 0.
(b) p has exactly one horizontal tangent iff p has only one real root iff D = 0.
(c) p has no horizontal tangent iff p has no real roots iff D < 0.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
78 SECTION 3.2
57. Let f(x) = x3 − x. The secant line through (−1, f(−1)) = (−1, 0) and (2, f(2)) = (2, 6) has slope
m =6 − 0
2 − (−1)= 2. Now, f ′(x) = 3x2 − 1 and 3c2 − 1 = 2 implies c = −1, 1.
58. msec =f(3) − f(1)
3 − 1=
34 − 1
2
2= 1
8
f ′(x) =(x + 1)(1) − x(1)
(x + 1)2=
1(x + 1)2
; f ′(c) = 18 =⇒ c = −1 ± 2
√2
59. Let f(x) = 1/x, x > 0. Then f ′(x) = −1/x2. An equation for the tangent line to the graph of f at
the point (a, f(a)), a > 0, is y = (−1/a2)x + 2/a. The y-intercept is 2/a and the x-intercept is 2a.
The area of the triangle formed by this line and the coordinate axes is: A = 12 (2/a)(2a) = 2 square
units.
60. Let (x, y) be the point on the graph that the tangent line passes through. f ′(x) = 3x2, so x3 − 8 =
3x2(x− 2). Thus x = 2 or x = −1. The lines are y − 8 = 12(x− 2) and y + 1 = 3(x + 1).
61. Let (x, y) be the point on the graph that the tangent line passes through. f ′(x) = 3x2 − 1, so x3 −x− 2 = (3x2 − 1)(x + 2). Thus x = 0 or x = −3. The lines are y = −x and y + 24 = 26(x + 3).
62. (a) f(c) = c3; f ′(x) = 3x2 and f ′(c) = 3c2. Tangent line: y − c3 = 3c2(x− c) or y = 3c2x− 2c3.
(b) We solve the equation 3c2x− 2c3 = x3 :
x3 − 3c2x + 2c3 = 0 =⇒ (x− c)(x2 + cx− 2c2) = 0 =⇒ (x− c)2(x + 2c) = 0
Thus, the tangent line at x = c, c �= 0 intersects the graph at x = −2c.
63. Since f and f + g are differentiable, g = (f + g) − f is differentiable. The functions f(x) = |x| and
g(x) = −|x| are not differentiable at x = 0 yet their sum f(x) + g(x) ≡ 0 is differentiable for all x.
64. No. If f and fg are differentiable, then g =fg
fwill be differentiable where f(x) �= 0.
65. Since (f
g
)(x) =
f(x)g(x)
= f(x) · 1g(x)
,
it follows from the product and reciprocal rules that(f
g
)′(x) =
(f · 1
g
)′(x) = f(x)
(− g′(x)
[g(x)]2
)+ f ′(x) · 1
g(x)=
g(x)f ′(x) − f(x)g′(x)[g(x)]2
.
66. (fgh)′(x) = [(fg)(x) · h(x)]′ = (fg)(x)h′(x) + h(x)[(fg)(x)]′
= f(x)g(x)h′(x) + h(x)[f(x)g′(x) + g(x)f ′(x)]
= f ′(x)g(x)h(x) + f(x)g′(x)h(x) + f(x)g(x)h′(x)
67. F ′(x) = 2x(
1 +1x
)(2x3 − x + 1) + (x2 + 1)
(−1x2
)(2x3 − x + 1) + (x2 + 1)
(1 +
1x
)(6x2 − 1)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.2 79
68. G′(x) =1
2√x
(1
1 + 2x
)(x2 + x− 1) +
√x
( −2(1 + 2x)2
)(x2 + x− 1) +
√x
(1
1 + 2x
)(2x + 1)
69. g(x) = [f(x)]2 = f(x) · f(x)
g′(x) = f(x)f ′(x) + f(x)f ′(x) = 2f(x)f ′(x)
70. g′(x) = 2(x3 − 2x2 + x + 2)(3x2 − 4x + 1)
71. (a) f ′(x) = 0 at x = 0, −2 (b) f ′(x) > 0 on (−∞,−2) ∪ (0,∞)
(c) f ′(x) < 0 on (−2,−1) ∪ (−1, 0)
72. (a) f ′(x) = 0 at x = 0, 1, 52 (b) f ′(x) > 0 on (−∞, 0) ∪
(1, 5
2
)∪ (5/2,∞)
(c) f ′(x) < 0 on (0, 1)
73. (a) f ′(x) �= 0 for all x �= 0 (b) f ′(x) > 0 on (0,∞)
(c) f ′(x) < 0 on (−∞, 0)
74. (a) f ′(x) = 0 at x = − 3√
4 ∼= −1.587 (b) f ′(x) > 0 on(− 3√
4, 0)
(c) f ′(x) < 0 on(−∞,− 3
√4)∪ (0,∞)
75. (a)sin(0 + 0.001) − sin 0
0.001∼= 0.99999
sin(0 − 0.001) − sin 0−0.001
∼= 0.99999
sin[(π/6) + 0.001] − sin(π/6)0.001
∼= 0.86578sin[(π/6) − 0.001] − sin(π/6)
−0.001∼= 0.86628
sin[(π/4) + 0.001] − sin(π/4)0.001
∼= 0.70675sin[(π/4) − 0.001] − sin(π/4)
−0.001∼= 0.70746
sin[(π/3) + 0.001] − sin(π/3)0.001
∼= 0.49957sin[(π/3) − 0.001] − sin(π/3)
−0.001∼= 0.50043
sin[(π/2) + 0.001] − sin(π/2)0.001
∼= −0.0005sin[(π/2) − 0.001] − sin(π/2)
−0.001∼= 0.0005
(b) cos 0 = 1, cos(π/6) ∼= 0.866025, cos(π/4) ∼= 0.707107, cos(π/3) = 0.5, cos(π/2) = 0
(c) If f(x) = sinx then f ′(x) = cosx.
76. (a)
x = −2, 0, 54
(b)
x1 = −2.732, x2 = −0.618,
x3 = 0.732, x4 = 1.618
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
80 SECTION 3.3
SECTION 3.3
1.dy
dx= 12x3 − 2x 2.
dy
dx= 2x− 8x−5
3.dy
dx= 1 +
1x2
4.dy
dx=
(1 − x)2 − 2x(−1)(1 − x)2
=2
(1 − x)2
5.dy
dx=
(1 + x2)(1) − x(2x)(1 + x2)2
=1 − x2
(1 + x2)26. y = x3 − x2 − 2x;
dy
dx= 3x2 − 2x− 2
7.dy
dx=
(1 − x)2x− x2(−1)(1 − x)2
=x(2 − x)(1 − x)2
8. y =2x− x2
3 + 3x;
dy
dx=
(3 + 3x)(2 − 2x) − (2x− x2)(3)(3 + 3x)2
=2 − 2x− x2
3(1 + x)2
9.dy
dx=
(x3 − 1)3x2 − (x3 + 1)3x2
(x3 − 1)2=
−6x2
(x3 − 1)2
10.dy
dx=
(1 + x)2x− x2(1)(1 + x)2
=x2 + 2x(1 + x)2
11.d
dx(2x− 5) = 2 12.
d
dx(5x + 2) = 5
13.d
dx[(3x2 − x−1)(2x + 5)] = (3x2 − x−1)2 + (2x + 5)(6x + x−2) = 18x2 + 30x + 5x−2
14.d
dx
[(2x2 + 3x−1)(2x− 3x−2)
]= (2x2 + 3x−1)(2 + 6x−3) + (2x− 3x−2)(4x− 3x−2) = 12x2 + 27x−4
15.d
dt
(t4
2t3 − 1
)=
(2t3 − 1)4t3 − t4(6t2)(2t3 − 1)2
=2t3(t3 − 2)(2t3 − 1)2
16.d
dt
(2t3 + 1
t4
)=
d
dt
(2t
+1t4
)= − 2
t2− 4
t5= − 2(t3 + 2)
t5
17.d
du
(2u
1 − 2u
)=
(1 − 2u)2 − 2u(−2)(1 − 2u)2
=2
(1 − 2u)2
18.d
du
(u2
u3 + 1
)=
(u3 + 1)(2u) − u2(3u2)(u3 + 1)2
=u(2 − u3)(u3 + 1)2
19.d
du
(u
u− 1− u
u + 1
)=
(u− 1)(1) − u
(u− 1)2− (u + 1)(1) − u
(u + 1)2
= − 1(u− 1)2
− 1(u + 1)2
= − 2(1 + u2)(u2 − 1)2
20.d
du
[u2(1 − u2)(1 − u3)
]=
d
du[u2 − u4 − u5 + u7] = 2u− 4u3 − 5u4 + 7u6
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.3 81
21.d
dx
(x3 + x2 + x + 1x3 − x2 + x− 1
)=
(x3 − x2 + x− 1)(3x2 + 2x + 1) − (x3 + x2 + x + 1)(3x2 − 2x + 1)(x3 − x2 + x− 1)2
=−2(x4 + 2x2 + 1)(x2 + 1)2(x− 1)2
=−2
(x− 1)2
22.d
dx
(x3 + x2 + x− 1x3 − x2 + x + 1
)=
(x3 − x2 + x + 1)(3x2 + 2x + 1) − (x3 + x2 + x− 1)(3x2 − 2x + 1)(x3 − x2 + x + 1)2
=−2x4 + 8x2 + 2
(x3 − x2 + x + 12)
23.dy
dx= (x + 1)
d
dx[(x + 2)(x + 3)] + (x + 2)(x + 3)
d
dx(x + 1) = (x + 1)(2x + 5) + (x + 2)(x + 3)
At x = 2,dy
dx= (3)(9) + (4)(5) = 47.
24.dy
dx= (x + 1)(x2 + 2)(3x2) + (x + 1)(x3 + 3)(2x) + (x2 + 2)(x3 + 3)(1)
At x = 2,dy
dx= 3(6)(12) + 3(11)(4) + (6)(11) = 414.
25.dy
dx=
(x + 2)d
dx[(x− 1)(x− 2)] − (x− 1)(x− 2)(1)
(x + 2)2
=(x + 2)(2x− 3) − (x− 1)(x− 2)
(x + 2)2
At x = 2,dy
dx=
4(1) − 1(0)16
=14.
26. y =x4 − x2 − 2
x2 + 2;
dy
dx=
(x2 + 2)(4x3 − 2x) − (x4 − x2 − 2)(2x)(x2 + 2)2
At x = 2,dy
dx=
6(28) − (10)436
=329
27. f ′(x) = 21x2 − 30x4, f ′′(x) = 42x− 120x3 28. f ′(x) = 10x4 − 24x3 + 2, f ′′(x) = 40x3 − 72x2
29. f ′(x) = 1 + 3x−2, f ′′(x) = −6x−3 30. f ′(x) = 2x + 2x−3, f ′′(x) = 2 − 6x−4
31. f(x) = 2x2 − 2x−2 − 3, f ′(x) = 4x + 4x−3, f ′′(x) = 4 − 12x−4
32. f(x) = 4x− 9x−1, f ′(x) = 4 + 9x−2, f ′′(x) = − 18x−3
33.dy
dx= x2 + x + 1
d2y
dx2= 2x + 1
d3y
dx3= 2
34.dy
dx= 10 + 50x
d2y
dx2= 50
d3y
dx3= 0
35.dy
dx= 8x− 20
d2y
dx2= 8
d3y
dx3= 0
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
82 SECTION 3.3
36.dy
dx= 1
2 x2 − 1
2 x + 1
d2y
dx2= x− 1
2
d3y
dx3= 1
37.dy
dx= 3x2 + 3x−4
d2y
dx2= 6x− 12x−5
d3y
dx3= 6 + 60x−6
38.dy
dx= 3x2 − 2x−2
d2y
dx2= 6x + 4x−3
d3y
dx3= 6 − 12x−4
39.d
dx
[xd
dx(x− x2)
]=
d
dx[x(1 − 2x)] =
d
dx[x− 2x2] = 1 − 4x
40. d2
dx2
[(x2 − 3x)
d
dx(x + x−1)
]=
d2
dx2
[(x2 − 3x)(1 − x−2)
]
=d2
dx2[x2 − 3x− 1 + 3x−1]
=d
dx(2x− 3 − 3x−2) = 2 + 6x−3
41.d4
dx4[3x− x4] =
d3
dx3[3 − 4x3] =
d2
dx2[−12x2] =
d
dx[−24x] = −24
42.d5
dx5[ax4 + bx3 + cx2 + dx + e] =
d4
dx4[4ax3 + 3bx2 + 2cx + d]
=d3
dx3[12ax2 + 6bx + 2c]
=d2
dx2[24ax + 6b] =
d
dx[24a] = 0
43.d2
dx2
[(1 + 2x)
d2
dx2(5 − x3)
]=
d2
dx2[(1 + 2x)(−6x)] =
d2
dx2
[−6x− 12x2
]= −24
44.d3
dx3
[1x
d2
dx2[x4 − 5x2]
]=
d3
dx3
[1x
(12x2 − 10)]
=d3
dx3[12x− 10x−1] =
d2
dx2[12 + 10x−2]
=d
dx[−20x−3] = 60x−4
45. y = x4 − x3
3+ 2x2 + C 46. y =
x2
2+
1x2
+ 3x + C
47. y = x5 − 1x4
+ C 48. y =2x6
3+
53x3
− 2x + C
49. Let p(x) = ax2 + bx + c. Then p′(x) = 2ax + b and p′′(x) = 2a. Now
p′′(1) = 2a = 4 =⇒ a = 2; p′(1) = 2(2)(1) + b = −2 =⇒ b = −6;
p(1) = 2(1)2 − 6(1) + c = 3 =⇒ c = 7
Thus p(x) = 2x2 − 6x + 7.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.3 83
50. p(x) = ax3 + bx2 + cx + d p′′′(−1) = 6 =⇒ a = 1
p′(x) = 3ax2 + 2bx + c p′′(−1) = −2 =⇒ b = 2
p′′(x) = 6ax + 2b p′(−1) = 3 =⇒ c = 4
p′′′(x) = 6a p(−1) = 0 =⇒ d = 3
Therefore, p(x) = x3 + 2x2 + 4x + 3.
51. (a) If k = n, f (n)(x) = n! (b) If k > n, f (k)(x) = 0.
(c) If k < n, f (k)(x) = n(n− 1)(n− 2) · · · (n− k + 1)xn−k.
52. (a)dn
dxn= n! an (b)
dk
dxk= 0 if k > n.
53. Let f(x) =
{x2 x ≥ 0
0 x ≤ 0
(a) f ′+(0) = lim
h→0+
f(0 + h) − f(0)h
= limh→0+
h2 − 0h
= 0,
f ′−(0) = lim
h→0−
f(0 + h) − f(0)h
= limh→0−
0h
= 0
Therefore, f is differentiable at 0 and f ′(0) = 0.
(b) f ′(x) =
{2x x ≥ 0
0 x ≤ 0(d)
(c) f ′′+(0) = lim
h→0+
f ′(0 + h) − f ′(0)h
= limh→0+
2h− 0h
= 2,
f ′′−(0) = lim
h→0−
f ′(0 + h) − f ′(0)h
= limh→0−
0h
= 0
Since f ′′+(0) �= f ′′
−(0), f ′′(0) does not exist.
54. Let g(x) =
{x3 x ≥ 0
0 x < 0
(a) g′+(0) = limh→0+
g(0 + h) − g(0)h
= limh→0+
h3 − 0h
= 0,
g′−(0) = limh→0−
g(0 + h) − g(0)h
= limh→0−
0h
= 0
Therefore, g is differentiable at 0 and g′(0) = 0.
g′(x) =
{3x2 x ≥ 0
0 x < 0
g′′+(0) = limh→0+
g′(0 + h) − g, (0)h
= limh→0+
3h2 − 0h
= 0,
g′′−(0) = limh→0−
g′(0 + h) − g′(0)h
= limh→0−
0h
= 0
Therefore, g′ is differentiable at 0 and g′′(0) = 0.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
84 SECTION 3.3
(b) g′(x) =
{3x2 x ≥ 0
0 x < 0
(d)g′′(x) =
{6x x ≥ 0
0 x < 0
(c) g′′′+(0) = limh→0+
g′′(0 + h) − g′′(0)h
= limh→0+
6h− 0h
= 6
g′′′− (0) = limh→0−
g′′(0 + h) − g′′(0)h
= limh→0−
0h
= 0
Since g′′′+(0) �= g′′′− (0), g′′′(0) does not exist.
55. It suffices to give a single counterexample. For instance, if
f(x) = g(x) = x, then (fg)(x) = x2 so that (fg)′′(x) = 2 but
f(x)g′′(x) + f ′′(x)g(x) = x · 0 + 0 · x = 0.
56.d
dx[f(x)g′(x) − f ′(x)g(x)] = [f(x)g′′(x) + f ′(x)g′(x)] − [f ′(x)g′(x) + f ′′(x)g(x)]
= f(x)g′′(x) − f ′′(x)g(x)
57. f ′′(x) = 6x; (a) x = 0 (b) x > 0 (c) x < 0
58. f ′′(x) = 12x2; (a) x = 0 (b) all x �= 0 (c) none
59. f ′′(x) = 12x2 + 12x− 24; (a) x = −2, 1 (b) x < −2, x > 1 (c) −2 < x < 1
60. f ′′(x) = 12x2 + 18x− 12; (a) x = −2, 12 (b) x < −2, x > 1
2 (c) −2 < x < 12
61. The result is true for n = 1:d1y
dx1=
dy
dx= −x−2 = (−1)11!x−1−1.
If the result is true for n = k:dky
dxk= (−1)kk!x−k−1
then the result is true for n = k + 1:
dk+1y
dxk+1=
d
dx
[dky
dxk
]=
d
dx
[(−1)kk!x−(k+1)
]= (−1)(k+1)(k + 1)!x−(k+1)−1.
62. y′ = −2x−3, y′′ = 6x−4, y′′′ = −24x−5; y(n) = (−1)n(n + 1)!x−(n+2)
Let S be the set of positive integers for which the result holds. Then 1 ∈ S. Assume that the positive
integer k ∈ S. Now,
y(k+1) =d
dxy(k) =
d
dx
[(−1)k(k + 1)!x−(k+2)
]= −(−1)k(k + 2)(k + 1)!x−(k+2)−1 = (−1)k+1(k + 2)!x−(k+3)
Thus, k + 1 ∈ S, and S is the set of positive integers.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.3 85
63.d
dx(uvw) = uv
dw
dx+ uw
dv
dx+ vw
du
dx
64. (a)dn
dxn[xn] = n! (b)
dn+1
dxn+1[xn] = 0
65. By the reciprocal rule,d
dx
[1
1 − x
]=
−(−1)(1 − x)2
=1
(1 − x)2.
d2
dx2
[1
1 − x
]=
d
dx
[1
(1 − x)2
]
=d
dx
[1
1 − x· 11 − x
]=
11 − x
· 1(1 − x)2
+1
1 − x· 1(1 − x)2
=2
(1 − x)3
d3
dx3
[1
1 − x
]=
d
dx
[2
(1 − x)3
]
= 2d
dx
[1
1 − x· 1(1 − x)2
]= 2
11 − x
· 2(1 − x)3
+ 21
(1 − x)2· 1(1 − x)2
=6
(1 − x)4=
3!(1 − x)4
and so on.
In general,dn
dxn
[1
1 − x
]=
n!(1 − x)n+1
. You can use induction to prove this result.
66.dn
dxn
[1 − x
1 + x
]=
(−1)n 2 · n!(1 + x)n+1
67. (b) The lines tangent to the graph of f are parallel to the line x− 2y + 12 = 0 at the points(−1√2,
12√
2
)and
(1√2,−12√
2
).
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
86 SECTION 3.3
68. (b) The normals to the graph of f are perpendicular to l at the points where
x = −12, x =
1 −√
54
, x =1 +
√5
4
69. (a) f(x) = x3 + x2 − 4x + 1; f ′(x) = 3x2 + 2x− 4.
(b) (c) The graph is “falling” when f ′(x) < 0;
the graph is “rising” when f ′(x) > 0.
70. (a) f(x) = x4 − x3 − 5x2 − x− 2; f ′(x) = 4x3 − 3x2 − 10x− 1.
(b)
x
y
ff’
-20
-10
10
20
-2 -1 1 2 3
(c) The graph is “falling” when f ′(x) < 0;
the graph is “rising” when f ′(x) > 0.
71. (a) f(x) = 12 x
3 − 3x2 + 3x + 3; f ′(x) = 32 x
2 − 6x + 3
(b) (c) The line tangent to the graph is horizontal; the
graph turns from rising to falling, or from
falling to rising.
(d) x1∼= 0.586, x2
∼= 3.414
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.4 87
72. (a) f(x) = 12 x
3 − 3x2 + 4x + 1; f ′(x) = 32 x
2 − 6x + 4. (b)
(c) The line tangent to the graph is horizontal; the
graph turns from rising to falling, or from
falling to rising.
(d) x1∼= 0.845, x2
∼= 3.155
SECTION 3.4
1. A = πr2,dA
dr= 2πr. When r = 2,
dA
dr= 4π. 2. V = s3,
dV
ds= 3s2. When s = 4,
dV
ds= 48.
3. A =12z2,
dA
dz= z. When z = 4,
dA
dz= 4. 4.
dy
dx= −x−2. When x = −1,
dy
dx= −1.
5. y =1
x(1 + x),dy
dx=
−(2x + 1)x2(1 + x)2
. At x = 2,dy
dx= − 5
36.
6.dy
dx= 3x2 − 24x + 45 = 3(x− 3)(x− 5);
dy
dx= 0 at x = 3, 5.
7. V =43πr3,
dV
dr= 4πr2 = the surface area of the ball.
8. S = 4πr2;dS
dr= 8πr and
dS
dr= 8πr0 at r = r0.
dS
dr= 1 =⇒ r0 =
18π
.
9. y = 2x2 + x− 1,dy
dx= 4x + 1;
dy
dx= 4 at x = 3
4 . Therefore x0 = 34 .
10. (a) A = π
(d
2
)2
=π
4d2; A′ =
π
2d (b) A = π
(C
2π
)2
=C2
4π;
dA
dC=
C
2π
11. (a) w = s√
2, V = s3 =(
w√2
)3
=√
24
w3,dV
dw=
3√
24
w2.
(b) z2 = s2 + w2 = 3s2, z = s√
3. V = s3 =(
z√3
)3
=√
39
z3,dV
dz=
√3
3z2.
12. A = bh = (constant) =⇒ h =A
b;
dh
db= − A
b2= − bh
b2= − h
b
13. (a)dA
dθ=
12r2 (b)
dA
dr= rθ
(c) θ =2Ar2
sodθ
dr=
−4Ar3
=−4r3
(12r2θ
)=
−2θr
14. (a)dA
dh= 2πr (b)
dA
dr= 2π(2r + h)
(c) h =A
2πr− r;
dh
dr= − A
2πr2− 1 =
−2πr(r + h) − 2πr2
2πr2= − 2r + h
r
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
88 SECTION 3.5
15. y = ax2 + bx + c, z = bx2 + ax + c.dy
dx= 2ax + b,
dz
dx= 2bx + a.
dy
dx=
dz
dxiff 2ax + b = 2bx + a. With a �= b, this occurs only at x =
12.
16. Set k(x) = f(x)g(x)h(x). Then k′(x) = f(x)g(x)h′(x) + f(x)g′(x)h(x) + f ′(x)g(x)h(x) and
k′(1) = (0)(2)(0) + (0)(−1)(−2) + (1)(2)(−2) = −4
SECTION 3.5
1. y = x4 + 2x2 + 1, y′ = 4x3 + 4x = 4x(x2 + 1)
y = (x2 + 1)2, y′ = 2(x2 + 1)(2x) = 4x(x2 + 1)
2. y = x6 − 2x3 + 1, y′ = 6x5 − 6x2 = 6x2(x3 − 1)
y = (x3 − 1)2, y′ = 2(x3 − 1)(3x2) = 6x2(x3 − 1)
3. y = 8x3 + 12x2 + 6x + 1, y′ = 24x2 + 24x + 6 = 6(2x + 1)2
y = (2x + 1)3, y′ = 3(2x + 1)2(2) = 6(2x + 1)2
4. y = x6 + 3x4 + 3x2 + 1, f ′(x) = 6x5 + 12x3 + 6x = 6x(x2 + 1)2
y = (x2 + 1)3, y′ = 3(x2 + 1)2(2x) = 6x(x2 + 1)2
5. y = x2 + 2 + x−2, y′ = 2x− 2x−3 = 2x(1 − x−4)
y = (x + x−1)2, y′ = 2(x + x−1)(1 − x−2) = 2x(1 + x−2)(1 − x−2) = 2x(1 − x−4)
6. y = 9x4 − 12x3 + 4x2, y′ = 36x3 − 36x2 + 8x = 4x(3x− 2)(3x− 1)
y = (3x2 − 2x)2, y′ = 2(3x2 − 2x)(6x− 2) = 4x(3x− 2)(3x− 1)
7. f ′(x) = −1(1 − 2x)−2 d
dx(1 − 2x) = 2(1 − 2x)−2
8. f ′(x) = 5(1 + 2x)4d
dx(1 + 2x) = 10(1 + 2x)4
9. f ′(x) = 20(x5 − x10)19d
dx(x5 − x10) = 20(x5 − x10)19(5x4 − 10x9)
10. f ′(x) = 3(x2 + x−2)2d
dx(x2 + x−2) = 6(x2 + x−2)2(x− x−3)
11. f ′(x) = 4(x− 1
x
)3d
dx
(x− 1
x
)= 4
(x− 1
x
)3 (1 +
1x2
)
12. f(t) = (1 + t)−4; f ′(t) = −4(1 + t)−5 d
dt(1 + t) = −4(1 + t)−5
13. f ′(x) = 4(x− x3 + x5)3d
dx(x− x3 + x5) = 4(x− x3 + x5)3(1 − 3x2 + 5x4)
14. f ′(t)= 3(t− t2)2d
dt(t− t2) = 3(t− t2)2(1 − 2t)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.5 89
15. f ′(t)= 4(t−1 + t−2)3d
dt(t−1 + t−2) = 4(t−1 + t−2)3(−t−2 − 2t−3)
16. f ′(x) = 3(
4x + 35x− 2
)2d
dx
(4x + 35x− 2
)= 3
(4x + 35x− 2
)2 [(5x− 2)4 − (4x + 3)5
(5x− 2)2
]= − 69(4x + 3)2
(5x− 2)4
17. f ′(x) = 4(
3xx2 + 1
)3d
dx
(3x
x2 + 1
)= 4
(3x
x2 + 1
)3 [(x2 + 1)3 − 3x(2x)
(x2 + 1)2
]=
324x3(1 − x2)(x2 + 1)5
18. f ′(x) = 3[(2x + 1)2 + (x + 1)2
]2 d
dx
[(2x + 1)2 + (x + 1)2
]= 3
[(2x + 1)2 + (x + 1)2
]2 [2(2x + 1)(2) + 2(x + 1)(1)]
= 6[(2x + 1)2 + (x + 1)2
]2 (5x + 3)
19. f ′(x) = −(x3
3+
x2
2+
x
1
)−2d
dx
(x3
3+
x2
2+
x
1
)= −
(x3
3+
x2
2+ x
)−2
(x2 + x + 1)
20. f ′(x) = 2[(6x + x5)−1 + x]d
dx[(6x + x5)−1 + x] = 2[(6x + x5)−1 + x][1 − (6x + x5)−2(6 + 5x4)]
21.dy
dx=
dy
du
du
dx=
−2u(1 + u2)2
· (2)
At x = 0, we have u = 1 and thusdy
dx=
−44
= −1.
22.dy
dx=
dy
du
du
dx= (1 − u−2) · 4(3x + 1)3(3)
At x = 0, we have u = 1 and thusdy
dx= 0.
23.dy
dx=
dy
du
du
dx=
(1 − 4u)2 − 2u(−4)(1 − 4u)2
· 4(5x2 + 1)3(10x) =2
(1 − 4u)2· 40x(5x2 + 1)3
At x = 0, we have u = 1 and thusdy
dx=
29(0) = 0.
24.dy
dx=
dy
du
du
dx= (3u2 − 1) ·
( −2(1 + x)2
)
At x = 0, we have u = 1 and thusdy
dx= 2(−2) = −4.
25.dy
dt=
dy
du
du
dx
dx
dt=
(1 + u2)(−7) − (1 − 7u)(2u)(1 + u2)2
(2x)(2)
=7u2 − 2u− 7
(1 + u2)2(4x) =
4x(7x4 + 12x2 − 2)(x4 + 2x2 + 2)2
=4(2t− 5)[7(2t− 5)4 + 12(2t− 5)2 − 2]
[(2t− 5)4 + 2(2t− 5)2 + 2]2
26.dy
dt=
dy
du
du
dx
dx
dt= 2u
((1 + x2)(−7) − (1 − 7x)(2x)
(1 + x2)2
)(5)
= 10u · 7x2 − 2x− 7(1 + x2)2
=10[1 − 7(5t + 2)][7(5t + 2)2 − 2(5t + 2) − 7]
[1 + (5t + 2)2]2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
90 SECTION 3.5
27.dy
dx=
dy
ds
ds
dt
dt
dx= 2(s + 3) · 1
2√t− 3
· (2x)
At x = 2, we have t = 4 so that s = 1 and thusdy
dx= 2(4)
12 · 1(4) = 16.
28.dy
dx=
dy
ds
ds
dt
dt
dx=
2(1 − s)2
(1 +
1t2
)1
2√x
At x = 2, we have t =√
2 and s =√
2/2. Thusdy
dx=
2(1 −
√2/2)2
(1 +
12
)1
2√
2=
33√
2 − 4.
29. (f ◦ g)′(0) = f ′(g(0))g′(0) = f ′(2)g′(0) = (1)(1) = 1
30. (f ◦ g)′(1) = f ′(g(1))g′(1) = f ′(1)g′(1) = (1)(0) = 0
31. (f ◦ g)′(2) = f ′(g(2))g′(2) = f ′(2)g′(2) = (1)(1) = 1
32. (g ◦ f)′(0) = g′(f(0))f ′(0) = g′(1)f ′(0) = (0)(2) = 0
33. (g ◦ f)′(1) = g′(f(1))f ′(1) = g′(0)f ′(1) = (1)(1) = 1
34. (g ◦ f)′(2) = g′(f(2))f ′(2) = g′(1)f ′(2) = (0)(1) = 0
35. (f ◦ h)′(0) = f ′(h(0))h′(0) = f ′(1)h′(0) = (1)(2) = 2
36. (f ◦ h ◦ g)′(1) = f ′(h(g(1))) h′(g(1))g′(1) = f ′(2)h′(1)g′(1) = (1)(1)(0) = 0
37. (g ◦ f ◦ h)′(2) = g′(f(h(2))) f ′(h(2))h′(2) = g′(1)f ′(0)h′(2) = (0)(2)(2) = 0
38. (g ◦ h ◦ f)′(0) = g′(h(f(0))) h′(f(0))f ′(0) = g′(2)h′(1)f ′(0) = (1)(1)(2) = 2
39. f ′(x) = 4(x3 + x)3(3x2 + 1)
f ′′(x) = 3(4)(x3 + x)2(3x2 + 1)2 + 4(x3 + x)3(6x) = 12(x3 + x)2[(3x2 + 1)2 + 2x(x3 + x)]
40. f ′(x) = 10(x2 − 5x + 2)9(2x− 5)
f ′′(x) = 9(10)(x2 − 5x + 2)8(2x− 5)2 + 10(x2 − 5x + 2)9(2)
= (10)(x2 − 5x + 2)8[9(2x− 5)2 + 2(x2 − 5x + 2)
]
41. f ′(x) = 3(
x
1 − x
)2
· 1(1 − x)2
=3x2
(1 − x)4
f ′′(x) =6x(1 − x)4 − 3x2(4)(1 − x)3(−1)
(1 − x)8=
6x(1 + x)(1 − x)5
42. f ′(x) =1
2√x2 + 1
(2x) =x√
x2 + 1
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.5 91
f ′′(x) =
√x2 + 1(1) − x
x√x2 + 1(√
x2 + 1)2 =
1
(x2 + 1)3/2
43. 2xf ′(x2 + 1) 44. f ′(x− 1x + 1
)d
dx
(x− 1x + 1
)=
2(x + 1)2
f ′(x− 1x + 1
)
45. 2f(x)f ′(x) 46.[f(x) + 1]f ′(x) − [f(x) − 1]f ′(x)
[f(x) + 1]2=
2f ′(x)[f(x) + 1]2
47. f ′(x) = −4x(1 + x2)−3; (a) x = 0 (b) x < 0 (c) x > 0
48. f ′(x) = 2(1 − x2)(−2x) = −4x(1 − x2); (a) x = −1, 0, 1 (b) − 1 < x < 0, x > 1
(c) x < −1, 0 < x < 1
49. f ′(x) =1 − x2
(1 + x2)2; (a) x = ±1 (b) −1 < x < 1 (c) x < −1, x > 1
50. f ′(x) = (1 − x2)3 + x(3)(1 − x2)2(−2x) = (1 − x2)2(1 − 7x2);
(a) x = ±1, x = ± 17
√7 (b) − 1
7
√7 < x < 1
7
√7
(c) x < −1, −1 < x < − 17
√7, 1
7
√7 < x < 1, x > 1
51.n!
(1 − x)n+152.
(−1)n+1n!(1 + x)n+1
53. n! bn 54.(−1)nn! abn
(bx + c)n+1
55. y = (x2 + 1)3 + C 56. y =(x2 − 1)2
2+ C
57. y = (x3 − 2)2 + C 58. y =(x3 + 2)3
3+ C
59. L′(x) =1
x2 + 1· 2x =
2xx2 + 1
60. H ′(x) = 2f(x)f ′(x) − 2g(x)g′(x) = 2f(x)g(x) − 2g(x)f(x) = 0
61. T ′(x) = 2f(x) · f ′(x) + 2g(x) · g′(x) = 2f(x) · g(x) − 2g(x) · f(x) = 0
62. (a) Suppose f is even: [f(x)]′ = [f(−x)]′ = f ′(−x)(−1) = −f ′(−x); thus f ′(−x) = −f ′(x).
(b) Suppose f is odd: [f(x)]′ = −[f(−x)]′ = −f ′(−x)(−1) = f ′(−x); thus f ′(−x) = f ′(x).
63. Suppose P (x) = (x− a)2q(x), where q(a) �= 0. Then
P ′(x) = 2(x− a)q(x) + (x− a)2q′(x) and P ′′(x) = 2q(x) + 4(x− a)q′(x) + (x− a)2q′′(x),
and it follows that P (a) = P ′(a) = 0, and P ′′(a) �= 0.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
92 SECTION 3.5
Now suppose that P (a) = P ′(a) = 0 and P ′′(a) �= 0.
P (a) = 0 =⇒ P (x) = (x− a)g(x) for some polynomial g.
Then P ′(x) = g(x) + (x− a)g′(x) and
P ′(a) = 0 =⇒ g(a) = 0 and so g(x) = (x− a)q(x) for some polynomial q.
Therefore, P (x) = (x− a)2q(x). Finally, P ′′(a) �= 0 implies q(a) �= 0.
64. Suppose P (x) = (x− a)3q(x), where q(a) �= 0. Then
P ′(x) = 3(x− a)2q(x) + (x− a)3q′(x)
P ′′(x) = 6(x− a)q(x) + 6(x− a)2q′(x) + (x− a)3q′′(x)
P ′′′(x) = 6q(x) + 18(x− a)q′(x) + 9(x− a)2q′′(x) + (x− a)3q′′′(x)
and it follows that P (a) = P ′(a) = P ′′(a) = 0, P ′′′(a) �= 0.
Now suppose that P (a) = P ′(a) = P ′′(a) = 0 and P ′′′(a) �= 0.
P (a) = 0 =⇒ P (x) = (x− a)g(x) for some polynomial g.
Then P ′(x) = g(x) + (x− a)g′(x) and
P ′(a) = 0 =⇒ g(a) = 0 and so g(x) = (x− a)h(x) for some polynomial h.
Therefore, P (x) = (x− a)2h(x). Now P ′′(x) = 2h(x) + 4(x− a)h′(x) + (x− a)2h′′(x) and
P ′′(a) = 0 =⇒ h(a) = 0 and so h(x) = (x− a)q(x) for some polynomial q.
Therefore, P (x) = (x− a)3q(x). Finally, P ′′′(a) �= 0 implies q(a) �= 0.
65. Let P be a polynomial function of degree n. The number a is a root of P of multiplicity k, (k < n) if
and only if P (a) = P ′(a) = · · · = P (k−1)(a) = 0 and P (k)(a) �= 0.
66. A =√
34
x2, where x =2√
33
h. Now
dA
dh=
dA
dx
dx
dh=
√3
2x · 2
√3
3=
2√
33
h;dA
dh= 4 when h = 2
√3
67. V = 43 πr
3 anddr
dt= 2 cm/sec. By the chain rule,
dV
dt=
dV
dr
dr
dt= 4πr2 dr
dt= 8πr2.
At the instant the radius is 10 centimeters, the volume is increasing at the ratedV
dt= 8π(10)2 = 800π cm3/sec.
68. V = 43 πr
3, S = 4πr2, anddV
dt= 200.
dS
dt=
dS
dr· dr
dV· dVdt
= 8πr · 14πr2
· 200
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.5 93
=400r
= 80 when r = 5
The surface area is increasing 80 cm2/sec. at the instant the radius is 5 centimeters.
69. (a)dF
dt=
dF
dr· drdt
=2kr3
· (49 − 9.8t) =2k
(49t− 4.9t2)3(49 − 9.8t), 0 ≤ t ≤ 10
(b)dF
dt(3) =
2k(102.9)3
(19.6);dF
dt(7) = − 2k
(102.9)3(19.6).
70. (a) f(9) = −2, f ′(9) = − 112 ; tangent T : y = − 1
12x− 54 .
(b)
2 4 6 8 10x
-2
-1
y
f
T
(c) (7.4, 10.8)
71. (a) f(1) = 12 , f ′(1) = − 1
2 ; tangent T : y = − 12x + 1.
(b)
1 2x
1
y
f
T
(c) (0.8, 1.2)
72.d
dx
[x2 d4
dx4
(x2 + 1
)4]
= 288(35x5 + 20x3 + x
)
73. (a)d
dx
[f
(1x
)]= −f ′(1/x)
x2(b)
d
dx
[f
(x2 − 1x2 + 1
)]=
4x f ′ [(x2 − 1)/(x2 + 1)]
(1 + x2)2
(c)d
dx
[f(x)
1 + f(x)
]=
f ′(x)[1 + f(x)]2
74. (a)d
dx[u1 (u2(x))] = u′
1 (u2(x))u′2(x) (b)
d
dx[u1 (u2(u3(x)))] = u′
1 [u2(u3(x)]u′2 (u3(x))
u′3(x)
(c)d
dx[u1 (u2[u3(u4(x))])] = u′
1 [u2(u3[u4(x)])]u′2 (u3[u4(x)])u′
3[u4(x)]u′4(x)
75.d2
dx2[f(g(x))] = [g′(x)]2 f ′′[g(x)] + f ′[g(x)]g′′(x)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
94 SECTION 3.6
SECTION 3.6
1.dy
dx= −3 sinx− 4 secx tanx 2.
dy
dx= 2x secx + x2 secx tanx
3.dy
dx= 3x2 cscx− x3 cscx cotx 4.
dy
dx= 2 sinx cosx
5.dy
dt= −2 cos t sin t 6.
dy
dt= 6t tan t + 3t2 sec2 t
7.dy
du= 4 sin3
√u
d
du(sin
√u ) = 4 sin3
√u cos
√u
d
du(√u ) = 2u−1/2 sin3
√u cos
√u
8.dy
du= cscu2 − 2u2 cscu2 cotu2 9.
dy
dx= sec2 x2 d
dx(x2) = 2x sec2 x2
10.dy
dx= − 1
2√x
sin√x 11.
dy
dx= 4[x + cotπx]3[1 − π csc2 πx]
12.dy
dx= 3(x2 − sec 2x)2(2x− 2 sec 2x tan 2x) = 6(x2 − sec 2x)2(x− sec 2x tan 2x)
13.dy
dx= cosx,
d2y
dx2= − sinx 14.
dy
dx= − sinx,
d2y
dx2= − cosx
15.dy
dx=
(1 + sinx)(− sinx) − cosx (cosx)(1 + sinx)2
=− sinx− (sin2 x + cos2 x)
(1 + sinx)2= −(1 + sinx)−1
d2y
dx2= (1 + sinx)−2 d
dx(1 + sinx) = cosx (1 + sinx)−2
16.dy
dx= 3 tan2(2πx) sec2(2πx)(2π) = 6π tan2(2πx) sec2(2πx)
d2y
dx2= 6π(2) tan(2πx) sec2(2πx) sec2(2πx)(2π) + 6π tan2(2πx)(2) sec(2πx)[sec(2πx) tan(2πx)](2π)
= 24π2 tan(2πx) sec2(2πx)[sec2(2πx) + tan2(2πx)]
17.dy
du= 3 cos2 2u
d
du(cos 2u) = −6 cos2 2u sin 2u
d2y
du2= −6[cos2 2u
d
du(sin 2u) + sin 2u
d
du(cos2 2u)]
= −6[2 cos3 2u + sin 2u (−4 cos 2u sin 2u)] = 12 cos 2u [2 sin2 2u− cos2 2u]
18.dy
dt= 5 sin4(3t) cos(3t)(3) = 15 sin4(3t) cos(3t)
d2y
dt2= 15(4) sin3(3t) 3 cos2(3t) + 15 sin4(3t)[−3 sin(3t)] = 45 sin3(3t)[4 cos2(3t) − sin2(3t)]
19.dy
dt= 2 sec2 2t,
d2y
dt2= 4 sec 2t
d
dt(sec 2t) = 8 sec2 2t tan 2t
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.6 95
20.dy
du= −4 csc2 4u;
d2y
du2= −4(2) csc(4u)[− csc(4u) cot(4u)(4)] = 32 csc2(4u) cot(4u)
21.dy
dx= x2(3 cos 3x) + 2x sin 3x
d2y
dx2= [x2(−9 sin 3x) + 2x(3 cos 3x)] + [2x(3 cos 3x) + 2(sin 3x)]
= (2 − 9x2) sin 3x + 12x cos 3x
22.dy
dx=
(1 − cosx) cosx− sinx (−[− sinx])(1 − cosx)2
=cosx− 1
(1 − cosx)2=
−11 − cosx
d2y
dx2= sinx(1 − cosx)−2
23. y = sin2 x + cos2 x = 1 sody
dx=
d2y
dx2= 0
24. y = sec2 x− tan2 x = 1 sody
dx=
d2y
dx2= 0
25.d4
dx4(sinx) =
d3
dx3(cosx) =
d2
dx2(− sinx) =
d
dx(− cosx) = sinx
26.d4
dx4(cosx) =
d3
dx3(− sinx) =
d2
dx2(− cosx) =
d
dx(sinx) = cosx
27.d
dt
[t2
d2
dt2(t cos 3t)
]=
d
dt
[t2
d
dt(cos 3t− 3t sin 3t)
]
=d
dt[t2(−3 sin 3t− 3 sin 3t− 9t cos 3t)]
=d
dt[−6t2 sin 3t− 9t3 cos 3t]
= (−18t2 cos 3t− 12t sin 3t) + (27t3 sin 3t− 27t2 cos 3t)
= (27t3 − 12t) sin 3t− 45t2 cos 3t
28.d
dt
[td
dt(cos t2)
]=
d
dt
[−t sin t2(2t)
]=
d
dt
[−2t2 sin t2
]= −4t sin t2 − 2t2 cos t2(2t) = −4t(sin t2 + t2 cos t2)
29.d
dx[f(sin 3x)] = f ′(sin 3x)
d
dx(sin 3x) = 3 cos 3xf ′(sin 3x)
30.d
dx[sin f(3x)] = cos[f(3x)]f ′(3x)(3) = 3f ′(3x) cos[f(3x)]
31.dy
dx= cosx; slope of tangent at (0, 0) is 1; tangent: y = x.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
96 SECTION 3.6
32.dy
dx= sec2 x; slope of tangent at (π/6,
√3/3) is sec2(π/6) = 4/3;
tangent: y − 13
√3 = 4
3
(x− 1
6π)
33.dy
dx= − csc2 x; slope of tangent at (
π
6,√
3) is −4, an equation for
tangent: y −√
3 = −4(x− π
6).
34.dy
dx= − sinx; slope of tangent at (0, 1) is 0; tangent: y = 1.
35.dy
dx= secx tanx, slope of tangent at (
π
4,√
2) is√
2, an equation for
tangent is y −√
2 =√
2(x− π
4).
36.dy
dx= − cscx cotx, slope of tangent at (π/3, 2
√3/3) is − 2/3;
tangent: y − 23
√3 = − 2
3
(x− 1
3π).
37.dy
dx= − sinx; x = π 38.
dy
dx= cosx; x = 1
2π, x = 32π
39.dy
dx= cosx−
√3 sinx;
dy
dx= 0 gives tanx =
1√3; x =
π
6,
7π6
40.dy
dx= − sinx−
√3 cosx;
dy
dx= 0 gives tanx = −
√3; x =
2π3,
5π3
41.dy
dx= 2 sinx cosx = sin 2x; x =
π
2, π,
3π2
42.dy
dx= −2 sinx cosx = − sin 2x; x =
π
2, π,
3π2
43.dy
dx= sec2 x− 2;
dy
dx= 0 gives secx = ±
√2; x =
π
4,
3π4,
5π4,
7π4
44.dy
dx= −3 csc2 x + 4;
dy
dx= 0 gives cscx = ± 2√
3; x =
π
3,
2π3,
4π3,
5π3
45.dy
dx= 2 secx tanx + sec2 x; since secx is never zero,
dy
dx= 0 gives
2 tanx + secx = 0 so that sinx = −1/2; x =7π6,
11π6
46.dy
dx= − csc2 x + 2 cscx cotx; since cscx is never zero,
dy
dx= 0 gives
2 cotx− cscx = 0 so that cosx = 1/2; x =π
3,
5π3
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.6 97
47. f(x) = x + 2 cos x, f ′(x) = 1 − 2 sin x.
(a) xf ′(x) = 0 : 1 − 2 sin x = 0, sin x = 1/2, x = π/6, 5π/6.
f ′:0 π
65 π6
2 π
-+ +
(b) f ′(x) > 0 on (0, π/6) ∪ (5π/6, 2π), (c) f ′(x) < 0 on (π/6, 5π/6).
48. f(x) = x−√
2 sin x, f ′(x) = 1 −√
2 cos x.
(a) f ′(x) = 0 : 1 −√
2 cos x = 0, cos x =√
2/2, x = π/4, 7π/4.
f ′:0 π
47 π4
2π
+−
(b) f ′(x) > 0 on (π/4, 7π/4), (c) f ′(x) < 0 on (0, π/4) ∪ (7π/4, 2π).
49. f(x) = sin x + cos x, f ′(x) = cos x− sin x.
(a) f ′(x) = 0 : cos x− sin x = 0, cos x = sin x, x = π/4, 5π/4.
f ′: 0 π4
5 π4
2 π
-+ +
(b) f ′(x) > 0 on (0, π/4) ∪ (5π/4, 2π), (c) f ′(x) < 0 on (π/4, 5π/4).
50. f(x) = sin x− cos x, f ′(x) = cos x + sin x.
(a) f ′(x) = 0 : cos x + sin x = 0, cos x = − sin x, x = 3π/4, 7π/4.
f ′: 0 3 π4
7
+-+
π4
2 π
(b) f ′(x) > 0 on (0, 3π/4) ∪ (7π/4, 2π), (c) f ′(x) < 0 on (3π/4, 7π/4).
51. (a)dy
dt=
dy
du
du
dx
dx
dt= (2u)(secx tanx)π = 2π sec2 πt tanπt
(b) y = sec2 πt− 1,dy
dt= 2 secπt (secπt tanπt)π = 2π sec2 πt tanπt
52. (a)dy
dt=
dy
du
du
dx
dx
dt= 3
[12(1 + u)
]2 (12
)(− sinx)(2) = 3
[12(1 + cos 2t)
]2
(− sin 2t)
= 3(cos4 t)(−2 sin t cos t) = −6 cos5 t sin t
(b) y =[12(1 + cos 2t)
]3
= cos6 t;dy
dt= 6 cos5 t(− sin t) = −6 cos5 t sin t
53. (a)dy
dt=
dy
du
du
dx
dx
dt= 4
[12(1 − u)
]3 (−1
2
)(− sinx)(2) = 4
[12(1 − cos 2t)
]3
sin 2t
= 4 sin6 t (2 sin t cos t) = 8 sin7 t cos t
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
98 SECTION 3.6
(b) y =[12(1 − cos 2t)
]4
= sin8 t,dy
dt= 8 sin7 t cos t
54. (a)dy
dt=
dy
du
du
dx
dx
dt= (−2u)(− cscx cotx)(3) = (−2 csc(3t)(− csc(3t) cot(3t)3
= 6 csc2(3t) cot(3t)
(b) y = 1 − csc2(3t);dy
dt= −2 csc(3t)[− csc(3t) cot(3t)](3) = 6 csc2(3t) cot(3t)
55.dn
dxn(cosx) =
{(−1)(n+1)/2 sinx, n odd
(−1)n/2 cosx, n even
56. (a)d
dx(cotx) =
d
dx
(cosxsinx
)=
sinx(− sinx) − cosx(cosx)sin2 x
=−1
sin2 x= − csc2 x
(b)d
dx(secx) =
d
dx
(1
cosx
)=
−1cos2 x
(− sinx) =1
cosx· sinx
cosx= secx tanx
(c)d
dx(cscx) =
d
dx
(1
sinx
)=
−1sin2 x
(cosx) =−1
sinx· cosx
sinx= − cscx cotx
57.d
dx(cosx) =
d
dx
[sin
(π
2− x
)]= − cos
(π
2− x
)= − sinx
58. Differentiating both sides, 2 cos 2x = 2(cos2 x− sin2 x). Thus cos 2x = cos2 x− sin2 x.
59. f ′(0) = limh→0
sin(0 + h) − sin 0h
= limh→0
sinh
h= lim
x→0
sinx
x
60. f ′(0) = limh→0
cos(0 + h) − cos 0h
= limh→0
cosh− 1h
= limx→0
cosx− 1x
61. f(x) = 2 sinx + 3 cosx + C 62. f(x) = tanx + cotx + C
63. f(x) = sin 2x + secx + C 64. f(x) =− cos 3x
3+
csc 2x2
+ C
65. f(x) = sin(x2) + cos 2x + C 66. f(x) =tan(x3)
3+ sec 2x + C
67. (a) f ′(x) = sin(1/x) + x cos(1/x)(−1/x2)
= sin(1/x) − (1/x) cos(1/x)
g′(x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2)
= 2x sin(1/x) − cos(1/x)
(b) limx→0
g′(x) = limx→0
[2x sin(1/x) − cos(1/x)] = − limx→0
cos(1/x) does not exist
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.6 99
68. (a) f must be continuous at 0:
limx→0+
f(x) = limx→0+
cosx = 1, limx→0−
f(x) = limx→0−
(ax + b) = b; thus b = 1
Differentiable at 0 :
limh→0+
f(h) − f(0)h
= limh→0+
cosh− 1h
= 0,
limh→0−
f(h) − f(0)h
= limh→0+
ah + 1 − 1h
= a
Therefore, f is differentiable at 0 if a = 0 and b = 1.
(b)
1x
1
y
f
69. (a) Continuity:
limx→2π/3−
sinx =√
32
, limx→2π/3+
(ax + b) =2πa3
+ b; thus2πa3
+ b =√
32
Differentiability:
limx→2π/3−
cosx = − 12, lim
x→2π/3+(a) = a; thus a = − 1
2
Therefore, f is differentiable at 2π/3 if a = − 12 and b = 1
2
√3 + 1
3π
(b)
2π3
x
1 g
y
70. (a) Continuity:
limx→π/3−
(1 + a cos x) = 1 + 12 a, lim
x→π/3+[b + sin (x/2)] = b + 1
2 which implies b = 12 + 1
2 a.
Differentiability:
limx→π/3−
(−a sin x) = − 12
√3 a, lim
x→π/3+[ 12 cos (x/2)] = 1
4
√3 which implies a = − 1
2 .
Therefore, f is differentiable at π/3 if a = − 12 and b = 1
4 .
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
100 SECTION 3.6
(b)
π3
x
1f
y
71. Let y(t) = A sinωt + B cosωt. Then
y′(t) = ωA cosωt− ωB sinωt and y′′(t) = −ω2A sinωt− ω2B cosωt
Thus,
d2y
dt2+ ω2y = 0.
72. (a) θ = a sin(ωt + φ); θ′ = aω cos(ωt + φ); θ′′ = −aω2 sin(ωt + φ)
Thus, θ satisfies the equation.
(b)
θ = a sin(ωt + φ0)
= a sin(ωt) cosφ0 − a cos(ωt) sinφ0
= A sin(ωt) + B cos(ωt) where A = −a sinφ0, B = a cosφ0
73. A = 12 c
2 sinx;dA
dx=
12c2 cosx
74. c =√a2 + b2 − 2ab cosx;
dc
dx=
12√a2 + b2 − 2ab cosx
(2ab sinx) =ab sinx√
a2 + b2 − 2ab cosx
75. (a) f (4p)(x) = k4p cos kx, f (4p+1)(x) = −k4p+1 sin kx, f (4p+2)(x) = −k4p+2 cos kx,
f (4p+3)(x) = k4p+3 sin kx, p = 0, 1, 2, . . .
(b) m = k2, k = 1, 2, 3, . . .
76. f(x) = A cos√
2x + B sin√
2x; f ′(x) = −A√
2 sin√
2x + B√
2 cos√
2x
f(0) = 2 =⇒ A = 2; f ′(0) = −3 =⇒ B = − 3√2
= −3√
22
77. f has horizontal at the points with x-coordinate 12π,
32π,
∼= 3.39, ∼= 6.03
78. f(x) = sinx− sin2 x has horizontal tangents at:(π6 ,
14
),
(π2 , 0
),
(5π6 , 1
4
),
(3π2 ,−2
)79. f(x) = sin x, f ′(x) = cos x; f(0) = 0, f ′(0) = 1. Therefore an equation for the tangent line T is
y = x.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.7 101
The graphs of f and T are:
π2
x
-1
-
1f
T
π2
y
| sin x− x| < 0.01 on (−0.39, 0.39).
80. f(x) = tan x, f ′(x) = sec2 x; f(π/4) = 1, f ′(π/4) = 2. Therefore an equation for the tangent line
T is y = 2x + 1 − 12π.
The graphs of f and T are:
π4 2
x
-1
1
y
f
T
π
| tan x− (2x + 1 − 12π)| < 0.01 on (0.712, 0.852).
SECTION 3.7
1. x2 + y2 = 4
2x + 2ydy
dx= 0
dy
dx=
−x
y
2. x3 + y3 − 3xy = 0
3x2 + 3y2 dy
dx− 3
(y + x
dy
dx
)= 0
dy
dx=
y − x2
y2 − x
3. 4x2 + 9y2 = 36
8x + 18ydy
dx= 0
dy
dx=
−4x9y
4.√x +
√y = 4
12√x
+1
2√y
dy
dx= 0
dy
dx= −
√y√x
5. x4 + 4x3y + y4 = 1
4x3 + 12x2y + 4x3 dy
dx+ 4y3 dy
dx= 0
dy
dx= −x3 + 3x2y
x3 + y3
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
102 SECTION 3.7
6. x2 − x2y + xy2 + y2 = 1
2x− 2xy − x2 dy
dx+ y2 + 2xy
dy
dx+ 2y
dy
dx= 0
dy
dx=
2xy − 2x− y2
2xy + 2y − x2
7. (x− y)2 − y = 0
2(x− y)(
1 − dy
dx
)− dy
dx= 0
dy
dx=
2(x− y)2(x− y) + 1
8. (y + 3x)2 − 4x = 0
2(y + 3x)(dy
dx+ 3
)− 4 = 0
dy
dx= −3 +
2y + 3x
9. sin (x + y) = xy
cos (x + y)(
1 +dy
dx
)= x
dy
dx+ y
dy
dx=
y − cos (x + y)cos (x + y) − x
10. tanxy = xy; sec2(xy)(y + x
dy
dx
)= y + x
dy
dx;
dy
dx= − y
x
11. y2 + 2xy = 16
2ydy
dx+ 2x
dy
dx+ 2y = 0
(x + y)dy
dx+ y = 0.
Differentiating a second time, we have
(x + y)d2y
dx2+
dy
dx
(2 +
dy
dx
)= 0.
Substitutingdy
dx=
−y
x + y, we have
(x + y)d2y
dx2− y
(x + y)
(2x + y
x + y
)= 0,
d2y
dx2=
2xy + y2
(x + y)3=
16(x + y)3
.
12. x2 − 2xy + 4y2 = 3
2x− 2y − 2xdy
dx+ 8y
dy
dx= 0
x− y + (4y − x)dy
dx= 0.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.7 103
Differentiating a second time, we have
1 − dy
dx+
(4dy
dx− 1
)dy
dx+ (4y − x)
d2y
dx2= 0.
Substitutingdy
dx=
y − x
4y − x, we have
d2y
dx2= − 3(x2 − 2xy + 4y2)
(4y − x)3=
−9(4y − x)3
.
13. y2 + xy − x2 = 9
2ydy
dx+ x
dy
dx+ y − 2x = 0.
Differentiating a second time, we have[2
(dy
dx
)2
+ 2yd2y
dx2
]+
[xd2y
dx2+
dy
dx
]+
dy
dx− 2 = 0
(2y + x)d2y
dx2+ 2
[(dy
dx
)2
+dy
dx− 1
]= 0.
Substitutingdy
dx=
2x− y
2y + x, we have
(2y + x)d2y
dx2+ 2
[(2x− y)2 + (2x− y)(2y + x) − (2y + x)2
(2y + x)2
]= 0
d2y
dx2=
10(y2 + xy − x2)(2y + x)3
=90
(2y + x)3.
14. x2 − 3xy = 18
2x− 3y − 3xdy
dx= 0.
Differentiating a second time, we have
2 − 3dy
dx− 3
dy
dx− 3x
d2y
dx2= 0
Substitutingdy
dx=
2x− 3y3x
, we have
d2y
dx2= − 6(x− 3y)
9x2= − 6(x2 − 3xy)
9x3= − 6(18)
9x3= − 12
x3
15. 4 tan y = x3
4 sec2 ydy
dx= 3x2
dy
dx=
34x2 cos2 y
d2y
dx2=
32x cos2 y +
34x2
(2 cos y(− sin y)
dy
dx
)
=32x cos2 y − 9
8x4 sin y cos3 y
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
104 SECTION 3.7
16. sin2 x + cos2 y = 1
2 sinx cosx− 2 cos y sin ydy
dx= 0
sin 2x− sin 2ydy
dx= 0
dy
dx=
sin 2xsin 2y
d2y
dx2=
sin 2y(cos 2x)2 − sin 2x(cos 2y)2dy
dxsin2 2y
Substitutingdy
dx=
sin 2xsin 2y
and using the double angle formulas, we find that
d2y
dx2=
8[cos2 x cos2 y(sin2 y − sin2 x) − sin2 x sin2 y(cos2 y − cos2 x)
]sin3 2y
= 0
since sin2 x = sin2 y and cos2 x = cos2 y from the original equation.
17. x2 − 4y2 = 9, 2x− 8ydy
dx= 0.
At (5, 2), we getdy
dx=
58. Then,
2 − 8
[yd2y
dx2+
(dy
dx
)2]
= 0.
At (5, 2) we get
2 − 8[2d2y
dx2+
2564
]= 0 so that
d2y
dx2= − 9
128.
18. x2 + 4xy + y3 + 5 = 0, 2x + 4y + 4xdy
dx+ 3y2 dy
dx= 0.
At (2, −1), we get 4 − 4 + 8dy
dx+ 3
dy
dx= 0 so
dy
dx= 0. Differentiating again,
2 + 4dy
dx+ 4
dy
dx+ 4x
d2y
dx2+ 6y
(dy
dx
)2
+ 3y2 d2y
dx2= 0
At (2, −1) we get 2 + 11d2y
dx2= 0 so
d2y
dx2= − 2
11
19. cos (x + 2y) = 0 − sin (x + 2y)(
1 + 2dy
dx
)= 0.
At (π/6, π/6), we getdy
dx= −1/2. Then,
− cos (x + 2y)(
1 + 2dy
dx
)2
− sin (x + 2y)(
2d2y
dx2
)= 0.
At (π/6, π/6), we get
− cosπ
2(0)2 − sin
π
2
(2d2y
dx2
)= 0 so that
d2y
dx2= 0.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.7 105
20. x = sin2 y 1 = 2 sin y cos ydy
dx= sin 2y
dy
dx.
At (1/2, π/4), we getdy
dx= 1. Differentiating again,
0 = 2 cos 2y(dy
dx
)2
+ sin 2yd2y
dx2.
At (1/2, π/4), we getd2y
dx2= 0.
21. 2x + 3y = 5
2 + 3dy
dx= 0
slope of tangent at (−2, 3) : −2/3
tangent: y − 3 = − 23 (x + 2)
normal: y − 3 = 32 (x + 2)
22. 9x2 + 4y2 = 72
18x + 8ydy
dx= 0
slope of tangent at (2, 3) : − 32
tangent: y − 3 = − 32 (x− 2)
normal: y − 3 = 23 (x− 2)
23. x2 + xy + 2y2 = 28
2x + xdy
dx+ y + 4y
dy
dx= 0
slope of tangent at (−2,−3) : −1/2
tangent: y + 3 = − 12 (x + 2)
normal: y + 3 = 2(x + 2)
24. x3 − axy + 3ay2 = 3a3
3x2 − ay − axdy
dx+ 6ay
dy
dx= 0
slope of tangent at (a, a) : − 25
tangent: y − a = − 25 (x− a)
normal: y − a = 52 (x− a)
25. x = cos ydy
dx
1 = − sin ydy
dx
slope of tangent at(
12,π
3
):
−2√3
tangent: y − π
3= − 2√
3
(x− 1
2
)
normal: y − π
3=
√3
2
(x− 1
2
)
26. tanxy = x
sec2(xy)(y + x
dy
dx
)= 1
slope of tangent at (1, π/4) :2 − π
4
tangent: y − π
4=
2 − π
4(x− 1)
normal: y − π
4= − 4
2 − π(x− 1)
27.dy
dx=
12(x3 + 1)−1/2 d
dx(x3 + 1) =
32x2(x3 + 1)−1/2 28.
dy
dx= 1
3 (x + 1)−2/3
29.dy
dx=
14(2x2 + 1)−3/4 d
dx(2x2 + 1) = x(2x2 + 1)−3/4
30.dy
dx= 1
3 (x + 1)−2/3(x + 2)2/3 + (x + 1)1/3(
23
)(x + 2)−1/3 =
3x + 43(x + 1)2/3(x + 2)1/3
31.dy
dx=
√2 − x2
[ −x√3 − x2
]+
√3 − x2
[ −x√2 − x2
]=
x(2x2 − 5)√2 − x2
√3 − x2
32.dy
dx= 3
2 (x4 − x + 1)1/2(4x3 − 1) = 32
√x4 − x + 1 (4x3 − 1)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
106 SECTION 3.7
33.d
dx
(√x +
1√x
)=
d
dx(x1/2 + x−1/2) =
12x−1/2 − 1
2x−3/2 =
12x−3/2(x− 1)
34.d
dx
(√3x + 12x + 5
)=
12
(3x + 12x + 5
)−1/2 ((2x + 5)3 − (3x + 1)2
(2x + 5)2
)=
132(2x + 5)2
√2x + 53x + 1
35.d
dx
(x√
x2 + 1
)=
d
dx
(x(x2 + 1)−1/2
)= x
(−1
2(x2 + 1)−3/2(2x)
)+ (x2 + 1)−1/2 = (x2 + 1)−3/2
36.d
dx
(√x2 + 1x
)=
x(
12
)(x2 + 1)−1/2(2x) − (x2 + 1)1/2
x2=
x2
√x2 + 1
−√x2 + 1
x2=
−1x2
√x2 + 1
37. (a) (b) (c)
38. y = (a2 + x2)1/2;dy
dx=
12(a2 + x2)−1/2(2x) = x(a2 + x2)−1/2;
d2y
dx2= (a2 + x2)−1/2 − 1
2x(a2 + x2)−3/2(2x) =
a2
(a2 + x2)3/2
39. y = (a + bx)1/3;dy
dx=
b
3(a + bx)−2/3;
d2y
dx2=
−2b2
9(a + bx)−5/3
40. y = x(a2 − x2)1/2;dy
dx= (a2 − x2)1/2 + 1
2 x(a2 − x2)−1/2(−2x) = (a2 − x2)1/2 − x2(a2 − x2)−1/2
d2y
dx2= 1
2 (a2 − x2)−1/2(−2x) − 2x(a2 − x2)−1/2 − x2(− 1
2
)(a2 − x2)−3/2(−2x) =
x(2x2 − 3a2)(a2 − x2)3/2
41. y =√x tan
√x;
dy
dx=
12√x
tan√x +
√x sec2
√x
(1
2√x
)=
12√x
tan√x +
12
sec2√x
d2y
dx2=
2√x sec2
√x (1/2
√x) − tan
√x(1/
√x)
4x+ sec
√x sec
√x tan
√x(1/2
√x)
=√x sec2
√x− tan
√x + 2x sec2
√x tan
√x
4x√x
42. y =√x sin
√x;
dy
dx=
12√x
sin√x +
√x cos
√x
(1
2√x
)=
12√x
sin√x +
12
cos√x
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.7 107
d2y
dx2=
2√x cos
√x(1/2
√x) − sin
√x(1/
√x)
4x− 1
2sin
√x(1/2
√x)
=√x cos
√x− sin
√x
4x3/2− sin
√x
4√x
43. Differentiation of x2 + y2 = r2 gives 2x + 2ydy
dx= 0 so that the slope of the normal line is
−1dy/dx
=y
x(x �= 0).
Let (x0, y0) be a point on the circle. Clearly, if x0 = 0, the normal line, x = 0, passes through the
origin. If x0 �= 0, the normal line is
y − y0 =y0
x0(x− x0), which simplifies to y =
y0
x0x,
a line through the origin.
44. y2 = x; 2ydy
dx= 1;
dy
dx=
12y
When x = a, y = ±√a.
Slope of tangent at (a,√a) :
12√a
Equation of tangent line: y −√a =
12√a(x− a)
x-intercept: −√a =
12√a(x− a) =⇒ x = −a
Slope of tangent at (a,−√a) : − 1
2√a
Equation of tangent line: y +√a = − 1
2√a(x− a)
x-intercept:√a = − 1
2√a(x− a) =⇒ x = −a
45. For the parabola y2 = 2px + p2, we have 2ydy
dx= 2p and the slope of a tangent is given by m1 = p/y.
For the parabola y2 = p2 − 2px, we obtain m2 = −p/y as the slope of a tangent. The parabolas
intersect at the points (0, ±p). At each of these points m1m2 = −1; the parabolas intersect at right
angles.
46. For y = 2x, the slope is m1 = 2. For x2 − xy + 2y2 = 28, we have
2x− y − xdy
dx+ 4y
dy
dx= 0 so
dy
dx= m2 =
y − 2x4y − x
At a point of intersection of the line and the curve, we have m2 = 0 since y = 2x. Thus
tanα = | −m1| = 2 =⇒ α ∼= 1.107(radians) ∼= 63.4◦
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
108 SECTION 3.7
47. For y = x2 we have m1 =dy
dx= 2x; for x = y3 we have 3y2 dy
dx= 1 or m2 =
dy
dx= 1/3y2.
At (1, 1), m1 = 2, m2 = 1/3 and
tanα =∣∣∣∣ m1 −m2
1 −m1m2
∣∣∣∣ =∣∣∣∣ 2 − (1/3)1 + 2(1/3)
∣∣∣∣ = 1 ⇒ α =π
4
At (0, 0), m1 = 0 and m2 is undefined. Thus α = π/2.
48. (x− 1)2 + y2 = 10; 2(x− 1) + 2ydy
dx= 0, m1 =
dy
dx= − x− 1
y
x2 + (y − 2)2 = 5; 2x + 2(y − 2)dy
dx= 0, m2 =
dy
dx= − x
y − 2
The circles intersect at the points (−2, 1) and (2, 3).
At (−2, 1) : m1 = 3, m2 = −2 and tanα =∣∣∣∣ 3 + 21 + (3)(−2)
∣∣∣∣ = 1. Thus α =π
4.
At (2, 3) : m1 = −1/3, m2 = −2 and tanα =∣∣∣∣ (−1/3) + 21 + (−1/3)(−2)
∣∣∣∣ = 1. Thus α =π
4.
49. The hyperbola and the ellipse intersect at the points (±3,±2). For the hyperbola,dy
dx=
x
yand for
the ellipsedy
dx= − 4x
9y. The product of these slopes is − 4x2
9y2. This product is −1 at each of the points
of intersection. Therefore the hyperbola and ellipse are orthogonal.
50. The curves intersect at the points (±1, 1). For the ellipse,dy
dx= − 3x
2yand for y3 = x2 we have
dy
dx=
2x3y2
. The product of these slopes is − 6x2
6y3= − x2
y3. This product is −1 at each of the points of
intersection. Therefore the curves are orthogonal.
51. For the circles,dy
dx= − x
y, y �= 0, and for the straight lines,
dy
dx= m =
y
x, x �= 0. Since the product
of the slopes is −1, it follows that the two families are orthogonal trajectories.
52. For the parabolas, m1 =dy
dx=
12ay
, y �= 0, and for the ellipses, m2 =dy
dx= − 2x
y, y �= 0. Let (x0, y0)
be a point of intersection of a parabola and an ellipse. Then
m1 ·m2 =1
2ay0·(− 2x0
y0
)= − x0
ay20
= −1 since x0 = ay20 .
53. The line x + 2y + 3 = 0 has slope m = −1/2. Thus, a line perpendicular to this line will have slope 2. A
tangent line to the ellipse 4x2 + y2 = 72 has slope m =dy
dx= − 4x
y. Setting − 4x
y= 2 gives y = −2x.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.7 109
Substituting into the equation for the ellipse, we have
4x2 + 4x2 = 72 ⇒ 8x2 = 72 ⇒ x = ±3
It now follows that y = ∓6 and the equations of the tangents are:
at (3,−6) : y + 6 = 2(x− 3) or y = 2x− 12;
at (−3, 6) : y − 6 = 2(x + 3) or y = 2x + 12.
54. The line 2x + 5y − 4 = 0 has slope m = −2/5. A tangent line to the hyperbola 4x2 − y2 = 36 has slope
dy
dx=
4xy. Therefore a normal line to the hyperbola will have slope m = − y
4x. Setting − y
4x= − 2
5
gives y = 8x/5. Substituting this into the equation for the hyperbola, we have
4x2 − 6425
x2 = 36 =⇒ x = ±5
It now follows that y = 8 when x = 5 and y = −8 when x = −5. The equations of the normals are:
at (5, 8) : y − 8 = − 25 (x− 5) or y = − 2
5x + 10;
at (−5,−8) : y + 8 = − 25 (x + 5) or y = − 2
5x− 10.
55. Differentiate the equation (x2 + y2)2 = x2 − y2 implicitly with respect to x :
2(x2 + y2)(
2x + 2ydy
dx
)= 2x− 2y
dy
dx
Now set dy/dx = 0. This gives
2x(x2 + y2) = x
x2 + y2 =12
(x �= 0)
Substituting this result into the original equation, we get
x2 − y2 =14
Nowx2 + y2 = 1/2
x2 − y2 = 1/4⇒ x = ±
√6
4, y = ±
√2
4
Thus, the points on the curve at which the tangent line is horizontal are:
(√
6/4,√
2/4), (√
6/4,−√
2/4), (−√
6/4,√
2/4), (−√
6/4,−√
2/4).
56. (a) x2/3 + y2/3 = a2/3; 23x
−1/3 + 23y
−1/3 dy
dx= 0, and
dy
dx= −
(y
x
)1/3
Thus, the slope at (x1, y1), x1 �= 0 is: m = −(y1
x1
)1/3
(b) m = 0 : −(y1
x1
)1/3
= 0 =⇒ y1 = 0 and x1 = ± a; (a, 0), (−a, 0)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
110 SECTION 3.7
m = 1 : −(y1
x1
)1/3
= 1 =⇒ y1 = −x1 and x1 = ± 14 a
√2;
(14a√
2,− 14a√
2),
(− 1
4a√
2,14a√
2)
m = −1 : −(y1
x1
)1/3
= −1 =⇒ y1 = x1 and x1 = ± 14 a
√2;
(14a√
2,14a√
2),
(− 1
4a√
2,− 14a√
2)
57. Differentiate the equation x1/2 + y1/2 = c1/2 implicitly with respect to x :12x−1/2 +
12y−1/2 dy
dx= 0 which implies
dy
dx= −
(y
x
)1/2
An equation for the tangent line to the graph at the point (x0, y0) is
y − y0 = −(y0
x0
)1/2
(x− x0)
The x- and y-intercepts of this line are
a = (x0y0)1/2 + x0 and b = (x0y0)1/2 + y0 respectively.
Now
a + b = 2(x0y0)1/2 + x0 + y0 =(x
1/20 + y
1/20
)2
= c.
58. The circle has equation x2 + (y − a)2 = 1 and 2x + 2(y − a)dy
dx= 0. Thus
dy
dx= − x
y − a.
A tangent line to the parabola has slopedy
dx= 4x. Now
4x = − x
y − a=⇒ x(4y − 4a + 1) = 0 =⇒ 4y − 4a + 1 = 0 since x �= 0
It follows that
y = a− 14
=⇒ x = ±√
154
=⇒ y =158
Points of intersection:
(±
√154
,158
)
59. (a) d(h) =3 3√h
h=
3h2/3
(b) d(h) → ∞ as h → 0− and as h → 0+
(c) The graph of f has a vertical tangent at (0, 0).
60. (a) d(h) =3 3√h2
h=
3h1/3
(b) d(h) → −∞ as h → 0−; and d(h) → ∞ as h → 0+
(c) The graph is said to have a “cusp” at (0, 0).
61. f ′(x) > 0 on (−∞,∞)
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
SECTION 3.7 111
62. (a) f ′(x) = 0 at x =√
33
; (b) f ′(x) > 0 on(√
3/3,∞); (c) f ′(x) < 0 on
(0,√
3/3)
63. x = t, y =√
4 − t2 x = t, y = −√
4 − t2
64.dy
dx
∣∣∣∣(2,1)
= − 32
65.dy
dx
∣∣∣∣(3,4)
= 3 66.dy
dx
∣∣∣∣(0,π/6)
= 0
67.dy
dx
∣∣∣∣(1,3
√3)
= −√
3
68. (b) x = 3 implies y = 3;dy
dx
∣∣∣∣(3,3)
= −1; tangent line: y − 3 = −(x− 3) or x + y = 6.
(c)
-2 2 3 4x
-2
2
3
4
y
69. (a) The graph of x4 = x2 − y2 is:
-0.5 0.5x
-0.5
0.05
y
(b) Differentiate the equation x4 = x2 − y2
implicitly with respect to x :
4x3 = 2x− 2ydy
dx
Now set dy/dx = 0. This gives 4x3 = 2x
which implies x = ±√
22 .
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
112 REVIEW EXERCISES
70. The graph of (2 − x)y2 = x3 is:
x
y
5
-15
-10
-5
10
15
5 10 15
REVIEW EXERCISES
1. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
[(x + h)3 − 4(x + h) + 3] − [x3 − 4x + 3]h
= limh→0
3x2h + 3xh2 + h3 − 4hh
= limh→0
(3x2 + 3xh + h2 − 4) = 3x2 − 4.
2. f ′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√1 + 2(x + h) −
√1 + 2x
h
= limh→0
√1 + 2(x + h) −
√1 + 2x
h·√
1 + 2(x + h) +√
1 + 2x√1 + 2(x + h) +
√1 + 2x
= limh→0
2h
h[√
1 + 2(x + h) +√
1 + 2x] =
1√1 + 2x
3.g′(x) = lim
h→0
f(x + h) − f(x)h
= limh→0
1x + h− 2
− 1x− 2
h
= limh→0
(x− 2) − (x + h− 2)h(x + h− 2)(x− 2)
= limh→0
−1(x + h− 2)(x− 2)
=−1
(x− 2)2.
4.F ′(x) = lim
h→0
F (x + h) − F (x)h
= limh→0
(x + h) sin (x + h) − x sin x
h
= limh→0
(x + h)(sin x cos h + cos x sin h) − x sin x
h
= limh→0
x sin x(cos h− 1) + h sin x cos h + x cos x sin h + h cos x sin h
h
= x sin x limh→0
cos h− 1h
+ limh→0
sin x cos h + x cos x limh→0
sin h
h+ lim
h→0cos x sin h
= sin x + x cos x.
5. y′ = 23x
−1/3
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
REVIEW EXERCISES 113
6. y′ = 2( 34 )x−1/4 − 4(− 1
4 )x−5/4 = 32 x
−1/4 + x−5/4.
7. y′ =(x3 − 1)(2x + 2) − (1 + 2x + x2)3x2
(x3 − 1)2= −x4 + 4x3 + 3x2 + 2x + 2
(x3 − 1)2.
8. f ′(t) = 3(2 − 3t2)2(−6t) = −18t(2 − 3t2)2.
9. f(x) = (a2 − x2)−1/2; f ′(x) = −12
(a2 − x2)−3/2(−2x) =x
(a2 − x2)3/2.
10. y′ = 2(a− b
x
) (b
x2
)=
2bx2
(a− b
x
).
11. y′ = 3(a +
b
x2
)2 (− 2bx3
)= − 6b
x3
(a +
b
x2
)2
.
12. y′ =√
2 + 3x +x
2(2 + 3x)−1/2(3) =
√2 + 3x +
3x2√
2 + 3x.
13. y′ = sec2√
2x + 1[12 (2x + 1)−1/2(2)
]=
sec2√
2x + 1√2x + 1
.
14. g′(x) = x2 [− sin (2x− 1)(2)] + 2x cos (2x− 1) = 2x cos (2x− 1) − 2x2 sin (2x− 1).
15. F ′(x) = (x + 2)2 12 (x2 + 2)−1/2(2x) +
√x2 + 2 (2)(x + 2) = 2(x + 2)
√x2 + 2 +
x(x + 2)2√x2 + 2
.
16. y′ =(a2 − x2)2x− (a2 + x2)(−2x)
(a2 − x2)2=
4a2x
(a2 − x2)2.
17. h′(t) = sec t2 + t(2t) sec t2 tan t2 + 6t2 = sec t2 + 2t2 tan t2 sec t2 + 6t2.
18. y′ =(1 + cosx)2 cos 2x− sin 2x(− sinx)
(1 + cosx)2=
2 cos 2x1 + cosx
+sin 2x sinx
(1 + cosx)2.
19. s′ =13
(2 − 3t2 + 3t
)−2/3
· (2 + 3t)(−3) − (2 − 3t)(3)(2 + 3t)2
= − 4(2 + 3t)4/3(2 − 3t)2/3
.
20. r′ = θ2( 12 )(3 − 4θ)−1/2(−4) + 2θ
√3 − 4θ = 2θ
√3 − 4θ − 2θ2
√3 − 4θ
.
21. f ′(θ) = −3 csc2(3θ + π).
22. y′ =(1 + x2)(sin 2x + 2x cos 2x) − x sin 2x(2x)
(1 + x2)2=
sin 2x + 2x cos 2x + 2x3 cos 2x− x2 sin 2x(1 + x2)2
.
23. f ′(x) = 13x
−2/3 + 12x
−1/2 =1
3x2/3+
12x1/2
; f ′(64) =148
+116
=112
.
24. f ′(x) =x
21√
8 − x2(−2x) +
√8 − x2 =
√8 − x2 − x2
√8 − x2
; f ′(2) = 0.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
114 REVIEW EXERCISES
25. f ′(x) = x2(2)(π) sinπx cosπx + 2x sin2 πx = 2x sin2 πx + 2πx2 sinπx cosπx; f ′(1/6) =112
+π√
372
.
26. f ′ = −3 csc2 3x; f ′(π/9) = −4.
27. f(x) = 2x3 − x2 + 3, f ′(x) = 6x2 − 2x; f ′(1) = 4.
Tangent line: y − 4 = 4(x− 1) or y = 4x; normal line: y − 4 = − 14 (x− 1) or y = − 1
4x + 174 .
28. f(x) =2x− 33x + 4
, f ′(x) =(3x + 4)2 − (2x− 3)3
(3x + 4)2=
17(3x + 4)2
; f ′(−1) = 17.
Tangent line: y + 5 = 17(x + 1); normal line: y + 5 = − 117 (x + 1).
29. f(x) = (x + 1) sin 2x, f ′(x) = 2(x + 1) cos 2x + sin 2x; f ′(0) = 2.
Tangent line: y = 2x; normal line: y = − 12x.
30. f(x) = x√
1 + x2, f ′(x) =√
1 + x2 +x2
√1 + x2
; f ′(1) = 32
√2.
Tangent line: y −√
2 = 32
√2 (x− 1); normal line: y −
√2 = − 1
3
√2 (x− 1)
31. f ′(x) = − sin(2 − x)(−1) = sin(2 − x), f ′′(x) = cos(2 − x)(−1) = − cos(2 − x).
32. f ′(x) = 32 (x2 + 4)1/2(2x) = 3x(x2 + 4)1/2,
f ′′ = 3x( 12 )(x2 + 4)−1/2(2x) + 3(x2 + 4)1/2 =
3x2
(x2 + 4)1/2+ 3(x2 + 4)1/2 =
6x2 + 12√x2 + 4
.
33. y′ = x cosx + sinx, y′′ = 2 cosx− x sinx.
34. g′(u) = 2 tanu sec2 u, g′′(u) = 2 tanu · 2 secu · secu tanu + 2 sec4 u = 4 sec2 u tan2 u + 2 sec4 u.
35. (−1)n n!bn 36.(−1)n n!abn
(bx + c)n+1
37. 3x2y + x3y′ + y3 + 3xy2y′ = 0, y′ = −y3 + 3x2y
x3 + 3xy2.
38. (1 + 2y′) sec2(x + 2y) = 2xy + x2y′, y′ =2xy − sec2(x + 2y)2 sec2(x + 2y) − x2
.
39. 6x2 + 3 cos y − 3xy′ sin y = 2y + 2xy′, y′ =6x2 + 3 cos y − 2y
2x + 3x sin y.
40. 2x + 3√y +
3xy′
2√y
=1y− xy′
y2, y′ =
2y2 − 4xy3 − 6y7/2
2xy + 3xy5/2.
41. 2x + 2y + 2xy′ − 6yy′ = 0, y′ =x + y
3y − x; at (3, 2) y′ =
53.
Tangent line: y − 2 = 53 (x− 3) or 5x− 3y = 9; normal line: y − 2 = − 3
5 (x− 3) or 3x + 5y = 19.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
REVIEW EXERCISES 115
42. 2y cos 2x + y′ sin 2x− xy′ cos y − sin y = 0, y′ =sin y − 2t cos 2xsin 2x− x cos y
; at ( 14π,
12π) y′ = 1.
Tangent line: y − 12π = x− 1
4π or y = x + 14π; normal line: y − 1
2π = −(x− 14π) or y = −x + 3
4π.
43. f ′(x) = 3(x− 4)(x− 2).
f ′:2 4
-+ +
(a) f ′(x) = 0 at x = 2, 4, (b) f ′(x) > 0 on (−∞, 2) ∪ (4,∞),
(c) f ′(x) < 0 on (2, 4).
44. f ′(x) =2 − 4x2
(1 + 2x2)2.
f ′:1
2
+-
-
-
1
2
(a) f ′(x) = 0 at x = ±√
2/2, (b) f ′(x) > 0 on (−√
2/2,√
2/2),
(c) f ′(x) < 0 on (−∞,−√
2/2) ∪ (√
2/2,∞).
45. f ′(x) = 1 + 2 cos 2x.
f ′:π
+ - + - +2 π 4 π 5 π
3 3 3 3
(a) f ′(x) = 0 at x = π/3, 2π/3, 4π/3, 5π/3,
(b) f ′(x) > 0 on (0, π/3) ∪ (2π/3, 4π/3) ∪ (5π/3, 2π),
(c) f ′(x) < 0 on (π/3, 2π/3) ∪ (4π/3, 5π/3).
46. f ′(x) =√
3 + 2 sinx.
f ′:0 4π
35π π3
2
+ -
(a) f ′(x) = 0 at x = 4π/3, 5π/3, (b) f ′(x) > 0 on (0, 4π/3) ∪ (5π/3, 2π),
(c) f ′(x) < 0 on (4π/3, 5π/3).
47. y′ =√x. (a) 1, (b) 3, (c) 1
3 .
48. y′ = 3x2. The tangent line to the curve y = x3 at the point (a, a3) has equation
y − a3 = 3a2(x− a) or y = 3a2x− 2a3.
Since this line must pass through (0, 2), we have
2 = −2a3 which implies a = −1.
There is one tangent line that passes through (0, 2), namely y = 3x + 2.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
116 REVIEW EXERCISES
49. y′ = 3x2 − 1. The tangent line to the curve y = x3 − x at the point (a, a3 − a) has equation
y − (a3 − a) = (3a2 − 1)(x− a) or y = (3a2 − 1)x− 2a3.
Since this line must past through (−2, 2), we have
2 = −6a2 + 2 − 2a3 which gives 6a2 + 2a3 = 0 and 2a2(3 + a) = 0.
Thus, a = 0, −3. There are two tangent lines that pass through (−2, 2) : y = −x, y = 26x + 54.
50. The curve y = Ax2 + Bx + C passes through the points (1, 3) and (2, 3). This implies that
A + B + C = 3 and 4A + 2B + C = 3.
y′ = 2Ax + B. The line x− y + 1 = 0 has slope 1. At x = 2, y′ = 4A + B. Therefore we have
4A + B = 1. The solution set of the system of equations
A + B + C = 3
4A + 2B + C = 3
4A + B = 1
is: A = 1, B = −3, C = 5; y = x2 − 3x + 5.
51. The curve y = Ax3 + Bx2 + Cx + D passes through the points (1, 1) and (−1,−9). This implies that
A + B + C + D = 1 and −A + B − C + D = −9.
y′ = 3Ax2 + 2Bx + C. The line y = 5x− 4 has slope 5; the line y = 9x has slope 9. At
x = 1, y′ = 3A + 2B + C; at x = −1, y′ = 3A− 2B + C. Therefore we have 3A + 2B + C = 5 and
3A− 2B + C = 9. The solution set of the system of equations
A + B + C + D = 1
−A + B − C + D = −9
3A + 2B + C = 5
3A− 2B + C = 9
is: A = 1, B = −1, C = 4, D = −3; y = x3 − x2 + 4x− 3.
52.1h
[1
(x + h)n− 1
xn
]=
1h
[xn − (x + h)n
(x + h)n xn
]
=1h
[xn − xn − nxn−1h− (terms of the form Cxk hm,m ≥ 2)
(x + h)n xn
]
=−nxn−1 − (terms of the form Cxk hm,m ≥ 1)
(x + h)n xn.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-03 JWDD027-Salas-v1 November 25, 2006 15:53
REVIEW EXERCISES 117
Therefore,
limh→0
1h
[1
(x + h)n− 1
xn
]= lim
h→0
−nxn−1 − (terms of the form Cxk hm,m ≥ 1)(x + h)n xn
=−nxn−1
x2n=
−n
xn+1.
53. Let f(x) = x2 − 2x + 1; f ′(x) = 2x− 2.
limh→0
(1 + h)2 − 2(1 + h) + 1h
= limh→0
(1 + h)2 − 2(1 + h) + 1 − 0h
= f ′(1) = 0
54. Let f(x) =√x; f ′(x) = 1/2
√x.
limh→0
√9 + h− 3
h= f ′(9) =
16
55. Let f(x) = sinx; f ′(x) = cosx.
limh→0
sin( 16π + h) − 1
2
h= f ′(
16π) =
√3
2
56. Let f(x) = x5; f ′(x) = 5x4.
limx→2
x5 − 32x− 2
= f ′(2) = 80
57. Let f(x) = sinx; f ′(x) = cosx.
limx→π
sinx
x− π= f ′(π) = −1
58. (a) From Figure A, M increases on [a, b], is constant on [b, e], and increases on [e,∞); M is differen-
tiable at b (f ′(b) = 0), M is not differentiable at e (f ′(e) �= 0).
(b) From Figure B, m is constant on [a, c], decreases on [c, d], and is constant on [d,∞); m is not
differentiable at c (f ′(c) �= 0), m is differentiable at d (f ′(d) = 0).
a b e a b e
Figure A Figure B