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



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:


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


f(2 + h) − f(2)h

= limh→0

(3 − h) = 3

13. f(x) = 2x3 + 1; c = 1:


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:


f(1 + h) − f(1)h

=5 − (1 + h)4 − (4)

h



SECTION 3.1 67

=5 − 1 − 4h− 6h2 − 4h3 − h4 − 4

h

=−4h− 6h2 − 4h3 − h4

h= −4 − 6h− 4h2 − h3


f(1 + h) − f(1)h

= limh→0

(−4 − 6h− 4h2 − h3

)= −4

15. f(x) =8

x + 4; c = −2:


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:


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


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.



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.



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.



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).



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.



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



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.



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



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)



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.



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.



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)



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



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



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



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.



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.



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.



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

).



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



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



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)



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)


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


dx=

29(0) = 0.

24.dy

dx=

dy

du

du

dx= (3u2 − 1) ·

( −2(1 + x)2

)


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



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



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.



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



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)



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



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.



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



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



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



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 .



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.



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



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.



SECTION 3.7 103


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.


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



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.



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)



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



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◦



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.



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)



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 (−∞,∞)



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 .



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



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.




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.




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.




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.




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

Calculus one and several variables 10E Salas solutions manual ch03

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