chapter 2 differentiation - east brunswick public … … · chapter 2 differentiation section 2.1...

C H A P T E R 2Differentiation

Section 2.1 The Derivative and the Tangent Line Problem . . . . . 95

Section 2.2 Basic Differentiation Rules and Rates of Change . . 109

Section 2.3 Product and Quotient Rules andHigher-Order Derivatives . . . . . . . . . . . . . . . 120

Section 2.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . 134

Section 2.5 Implicit Differentiation . . . . . . . . . . . . . . . . 147

Section 2.6 Related Rates . . . . . . . . . . . . . . . . . . . . . 161

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 173

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

C H A P T E R 2Differentiation

Section 2.1 The Derivative and the Tangent Line Problem

95

1. (a) At slope

At slope

(b) At slope

At slope 2.x2, y2,

52.x1, y1,

52.x2, y2,

0.x1, y1, 2. (a) At slope

At slope

(b) At slope

At slope 54.x2, y2,

43.x1, y1,

25.x2, y2,

23.x1, y1,

3. (a), (b)

(c)

x 1

1x 1 2

33

x 1 2

y f 4 f 1

4 1x 1 f 1)

6

5

4

3

2

654321

1

y

x

1)1f ) x

)1 3f ))f )4

11)f )

1)x

))f 44

y

4) , )5

2)

21) ,

f )

)

1

f ) )4 5

)4. (a)

Thus,

(b) The slope of the tangent line at equals This slope is steeper than the slope of the line through

and Thus,

f 4 f 14 1

< f1.

4, 5.1, 2

f1.1, 2

f 4 f 14 1

>f 4 f 3

4 3.

f 4 f 34 3

5 4.75

1 0.25

f 4 f 14 1

5 2

3 1

5. is a line. Slope 2f x 3 2x 6. is a line. Slope 32g x 32 x 1

7. Slope at

limx0

2 x 2

limx0

1 2x x2 1

x

limx0

1 x2 4 3

x

1, 3 limx0

g1 x g1

x8. Slope at

limx0

4 x 4

limx0

5 4 4x x2 1

x

limx0

5 2 x2 1

x

2, 1 limx0

g2 x g2

x

9. Slope at

limt0

3 t 3

limt0

3t t2 0

t

0, 0 limt0

f 0 t f 0

t10. Slope at

limt0

4 t 4

limt0

4 4t t2 4

t

limt0

2 t2 3 7

t

2, 7 limt0

h2 t h2

t

96 Chapter 2 Differentiation

14.

limx0

3 3

limx0

3xx

limx0

3x x 2 3x 2

x

fx limx0

f x x f x

x

f x 3x 2 15.

lims0

23 s

s

23

lims0

3 23s s 3 23s

s

hs lims0

hs s hs

s

hs 3 23

s

16.

limx0

12 12

limx0

9 12x x 9 12x

x

fx limx0

f x x f x

x

f x 9 12

x

11.

limx0

0 0

limx0

3 3

x

f x limx0

f x x f x

x

f x 3 12.

limx0

0

x 0

limx0

5 5

x

gx limx0

gx x gx

x

gx 5 13.

limx0

5 5

limx0

5x x 5x

x

fx limx0

f x x f x

x

f x 5x

17.

limx0

4x x 2x2 x

x lim

x0 4x 2 x 1 4x 1

limx0

2x2 4x x 2x2 x x 1 2x2 x 1

x

limx0

2x x2 x x 1 2x2 x 1

x

fx limx0

f x x f x

x

f x 2x 2 x 1

18.

limx0

2x x x2

x lim

x0 2x x 2x

limx0

1 x2 2x x x2 1 x2

x

limx0

1 x x2 1 x2

x

fx limx0

f x x f x

x

f x 1 x2

Section 2.1 The Derivative and the Tangent Line Problem 97

19.

limx0

3x2 3x x x2 12 3x2 12

limx0

3x2 x 3xx2 x3 12 x

x

limx0

x3 3x2 x 3xx2 x3 12x 12 x x3 12x

x

limx0

x x3 12x x x3 12x

x

fx limx0

f x x f x

x

f x x3 12x

20.

limx0

3x2 3x x x2 2x x 3x2 2x

limx0

3x2 x 3xx2 x3 2x x x2

x

limx0

x3 3x2 x 3xx2 x3 x2 2x x x2 x3 x2

x

limx0

x x3 x x2 x3 x2

x

fx limx0

f x x f x

x

f x x3 x2

21.

1

x 12

limx0

1

x x 1x 1

limx0

x

xx x 1x 1

limx0

x 1 x x 1xx x 1x 1

limx0

1x x 1

1

x 1x

fx limx0

f x x f x

x

f x 1x 1

22.

2x3

2xx 4

limx0

2x xx x2x2

limx0

2x x x2xx x2x2

limx0

x2 x x2xx x2x2

limx0

1x x2

1x2

x

fx limx0

f x x f x

x

f x 1x2

23.

1

x 1 x 1

1

2x 1 lim

x0

1

x x 1 x 1

limx0

x x 1 x 1

x x x 1 x 1

limx0

x x 1 x 1

x x x 1 x 1x x 1 x 1

fx limx0

f x x f x

x

f x x 1


24.

4

xxx x 2

xx

limx0

4

xx xx x x

limx0

4x 4x x

xxx xx x x

limx0

4x 4x x

xxx x x x xx x x

limx0

4x x

4x

x

fx limx0

f x x f x

x

f x 4x

25. (a)

At the slope of the tangent line isThe equation of the tangent line is

(b)

(2, 5)

5 5

2

8

y 4x 3.

y 5 4x 8

y 5 4x 2

m 22 4.2, 5,

limx0

2x x 2x

limx0

2x x x2

x

limx0

x x2 1 x2 1

x

fx limx0

f x x f x

x

f x x2 1

26. (a) (b)

At the slope of the tangent line is The equation of the tangent line is

y 4x 8.

y 4 4x 3

m 23 2 4.3, 4,

limx0

2x x 2 2x 2

limx0

2x x x2 2 x

x

limx0

x x2 2x x 1 x2 2x 1

x

fx limx0

f x x f x

x6 3

1

( 3, 4)

5 f x x2 2x 1

27. (a) (b)

At the slope of the tangent is The equation of the tangent line is

y 12x 16.

y 8 12x 2

m 322 12.2, 8,

limx0

3x2 3x x x2 3x2

limx0

3x2 x 3xx2 x3

x

limx0

x x3 x3

x

fx limx0

f x x f x

x5 5

4

(2, 8)

10 f x x3


28. (a) (b)

At the slope of the tangent line is The equation of the tangent line is

y 3x 1.

y 2 3x 1

m 312 3.1, 2,

limx0

3x2 3xx x2 3x2

limx0

x3 3x2x 3xx2 x3 1 x3 1

x

limx0

x x3 1 x3 1

x

fx limx0

f x x f x

x 6

4

6

4

(1, 2)

f x x3 1

29. (a)

At the slope of the tangent line is

The equation of the tangent line is

y 12

x 12

.

y 1 12

x 1

m 1

21

12

.

1, 1,

limx0

1

x x x

1

2x

limx0

x x x

xx x x

limx0

x x x

xx x xx x x

fx limx0

f x x f x

x

f x x (b)

5

1

1

3

(1, 1)

30. (a) (b)



y 14

x 34

y 2 14

x 5

m 1

25 1

14

5, 2,

limx0

1

x x 1 x 1

1

2x 1

limx0

x x 1 x 1

xx x 1 x 1

limx0

x x 1 x 1

x x x 1 x 1x x 1 x 1

fx limx0

f x x f x

x

(5, 2)

2

4

10

4f x x 1


31. (a) (b)



y 34

x 2.

y 5 34

x 4

m 1 4

16

34

.

4, 5,

x2 4

x2 1

4x2

limx0

x2 xx4

xx x

limx0

x2x xx2 4x

xxx x

limx0

x3 2x 2x xx2 x3 x 2x 4x

xxx x

limx0

xx xx x 4x x2x x 4x x

xxx x

limx0

x x 4

x x x 4x

x

fx limx0

f x x f x

x 12

6

12

10

(4, 5)

f x x 4x

32. (a)



(b)

6 3

3

(0, 1)

3

y x 1.

m 1

0 12 1.

0, 1,

1

x 12

limx0

1

x x 1x 1

limx0

x 1 x x 1xx x 1x 1

limx0

1x x 1

1

x 1x

fx limx0

f x x f x

x

f x 1x 1

33. From Exercise 27 we know that Since theslope of the given line is 3, we have

Therefore, at the points and the tangentlines are parallel to These lines haveequations

and

y 3x 2. y 3x 2

y 1 3x 1y 1 3x 1

3x y 1 0.1, 11, 1

x 1.

3x2 3

fx 3x2.


34. Using the limit definition of derivative, Since the slope of the given line is 3, we have

Therefore, at the points and the tangent lines are parallel to These lines have equations

y 3x y 3x 4.

y 3 3x 1 and y 1 3x 1

3x y 4 0.1, 11, 3

x2 1 x 1.

3x2 3

fx 3x2.

35. Using the limit definition of derivative,

Since the slope of the given line is , we have

Therefore, at the point the tangent line is parallel toThe equation of this line is

y 12

x 32

.

y 1 12

x 12

y 1 12

x 1

x 2y 6 0.1, 1

x 1.

1

2xx

12

12

fx 12xx

.

36. Using the limit definition of derivative,

Since the slope of the given line is we have

At the point the tangent line is parallel toThe equation of the tangent line is

y 12

x 2.

y 1 12

x 2

x 2y 7 0.2, 1,

1 x 1 x 2.

1 x 132

12x 132

12

12,

fx 12x 13 2.

37. Matches (b).f x x fx 1 38. Matches (d).f x x2 fx 2x

39. Matches (a).

decreasing slope as x

f x x fx 40. does not exist at Matches (c).x 0.f

41. because the tangent line passes through

g5 2 05 9

2

4

12

5, 2.g5 2 42. because the tangent line passes through

h1 6 43 1

24

12

1, 4.h1 4

43. The slope of the graph of is

2

2

1

3

4

1 22 1 33

f

y

x

1 fx 1.f 44. The slope of the graph of

is

12 1 2

1

2

1

2

x

y

f

0 fx 0.f 45. The slope of the graph of is

negative for positive for and 0 at

246 2 4 62

4

6

8

2

4

x

y

f

x 4.x > 4,x < 4,

f


46. The slope of the graph of is for 1 for andundefined at

x

y

f

1 2 3 4 5 6

1

2

3

x 4.x > 4,x < 4,

1f 47. Answers will vary. Sample answer:

x1

1

2

3

4

2

1

3

4

2 134 2 3 4

y

y x48. Answers will vary.

Sample answer:

x1

2

3

4

2

1

3

4

234 2 3 4

y

y x

49. and c 1f x 5 3x 50. and c 2f x x3 51. and c 6f x x2 52. and c 9f x 2x

53. and

123 2 31

2

3

x

y

f

1

2

f x 3x 2

< x < fx 3,f 0 2 54. for

for

246 2 4 6

2

4

6

8

10

12

x

y

f

f x x2 4

x > 0f x > 0x < 0,fx < 0f0 0;f 0 4, 55. if

23 1 2 31

2

3

1

2

3

x

y

f

f x x3x 0

fx > 0f0 0;f 0 0;

56. (a) If and is odd, then (b) If and is even, then fc f c 3.ffc 3fc fc 3.ffc 3

57. Let be a point of tangency on the graph of f. By the limit definition for the derivative,The slope of the line through and equals the derivative of f at

Therefore, the points of tangency are and andthe corresponding slopes are 2 and . The equations ofthe tangent lines are:

x1 2 3 62

2

4

3

5

7

6

1

(1, 3)(3, 3)

(2, 5)

y

y 2x 1 y 2x 9

y 5 2x 2 y 5 2x 2

23, 3,1, 3

0 x0 1x0 3 x0 1, 3

0 x02 4x0 3

5 4x0 x02 8 8x0 2x02 5 y0 2 x04 2x0

5 y02 x0

4 2x0

x0:x0, y02, 5

fx 4 2x.x0, y0 58. Let be a point of tangency on the graph of f. By

the limit definition for the derivative, Theslope of the line through and equals thederivative of f at

Therefore, the points of tangency are and and the corresponding slopes are 6 and The equationsof the tangent lines are:

x2 4 626 48

2

4

4

8

6

10(3, 9)

(1, 3)

( 1, 1)

y

y 6x 9 y 2x 1

y 3 6x 1 y 3 2x 1

2.(1, 1,3, 9

x0 3x0 1 0 x0 3, 1

x02 2x0 3 0

3 x02 2x0 2x0

2

3 y0 1 x02x0

3 y01 x0

2x0

x0:x0, y01, 3fx 2x.

x0, y0


59. (a)

(b)

(c) Because g is decreasing (falling) at

(d) Because g is increasing (rising) at

(e) Because and are both positive, is greater than , and

(f) No, it is not possible. All you can say is that g is decreasing (falling) at x 2.

g6 g4 > 0.g4g6g6g4

x 4.g4 73 ,

x 1.g1 83 ,

g3 0

g0 3

60. (a)

At and the tangent line is

or .



For this function, the slopes of the tangent lines arealways distinct for different values of x.

3

3

3

2

y 2x 1.f1 2x 1,y 0.f0 0 x 0,

y 2x 1y 1 2x 1f1 2x 1,

limx0

2x x 2x

limx0

x2x x

x

limx0

x2 2xx x2 x2

x

limx0

x x2 x2

x

fx limx0

f x x f x

x

f x x2 (b)


or .



For this function, the slopes of the tangent lines are sometimes the same.

3 3

2

2

y 1 3x 1 or y 3x 2.

x 1, g1 3

y 0.g0 0x 0,y 3x 2y 1 3x 1

g1 3x 1,

limx0

3x2 3xx x2 3x2

limx0

x3x2 3xx x2

x

limx0

x3 3x2x 3xx2 x3 x3

x

limx0

x x3 x3

x

gx limx0

g x x g x

x

61.

By the limit definition of the derivative we have fx 34 x2.

2

2

2

2

f x 14 x3

x 0 0.5 1 1.5 2

0 2

3 0 3271634

316

316

34

2716fx

2732

14

132

132

14

27322f x

0.511.52

62.

By the limit definition of the derivative we have fx x.

2 2

1

3f x 12 x2

x 1.5 0.5 0 0.5 1 1.5 2

2 1.125 0.5 0.125 0 0.125 0.5 1.125 2

1.5 0.5 0 0.5 1 1.5 212fx

f x

12


63.

The graph of is approximatelythe graph of fx 2 2x.

gx

3

1

2 4

g

f

2 2x 0.01

2x 0.01 x 0.012 2x x2100

gx f x 0.01 f x0.01

64.

The graph of is approximately

the graph of fx 32x

.

gx

1

1 8

f

g

8

3x 0.01 3x 100gx f x 0.01 f x

0.01

65.

Exact: f2 0f2 3.99 42.1 2

0.1

f 2.1 2.14 2.1 3.99f 2 24 2 4, 66.

Exact: f2 3f2 2.31525 22.1 2

3.1525

f 2.1 2.31525f 2 14

23 2,

67. and

As is nearly horizontal

and thus f 0.

x , f

5

5

2 5

f

f

fx 12x32

.f x 1x

68. and

9 9

6

f

f

6

fx 34

x2 3f x x3

4 3x

69.

(a)

(b) As the line approaches the tangent line to at 2, 3.fx 0,

x 0.1: Sx 1910x 2 3 1910

x 45

x 0.5: Sx 32x 2 3 32

x

5

1

2 7

f

S0.1

S1

S0.5

x 1: Sx x 2 3 x 1

4 2 x 32 3

xx 2 3 1 x 1

2

xx 2 3 x 2x 2 3

Sx x f 2 x f 2

xx 2 f 2

f x 4 x 32

70.

(a)

(b) As the line approaches the tangent line to at 2, 52.fx 0,

x 0.1: Sx 1621

x 2 52

1621

x 4142

x 0.5: Sx 45

x 2 52

45

x 9

106

4

6

f1S

0.1S

0.5S

4x 1: Sx 56

x 2 52

56

x 56

22 x2 2 52 x

22 x x x 2 52

2x 322 x x 2

52

Sx x f 2 x f 2

xx 2 f 2

2 x 12 x

52

xx 2 5

2

f x x 1x


71.

f2 limx2

f x f 2

x 2 lim

x2 x2 1 3

x 2 lim

x 2 x 2x 2

x 2 lim

x2x 2 4

f x x2 1, c 2

72.

limx1

x 1 g1 limx1

gx g1

x 1 lim

x1 x2 x 0

x 1 lim

x1 xx 1

x 1

g x xx 1 x2 x, c 1

73.

f2 limx2

f x f 2

x 2 lim

x2 x3 2x2 1 1

x 2 lim

x 2

x2x 2x 2

limx2

x2 4

f x x3 2x2 1, c 2

74.

f1 limx1

f x f 1

x 1 lim

x1 x3 2x 3

x 1 lim

x 1 x 1x2 x 3

x 1 lim

x1x2 x 3 5

f x x3 2x, c 1

75.

Does not exist.

As

As x 0, x

x

1x

.

x 0, x

x

1x .

g0 limx0

gx g0

x 0 lim

x0 x

x.

gx x, c 0

77.

Does not exist.

limx6

1

x 613 limx6 x 623 0

x 6f6 lim

x6 f x f 6

x 6

f x x 623, c 6

76.

f3 limx3

f x f 3

x 3 lim

x3 1x 13

x 3 lim

x3 3 x

3x

1x 3

limx3

13x

19

f x 1x, c 3

78.

Does not exist.

g3 limx3

gx g3

x 3 limx3 x 313 0

x 3 lim

x3

1x 323

g x x 313, c 3

80.

Does not exist.

f4 limx4

f x f 4

x 4 lim

x4 x 4 0

x 4 lim

x4 x 4

x 4

f x x 4, c 4

79.

Does not exist.

limx5

x 5

x 5

limx5

x 5 0x 5

h5 limx5

hx h5

x 5

h x x 5, c 5

81. is differentiable everywhere except at (Discontinuity)

x 1.f x


86. is differentiable everywhere except at (Discontinuity)x 0.f x

82. is differentiable everywhere except at (Sharp turns in the graph)

x 3.f x 83. is differentiable everywhere except at (Sharp turn in the graph)

x 3.f x

84. is differentiable everywhere except at (Discontinuities)

x 2.f x 85. is differentiable on the interval (At the tangent line is vertical.)x 1

1, .f x

90. is differentiable for all is not continuous at

3

5

3

4

x 1.fx 1.f 91.

The derivative from the left is

The derivative from the right is

The one-sided limits are not equal. Therefore, f is not differentiable at x 1.

limx1

f x f 1

x 1 lim

x1 x 1 0

x 1 1.

limx1

f x f 1

x 1 lim

x1 x 1 0

x 1 1.

f x x 1

92.

The derivative from the left does not exist because

(Vertical tangent)

The limit from the right does not exist since f is undefined for Therefore, f is not differentiable at x 1.x > 1.

limx1

f x f 1

x 1 lim

x1 1 x2 0

x 1 lim

x1 1 x2

x 11 x2

1 x2 lim

x1

1 x1 x2

.

f x 1 x2

87. is differentiable for all There is a sharpcorner at

5

2

1

7

x 3.x 3.

f x x 3 88. is differentiablefor all is not defined at

(Vertical asymptote)

66

2

6

x 1.fx 1.

f x 2xx 1

89. is differentiable for all There is a sharp cornerat

6

3

6

5

x 0.x 0.

f x x25

93.



These one-sided limits are equal. Therefore, f is differentiable at x 1. f1 0

limx1

x 1 0.

limx1

f x f 1

x 1 lim

x1 x 12 0

x 1

limx1

x 12 0.

limx1

f x f 1

x 1 lim

x1 x 13 0

x 1

f x x 13,x 12, x 1x > 1


94.



These one-sided limits are not equal. Therefore, f is not differentiable at x 1.

limx1

f x f 1

x 1 lim

x1 x2 1x 1

limx1

x 1 2.

limx1

f x f 1

x 1 lim

x1 x 1x 1

limx1

1 1.

f x x, x2, x 1x > 1

95. Note that f is continuous at



The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 4

limx2

f x f 2

x 2 lim

x2 4x 3 5

x 2 lim

x2 4 4.

limx2

f x f 2

x 2 lim

x2 x2 1 5

x 2 lim

x2 x 2 4.

f x x2 1,4x 3, x 2x > 2x 2.

96. Note that f is continuous at



The one-sided limits are equal. Therefore, f is differentiable at x 2. f2 12

limx2

2x 4

x 22x 2 limx2 2x 2

x 22x 2 limx2 2

2x 2

12

.

limx2

f x f 2

x 2 lim

x2 2x 2

x 22x 22x 2

limx2

f x f 2

x 2 lim

x2 12 x 1 2

x 2 lim

x2 12x 2

x 2

12

.

f x 12 x 1,2x,

x < 2x 2x 2.

97. (a) The distance from to the line is

.

(b)

The function d is not differentiable at This corresponds to the linewhich passes through the point 3, 1.y x 4,

m 1.

5

1

4 4

m3 11 4

m2 1

3m 3

m2 1

d Ax1 By1 C

A2 B2

x

2

3

1 2 3 4

1

ymx y 4 03, 1


(d) If then

Hence, if then which is consistent with the conjecture. However, this is not a proof since you must verify the conjecture for all integer values of n, n 2.

fx 4x3f x x4,

limx0

4x3 6x2x 4xx2 x34x3.

limx0

x4x3 6x2x 4xx2 x3

x

limx0

x 4 4x3x 6x2x2 4xx3 x4 x 4

x

limx0

x x4 x4

x

fx limx0

f x x f x

x

f x x4,

99. False. The slope is limx0

f 2 x f 2

x. 100. False. is continuous at but is not

differentiable at (Sharp turn in the graph)x 2.x 2,y x 2

101. False. If the derivative from the left of a point does not equal the derivative from the right of a point,then the derivative does not exist at that point. For example, if then the derivative from the left at is and the derivative from the right at is At the derivative does not exist.x 0,1.x 01x 0

f x x,

102. Truesee Theorem 2.1.

103.

Using the Squeeze Theorem, we have Thus, and is continuous at Using the alternative form of the derivative, we have

Since this limit does not exist ( oscillates between and 1), the function is not differentiable at

Using the Squeeze Theorem again, we have Thus,

and is continuous at Using the alternative form of the derivative again, we have

Therefore, g is differentiable at x 0, g0 0.

limx0

gx g0

x 0 lim

x0 x2 sin1x 0

x 0 lim

x0 x sin

1x

0.

x 0.g

limx0

x2 sin1x 0 g0x2 x2 sin1x x2, x 0.

gx x2 sin1x,0, x 0x 0x 0.1sin1x

limx0

f x f 0

x 0 lim

x0 x sin1x 0

x 0 lim

x0sin 1x.x 0.f

limx0

x sin1x 0 f 0x x sin1x x, x 0.

f x x sin1x,0, x 0x 0

98. (a) and

x1 2 3 413 24

1

3

2

f

f '

4

5

3

y

fx 2xf x x2 (b) and

g

g

x2 1

1

3

2

1 2

1

y

gx 3x2gx x3 (c) The derivative is a polynomial ofdegree 1 less than the originalfunction. If thenhx nx n1.

hx x n,

Section 2.2 Basic Differentiation Rules and Rates of Change 109

Section 2.2 Basic Differentiation Rules and Rates of Change

104.

As you zoom in, the graph of appears to be locally the graph of a horizontal line, whereas the graph ofalways has a sharp corner at is not differentiable at 0, 1.y2 0, 1.y2 x 1

y1 x2 1

3 3

1

3

1. (a)

y1 12

y 12 x12

y x12 (b)

y1 3

y 3x2

y x3 2. (a)

y1 12

y 12 x32

y x12 (b)

y1 1

y x 2

y x 1

3.

y 0

y 8 4.

f x 0

f x 2 5.

y 6x5

y x6 6.

y 8x7

y x 8

7.

y 7x 8 7x 8

y 1x7

x 7 8.

y 8x 9 8x 9

y 1x 8

x 8 9.

y 15

x 45 1

5x 45

y 5x x 15 10.

y 14

x 34 1

4x 34

y 4x x 14

11.

fx 1

f x x 1 12.

gx 3

gx 3x 1 13.

ft 4t 3

f t 2t2 3t 6 14.

y 2t 2

y t2 2t 3

15.

gx 2x 12x2 gx x2 4x3 16.

y 3x2

y 8 x3 17.

st 3t2 2

st t3 2t 4 18.

fx 6x2 2x 3

f x 2x3 x2 3x

19.

y

2 cos sin

y

2 sin cos 20.

gt sin t

gt cos t 21.

y 2x 12

sin x

y x2 12

cos x 22.

y cos x

y 5 sin x

23.

y 1x2

3 cos x

y 1x

3 sin x 24.

y 58

3x 4 2 sin x 158x 4

2 sin x

y 5

2x3 2 cos x 58

x3 2 cos x

25. y 5x 3

y 5x 3y 52

x 2y 5

2x2

27. y 98x 4

y 98

x4y 38

x 3y 3

2x3

26. y 2

3x2y

23

x 2 y 43

x 3 y 4

3x3

Function Rewrite Differentiate Simplify


28. y

3x2 y

9x 2 y

29

x 3 y 29x3


29. y 1

2x 32y

12

x 32y x 12y xx

30. y 4

x 3y 4x3

y 12x2 y 12x2

34.

y2 36

y 9x2

y 3x3 6, 2, 18 35.

y0 4

y 8x 4

4x2 4x 1

y 2x 12, 0, 1 36.

f5 0

fx 6x 30

3x2 30x 75

f x 35 x2, 5, 0

31.

f1 6

fx 6x 3 6x3

f x 3x2

3x 2, 1, 3 32.

f 35 53

ft 35t2

f t 3 35t

, 35, 2 33.

f0 0

fx 215

x2

f x 12

75

x3, 0, 12

37.

f0 41 1 3

f 4 cos 1

f 4 sin , 0, 0 38.

g 0

gt 3 sin t

gt 2 3 cos t, , 1 39.

fx 2x 6x 3 2x 6x3

f x x2 5 3x 2

40.

2x 3 6x3

f x 2x 3 6x 3 f x x2 3x 3x 2 41.

gt 2t 12t 4 2t 12t 4

gt t2 4t3

t2 4t 3 42.

1 2x3

fx 1 2x 3 f x x x 2

43.

fx 1 8x3

x3 8

x3

f x x3 3x2 4

x2 x 3 4x 2 44.

hx 2 1x2

2x2 1

x2

hx 2x2 3x 1

x 2x 3 x 1

47.

fx 12

x 12 2x 23 1

2x

2

x23

f x x 6 3x x 12 6x 13 48.

fx 13

x 23 15

x 45 1

3x 23

15x 45

f x 3x 5x x 13 x 15

45.

y 3x2 1

y xx2 1 x3 x 46.

y 36x 45x2

y 3x6x 5x 2 18x 2 15x3

49.

h(s 45

s15 23

s 13 4

5s 15

23s 13

hs s 45 s 23 50.

ft 23

t 13 13

t 23 2

3t 13

13t 23

f t t 23 t 13 4


51.

fx 3x 12 5 sin x 3x

5 sin x

f x 6x 5 cos x 6x 12 5 cos x 52.

fx 23

x 43 3 sin x 2

3x 43 3 sin x

f x 23x 3 cos x 2x 13 3 cos x

53. (a)

At

Tangent line:

(b) 3

1

2 2(1, 0)

2x y 2 0

y 0 2x 1

y 413 61 21, 0:

y 4x3 6x

y x 4 3x2 2 54. (a)

At

Tangent line:

(b)

5 5

7

5

4x y 2 0

y 2 4x 1

y 312 1 41, 2:

y 3x2 1

y x3 x

55. (a)

At

Tangent line:

(b)

7

1

2

5

(1, 2)

3x 2y 7 0

y 32

x 72

y 2 32

x 1

f1 32

1, 2:

fx 32

x 74 3

2x 74

f x 24x3 2x 34 56. (a)

At :

Tangent line:

(b)

3 3

2

12

0 11x y 5

y 6 11x 1

y 312 61 2 111, 6

y 3x2 6x 2

x3 3x2 2x

y x2 2xx 1

57.

Horizontal tangents: , 2, 142, 140, 2,

y 0 x 0, 2

4xx 2x 2

4xx2 4

y 4x3 16x

y x 4 8x2 2 58.

for all x.

Therefore, there are no horizontal tangents.

y 3x2 1 > 0

y x3 x

59.

cannot equal zero.

Therefore, there are no horizontal tangents.

y 2x 3 2x3

y 1x2

x 2 60.

At .

Horizontal tangent: 0, 1

y 1x 0,

y 2x 0 x 0

y x2 1


61.

At

Horizontal tangent: ,

y x :

cos x 1 x

y 1 cos x 0

0 x < 2 y x sin x, 62.

At

At

Horizontal tangents: 23 , 23 33 3, 3 33 ,y

23 33

x 23

:

y 3 3

3x

3:

sin x 32

x 3

or 23

y 3 2 sin x 0

0 x < 2 y 3x 2 cos x,

63.

Hence, and

For and for x 3, k 10.x 3, k 2

x 2 2x 4x 4x 9 x 2 9 x 3.

k 2x 4

2x k 4 Equate derivatives.

x 2 kx 4x 9 Equate functions. 64.

Hence, and k 4 8 7 k 3.x 2

2x 4 Equate derivatives.

k x2 4x 7 Equate functions.

65.

Hence, and

32

x 3 x 2 k 3.

34 x

2

x

34

x 3 34

x 34

x 3

k 34 x2

kx2

34

Equate derivatives.

kx

34

x 3 Equate functions.

68. The graph of a function f such that for all x and the rateof change the function is decreas-ing would, in general, look like the graph at the right.

i.e., f < 0

f > 0

x

y

69. gx f x 6 gx fx

67. (a) The slope appears to be steepest between A and B.

(b) The average rate of change between A and B is greater than the instantaneous rate of change at B.

(c)

x

f

CA

B

ED

y

70. gx 5 f x gx 5 fx

66. Equate functions.

Equate derivatives.

Hence, and

2x x x 4 2x x 4 x 4 k 4.k 2x

k

2x 1

kx x 4


73. Let and be the points of tangency on and respectively. The derivatives of these functions are:

and

Since and

and

Thus, the tangent line through and is

and

Thus, the tangent line through and is

y 1 3 12 1x 1 y 2x 1.1, 12, 3

y1 1x1 1x2 2 y2 3,

y 0 4 02 1x 1 y 4x 4.2, 41, 0

y1 4x1 2x2 1 y2 0,

x2 1 or 2

2x2 2x2 1 05

4

3

1

1

2

2

2 3x

), 4)2

, )) 01

y 2x22 6x2 4 0

2x22 12x2 14 4x2

2 18x2 18

x22 6x2 5 x22 6x2 9 2x2 62x2 3

x22 6x2 5 x2 32

x2 x2 3 2x2 6

m y2 y1x2 x1

x22 6x2 5 x12

x2 x1 2x2 6

5

4

3

1

1

2

2 3x

32) ,

1)1,)

)

yy2 x22 6x2 5:y1 x1

2

x1 x2 3

m 2x1 2x2 6

y 2x 6 m 2x2 6y 2x m 2x1

y x2 6x 5,y x2x2, y2x1, y1

71.

If f is linear then its derivative is a constant function.

fx a

f x ax b

3

3

1

21123

2

x

f

f

y 72.

If is quadratic, then its derivative is a linear function.

fx 2ax b

f x ax2 bx c

f

x2 1 1 3 4

1

2

3

4

f

f

y

74. is the slope of the line tangent to is the slope of the line tangent to Since

and

The points of intersection of and are

At Since these tangent lines are perpendicular at the points of intersection.m2 1m1,m2 1.x 1,

x 1x x2 1 x 1.

y 1xy x

y 1x y 1

x2 m2

1x2

.y x y 1 m1 1

y 1x.m2y x.m1


75.

Since for all and does not havea horizontal tangent line.

fxfx 0cos x 1, f x 3 cos x

f x 3x sin x 2 76.

Since Thus, does not have atangent line with a slope of 3.

ffx 5.5x 4 9x2 0,

f x 5x 4 9x2 5

f x x5 3x3 5x

77.

The point is on the graph of f.

Tangent line:

0 x 4y 4

4y 8 x 4

y 2 0 2

4 4x 4

4, 2

x 4, y 2

4 x 2x

4 x 2xx

4 x 2xy

1

2x

0 y4 x

f x 12

x12 1

2x

f x x, 4, 0 78.

The point is on the graph of f. The slope of the

tangent line is

Tangent line:

8x 25y 40 0

25y 20 8x 20

y 45

825x

52

f 52 825.52 , 45

x 52

, y 45

4x 10

10 2x 2x

10 2x x22x10 2x x2y

2x2

0 y5 x

fx 2x2

f x 2x, 5, 0

79.

1.243.33

0.77

3.64

f1 1 80.

10

1

19

16

f4 1

81. (a) One possible secant is between and :

(b)

is an approximation of the tangent line

CONTINUED

Tx.Sx

Tx 3x 4 8 3x 4

f x 32

x12 f 4 32

2 3

y Sx 2.981x 3.924

y 8 2.981x 4

y 8 8 7.7019

4 3.9x 4

2

2 12

(4, 8)

204, 83.9, 7.7019


81. CONTINUED

(c) As you move further away from the accuracy of the approximation T gets worse.

(d)

2

2 12T

f

20

4, 8,

0.5 0.1 0 0.1 0.5 1 2 3

1 2.828 5.196 6.548 7.702 8 8.302 9.546 11.180 14.697 18.520

2 5 6.5 7.7 8 8.3 9.5 11 14 171T4 x

f 4 x

123x

82. (a) Nearby point:

Secant line:

(Answers will vary.)

(d)

The accuracy decreases more rapidly than in Exercise 81 because is less linear than y x 32.y x3

3 3

2

(1, 1)

2

y 3.022x 1 1

y 1 1.0221024 11.0073138 1

x 1

1.0073138, 1.0221024 (b)

(c) The accuracy worsens as you move away from

3 3

2

fT

2

(1, 1)

1, 1.

Tx 3x 1 1 3x 2

fx 3x2

0.5 0.1 0 0.1 0.5 1 2 3

0 0.125 0.729 1 1.331 3.375 8 27 64

0.5 0.7 1 1.3 2.5 4 7 10258Tx

18f x

123x

83. False. Let and Thenbut f x gx.f x gx 2x,

x2 4.gx f x x2 84. True. If then fx gx 0 gx.f x gx c,

85. False. If then 2 is a constant.dydx 0.y 2, 86. True. If then 1.dydx 11y x 1 x,

87. True. If gx 3 f x, then gx 3 fx. 88. False. If then f x nx n1 nx n1

.f x 1x n

x n,

89.

Instantaneous rate of change is the constant 2. Average rate of change:

(These are the same because f is a line of slope 2.)

f 2 f 12 1

22 7 21 7

1 2

f t 2

f t 2t 7, 1, 2 90.

Instantaneous rate of change:

Average rate of change:

f 2.1 f 22.1 2

1.41 1

0.1 4.1

2.1, 1.41 f2.1 4.2

2, 1 f2 22 4

f t 2t

f t t2 3, 2, 2.1


91.



f 2 f 12 1

12 1

2 1

12

2, 12 f2 14

1, 1 f1 1

f x 1x2

f x 1x, 1, 2

93. (a)

(b)

(c)

When

When

(d)

(e)

81362 295.242 ftsec

v13624 3213624 t 2

136216

t 1362

4

9.226 sec

16t2 1362 0

v2 64 ftsect 2:

v1 32 ftsect 1:

vt st 32t

s2 s12 1

1298 1346 48 ftsec

vt 32t

st 16t2 1362 94.

86 ftsec

v2 322 22

t 2

2t 28t 27 0

16t2 22t 108 0

112 height after falling 108 ft

st 16t2 22t 220

v3 118 ftsec

vt 32t 22

st 16t2 22t 220

92.



f 6 f 06 0

12 06 0

3

0.955

6, 12 f

6 32

0.866

0, 0 f 0 1

f x cos x

f x sin x, 0, 6

95.

v10 9.810 120 22 msec

v5 9.85 120 71 msec

vt 9.8t 120

4.9t 2 120t

st 4.9t 2 v0 t s0 96.

when .

s0 4.9t2 4.96.82 226.6 m

t 6.8 4.9t 2 s0 0

st 4.9t2 v0 t s0

97. From to

mph for

Similarly, for Finally, from to

t

Time (in minutes)2 4 6 8 10

10

20

30

40

50

60

Vel

ocity

(in

mph

)

v

st t 4 vt 1 miin 60 mph.

10, 6,6, 24 < t < 6.vt 0

0 < t < 4vt 1260 30

st 12 t vt 12 mimin.4, 2,0, 0 98.

(The velocity has been converted to miles per hour.)

t

v

10

30

20

50

40

2 4 6 8 10

Time (in minutes)

Vel

ocity

(in

mph

)


99.

1 mimin2 min 2 mi

v 60 mph 1 mimin

0 mimin2 min 0 mi

v 0 mph 0 mimin

23 mimin6 min 4 mi

Time (in minutes)

Dis

tanc

e (i

n m

iles)

t2 4 6 8 10

2

4

6

8

10

(10, 6)

(8, 4)

(6, 4)

(0, 0)

sv 40 mph 23 mimin 100. This graph corresponds with Exercise 97.

t2 4 6 8 10

2

6

10

4

8

(0, 0)

(4, 2)(6, 2)

(10, 6)

Dis

tanc

e (i

n m

iles)

Time (in minutes)

s

101. (a) Using a graphing utility,

(b) Using a graphing utility,

(c)

(d)

1200

0

TB

R

80

T R B 0.00557v2 0.418v 0.02

B 0.00557v2 0.0014v 0.04.

R 0.417v 0.02.

(e)

For

For

For

(f ) For increasing speeds, the total stopping distance increases.

T100 1.53.v 100,

T80 1.31.v 80,

T40 0.86.v 40,

dTdv

0.01114v 0.418

102.

The driver who gets 15 miles per gallon would benefit more. The rate of change at is larger in absolute value than that at x 35.

x 15

dCdx

23,250

x2

15,000x 1.55 23,250

x

C gallons of fuel usedcost per gallon

x 10 15 20 25 30 35 40

C 2325 1550 1163 930 775 664 581

1519263758103233dCdx

103.

When per cm change in s.dVds

48 cm2s 4 cm,

dVds

3s2V s3, 104.

When

square meters per meter change in s.dAds

8

s 4 m,

dAds

2sA s2,


105. and

Average velocity:

instantaneous velocity at t t0 s t0

at0

2at0 t

2 t

12at02 2t0 t t2 12at02 2t0 t t2

2 t

st0 t st0 tt0 t t0 t

12at0 t2 c 12at0 t2 c

2t

st atst 12

at 2 c

106.

When dCdQ

$1.93.Q 350,

C351 C350 5083.095 5085 $1.91

dCdQ

1,008,000

Q2 6.3

C 1,008,000

Q 6.3Q

107.

(a) is the rate of change of gallons of gasoline sold when the price is $1.479 per gallon.

(b) is usually negative. As prices go up, sales go down.f1.479

f 1.479

N f p

108.dTdt

KT Ta

109.

Since the parabola passes through and we have:

Thus, From the tangent linewe know that the derivative is 1 at the point

Therefore, y 2x2 3x 1.

b a 1 3

a 2

1 a 1

1 2a1 a 1

y 2ax a 1

1, 0.y x 1,

y ax2 a 1x 1.

0 a12 b1 1 b a 11, 0:

1 a02 b0 c c 10, 1:

1, 0,0, 1

y ax2 bx c 110.

At the equation of the tangent line is

or

The x-intercept is

The y-intercept is

The area of the triangle is

x1 2 3

1

2

)a,(( , ) =a b 1a

y

A 12

bh 12

2a2a 2.0, 2a.2a, 0.

y xa2

2a

.y 1a

1a2

x a

a, b,

y 1x2

y 1x, x > 0


111.

Tangent lines through

The points of tangency are and At the slope is At the slope is

Tangent lines:

9x y 0 9x 4y 27 0

y 9x y 94 x 274

y 0 9x 0 and y 818 94 x 32

y 32 94.32, 818 ,y0 9.0, 0,32 , 818 .0, 0 x 0 or x 32

0 2x3 3x2 x22x 3

x3 9x 9 3x3 3x2 9x 9

y 9 3x2 9x 1

1, 9:

y 3x2 9

y x3 9x

112.

(a) Tangent lines through

The points of tangency are At the slope is At the slope is

Tangent lines: and

Restriction: a must be negative.

(b) Tangent lines through

The points of tangency are and At the slope is At the slope is

Tangent lines:

Restriction: None, a can be any real number.

y 0 y 4ax 4a2

y 0 0x 0 and y 4a2 4ax 2a

y2a 4a.2a, 4a2,y0 0.0, 0,2a, 4a2.0, 0

0 x2 2ax xx 2a

x2 2x2 2ax

y 0 2xx a

a, 0:

y 2a x a y 2a x a

y a 2ax a y a 2ax aya 2a.a, a,

ya 2a.a, a,a, a.a x

a x2

x2 a 2x2

y a 2xx 0

0, a:

y 2x

y x2


113.

f must be continuous at to be differentiable at

For f to be differentiable at the left derivative must equal the right derivative.

b 8a 4 43

a 13

12a 4

3a22 22

x 2,

fx 3ax2,2x, x < 2x > 2

limx2

f x limx2

x2 b 4 b

limx2

f x limx2

ax3 8a

x 2.x 2

f x ax3,x2 b, x 2x > 2

8a 4 b

8a 4 b

114.

Hence,

Answer: a 0, b 1

a 0.

f x sin x,a, x < 0 x > 0

f 0 b cos0 1 b 1

f x cos x,ax b, x < 0 x 0 115. is differentiable for all an integer.is differentiable for all

You can verify this by graphing and and observing the locations of the sharp turns.

f2f1

x 0.f2x sinx

x n, nf1x sin x

116. Let

0 sin x1 sin x

limx0

cos xcos x 1

x lim

x0 sin xsin xx

limx0

cos x cos x sin x sin x cos x

x

fx limx0

f x x f x

x

f x cos x.

Section 2.3 Product and Quotient Rules and Higher-Order Derivatives

1.

4x3 6x 2 2x 2

2x3 2x 2 2x 2 2x3 4x 2

gx x2 12x 2 x2 2x2x

gx x2 1x2 2x 2.

24x3 15x2 12

18x3 15x2 6x3 12

fx 6x 53x 2 x3 26

f x 6x 5x3 2

3.

7t 2 4

3t 23

2t 43 t 2 43t 23

ht t 132t t 2 413

t 23

ht 3tt 2 4 t 13t 2 4 4.

4 5s 2

2s 12

2s32 4 s2

2s12

gs s122s 4 s 212

s 12

gs s4 s2 s124 s2

Section 2.3 Product and Quotient Rules and Higher-Order Derivatives 121

5.

3x2 cos x x 3 sin x

fx x 3sin x cos x3x2

f x x 3 cos x 6.

gx x cos x sin x 12x x cos x 1

2x sin x

gx x sin x

7.

fx x2 11 x2x

x2 12 1 x2

x2 12

f x xx2 1

8.

gt 2t 72t t2 22

2t 72 2t2 14t 4

2t 72

gt t2 2

2t 7

9.

1 8x 3

3x 23x 3 12

x3 1 x9x2

3x 23x 3 12

hx x 3 113x 23 x133x2

x 3 12

hx 3x

x3 1

x 13

x3 110.

s 1

12s

s 12 s 2

2s 12

hs s 11 s12 s

12s 12

hs ss 1

11.

gx x2cos x sin x2x

x22 x cos x 2 sin x

x 3

gx sin xx2

12.

ft t3sin t cos t3t 2

t 32 t sin t 3 cos t

t 4

f t cos tt3

13.

f0 15 10x4 12x3 3x2 18x 15

fx x3 3x4x 3 2x2 3x 53x2 3

f x x3 3x2x2 3x 5 14.

f1 0 x 125x2 2x 2

3x2x 12 2x 12x2 x 1

fx x2 2x 13x2 x3 12x 2

f x x2 2x 1x3 1

15.

f1 1 6 41 32 14

x2 6x 4

x 32

2x2 6x x2 4

x 32

fx x 32x x2 41

x 32

f x x2 4

x 316.

f2 22 12 2

2

x 12

x 1 x 1

x 12

fx x 11 x 11x 12

f x x 1x 1

17.

f 4 22

4 22 28 4 fx xsin x cos x1 cos x x sin x

f x x cos x


18.

33 6

2

33 18

2

f6 632 12

236

x cos x sin x

x2

fx xcos x sin x1x2

f x sin xx


20. y 5x2 3

4y

54

x2 34

y 104

x y 5x

2

19. y x2 2x

3y

13

x2 23

x y 23

x 23

y 2x 2

3

21. y 7

3x3y

73

x3 y 7x4 y 7x4

23. y 4x32

xy 4x, x > 0 y 2x12 y 2

x

22. y 4

5x2y

45

x2 y 85

x3 y 8

5x3

24. y 3x2 5

7y

37

x2 57

y 6x7

y 67

x

25.

2

x 12 , x 1

2x2 4x 2

x2 12 2x 12x2 12

fx x2 12 2x 3 2x x22x

x2 12

f x 3 2x x2

x2 126.

x 4 6x2 4x 3

x2 12

fx x2 13x2 3 x3 3x 22x

x2 12

f x x3 3x 2

x2 1

27.

x2 6x 3

x 32

x2 6x 9 12

x 32

fx 1 x 34 4x1x 32

f x x1 4x 3 x 4x

x 328.

2x32x2 x 2x 12

x4 2x 12

x2 1

x 124x3

fx x 4x 1 x 1x 12

x 1x 14x3

f x x 41 2x 1 x 4

x 1x 1


29.

2x 5

2xx

2x 52x 32

fx x 12 52

x 32 x 32x 52

f x 2x 5x

2x 12 5x 12

30.

5

6x 16

1x 23

56

x 16 x 23

fx x 1312 x 12 x 12 313

x 23 f x 3xx 3 x 13x 12 3 Alternate solution:

5

6x16

1x23

fx 56

x16 x23

x56 3x13

f x 3x x 3

31.

hs 6s5 12s2 6s2s3 2

hs s3 22 s6 4s3 4 32.

hx 4x3 4x 4xx2 1

hx x2 12 x 4 2x2 1

33.

2x 2 2x 3

x2 3x2 2x2 2x 3

x2x 32

fx x2 3x2 2x 12x 3

x2 3x2 2x2 6x 4x2 8x 3

x2 3x2

f x 2 1xx 3

2x 1

xx 3 2x 1x2 3x

34.

gx 2 x 12x x21

x 12 2x2 2x 1 x2 2x

x 12 x2 2x 2

x 12

gx x22x 1

x 1 2x x2

x 1

35.

15x 4 48x 3 33x 2 32x 20

9x 4 36x3 41x 2 16x 20 6x 4 12x 3 8x 2 16x

9x2 4x2 4x 5 3x 4 3x 3 4x 2 4x 3x 4 15x3 4x2 20x

fx 9x2 4x 5x 1 3x3 4x1x 1 3x3 4xx 51

f x 3x3 4xx 5x 1

36.

6x5 4x3 3x2 1

2x5 x 4 3x3 x 1 2x5 2x2 2x5 x 4 x3 x2 x

2x 1x 4 x3 2x2 x 1 x2 x2x3 2x2 2x x2 x2x3 x2 2x 1

fx 2x 1x2 1x2 x 1 x2 x2xx2 x 1 x2 xx2 12x 1 f x x2 xx2 1x2 x 1


37.

4xc2

x2 c22

fx x2 c22x x2 c22x

x2 c22

f x x2 c2

x2 c238.

4xc2

c2 x22

fx c2 x22x c2 x22x

c2 x22

f x c2 x2

c2 x2

39.

t t cos t 2 sin t

ft t 2 cos t 2t sin t

f t t 2 sin t 40.

cos 1 sin

f 1sin cos 1

f 1 cos

41.

ft t sin t cos tt2

t sin t cos t

t2

f t cos tt

42.

fx x cos x sin xx2

f x sin xx

43.

fx 1 sec2 x tan2 x

f x x tan x 44.

y 1 csc2 x cot2 x

y x cot x

45.

gt 14

t 34 8 sec t tan t 1

4t 34 8 sec t tan t

gt 4t 8 sec t t 14 8 sec t 46.

hs 1s2

10 csc s cot s

hs 1s

10 csc s

47.

32

sec x tan x tan2 x 1

y 32

sec x tan x sec2 x 32

sec xtan x sec x

y 31 sin x

2 cos x

32

sec x tan x 48.

sec xx tan x 1

x2

y x sec x tan x sec x

x2

y sec x

x

49.

cos x cot2 x

cos xcsc2 x 1

cos xsin2 x

cos x

y csc x cot x cos x

y csc x sin x 50.

y x cos x sin x sin x x cos x

y x sin x cos x

51.

xx sec2 x 2 tan x

fx x2 sec2 x 2x tan x f x x2 tan x 52.

cos 2x

fx sin xsin x cos xcos x f x sin x cos x

53.

4x cos x 2 sin x x2 sin x

y 2x cos x 2 sin x x2sin x 2x cos x

y 2x sin x x2 cos x 54.

h 5 sec tan 5 sec sec2 tan

h 5 sec tan


55.

(Form of answer may vary.)gx 2x2 8x 1x 22

gx x 1x 22x 5 56.

(Form of answer may vary.)fx 2 x5 2x3 2x2 2

x2 12

f x x2 x 3x2 1 x2 x 1

57.

(Form of answer may vary.)g 1 sin cos sin 12

g 1 sin

58.

(Form of answer may vary.)

f 1cos 1

cos 1

1 cos 2

f sin 1 cos

59.

y 6 223

1 22 43

y 1 csc xcsc x cot x 1 csc xcsc x cot x

1 csc x2 2 csc x cot x1 csc x2

y 1 csc x1 csc x

60.

f1 0

fx 0

f x tan x cot x 1 61.

h sec tan 1 2

1

2

ht tsec t tan t sec t1t2

sec tt tan t 1

t 2

ht sec tt

62.

f4 sin

2 cos

2 1

sin 2x cos 2x

sin x cos x sin2 x sin x cos x cos2 x

f x sin xcos x sin x sin x cos x cos x

f x sin xsin x cos x 63. (a)

Tangent line:

(b)

10

10 10

10

(1, 3)

y 3 1x 1 y x 2

f1 1; Slope at 1, 3 4x3 6x2 6x 5

fx x3 3x 11 x 23x2 3

f x x3 3x 1x 2, 1, 3

64. (a)

Tangent line:

(b)

4 4

4

4

y 2 2x y 2x 2

f0 2; Slope at 0, 2

fx x 12x x2 21 3x2 2x 2

f x x 1x2 2, 0, 2 65. (a)

Tangent line:

(b)

3

3 6

6

(2, 2)

y 2 1x 2 y x 4

f2 12 12 1; Slope at 2, 2

fx x 11 x1x 12 1

x 12

f x xx 1

, 2, 2


67. (a)

Tangent line:

(b)

4

4

4( (, 1

4x 2y 2 0

y 1 2x

2

y 1 2x 4

f4 2; Slope at

4, 1

fx sec2 x

f x tan x, 4, 166. (a)

Tangent line:

(b)

3 6

4

4

y 13

29

x 2 y 29

x 19

f2 29

; Slope at 2, 13

fx x 11 x 11x 12 2

x 12

f x x 1x 1

, 2, 13

68. (a) (b)

Tangent line:

63x 3y 6 23 0

y 2 23x 3

f 3 23; Slope at

3, 2

fx sec x tan x

2

6

f x sec x, 3, 2

69.

2y x 4 0

y 12

x 2

y 1 12

x 2

f2 1624 42 12

fx x2 40 82x

x2 42 16x

x2 42

f x 8x2 4

; 2, 1 70.

2y x 6 0

y 12

x 3

y 32

12

x 3

f 3 5439 92 12

fx x2 90 272x

x2 92 54x

x2 92

f x 27x2 9

; 3, 32

71.

25y 12x 16 0

y 1225

x 1625

y 85

1225

x 2

f 2 256 164202

1225

f x x2 1616 16x2x

x2 162 25616x2

x2 162

f x 16xx2 16

; 2, 85 72.

25y 2x 16 0

y 2

25x

1625

y 45

2

25x 2

f2 24 16102

2

25

f x x2 64 4x2x

x2 62 244x2

x2 62

f x 4xx2 6

; 2, 45


76.

for

has horizontal tangents at and 7, 114.1, 12f

f 1 12

, f 7 114

x 1, 7;f x 0

x 7x 1

x2 72 x2 8x 7

x2 72x2 7 2x2 8x

x2 72 f x x2 71 x 42x

x2 72

f x x 4x2 7

77.

Slope:

2 2 4 6

4

6

6

(3, 2)(1, 0)

2y + x = 1

2y + x = 7y

x

f (x) = x + 1x 1

246

y 2 12

x 3 y 12

x 72

y 0 12

x 1 y 12

x 12

x 1, 3; f 1 0, f 3 2

x 1 2

x 12 4

2

x 12 12

12

2y x 6 y 12

x 3;

f x x 1 x 1x 12 2

x 12

f x x 1x 1

78.

Let be a point of tangency on the graph of

Two tangent lines:

y 2 1x 2 y x 4

y 1 4x 12 y 4x 1

f 12 4, f 2 1f 12 1, f 2 2;

x 22x 1 0 x 12

, 2

4x 2 10x 4 0

4x 5x 1 x 1

4x 5

x 1x 1 1

x 12

5 xx 1

1 x

1x 12

f.

x

y

(2, 2)

(1, 5)

, 1 12( )

24 2 4

2

6

y = 4x + 1

y = x + 4

f(x) = xx 1

x, y x, xx 1

f x x 1 xx 12 1

x 12

f x xx 1

73.

Horizontal tangents are at and 2, 4.

0, 0

fx 0 when x 0 or x 2.

x2 2xx 12

xx 2x 12

fx x 12x x21

x 12

f x x2

x 174.

Horizontal tangent is at 0, 0.

fx 0 when x 0.

2x

x2 12

fx x2 12x x22x

x2 12

f x x2

x2 175.

for

has a horizontal tangent at 1, 2.f

f 1 2

4 4x 0 x 1.fx 0

4 4x

x3

4x2 8x2 4x

x 4

fx x24 4x 22x

x 4

f x 4x 2x2


79.

and differ by a constant.gf

gx 5x 4x 2 3x

x 2 2x 4x 2 f x 2

gx x 25 5x 41x 22 6

x 22

fx x 23 3x1x 22 6

x 22 80.

and differ by a constant.gf

gx sin x 2xx

sin x 3x 5x

x f x 5

gx xcos x 2 sin x 2x1x2

x cos x sin x

x2

f x xcos x 3 sin x 3x1x2

x cos x sin x

x2

81. (a)

(b)

q4 31 7032

13

q x gx f x f xg xgx2

p 1 f 1g1 f 1g 1 14 612 1 p x f xgx f xg x 82. (a)

(b)

q7 42 4142

1216

34

q x gx f x f xg xgx2

p 4 12

8 10 4

p x f xgx f xg x

83.

6t 1

2t cm2sec

3t12 12

t12

A t 232 t12 12

t12

Area At 2t 1t 2t32 t12 84.

Vt 12

32

t 12 t 12 3t 24t 12 cubic inchessec

12

t 32 2t 12

V r 2h t 212t

85.

(a) When

(b) When

(c) When

As the order size increases, the cost per item decreases.

x 20: dCdx

$3.80

x 15: dCdx

$10.37

x 10: dCdx

$38.13

dCdx

100400x3 30

x 302

C 100200x2 x

x 30, 1 x 86.dPdV

k

V 2

P kV

87.

P2 31.55 bacteria per hour

2000 50 t2

50 t 2 2

500200 4t2

50 t 2 2

Pt 50050 t24 4t2t

50 t 22

Pt 5001 4t50 t 2 88.

dFdd

Fd 2 Gm1m2

d 3

F Gm1m2

d 2 Gm1m2 d

2


89. (a)

(b)

(c)

ddx

cot x ddx

cos xsin x

sin xsin x cos xcos xsin x2

sin2 x cos2 xsin2 x

1

sin2 x csc2 x

cot x cos xsin x

ddx

csc x ddx

1sin x

sin x0 1cos xsin x2

cos xsin x sin x

1

sin x

cos xsin x

csc x cot x

csc x 1

sin x

ddx

sec x ddx

1cos x

cos x0 1sin xcos x2

sin xcos x cos x

1

cos x

sin xcos x

sec x tan x

sec x 1

cos x

90.

x 34

, 74

sin3 x

cos3 x 1 tan3 x 1 tan x 1sec x tan x csc x cot x sec x tan x

csc x cot x 1

1

cos x

sin xcos x

1sin x

cos xsin x

1

fx gx

gx csc x, 0, 2

f x sec x

91. (a)

(c)

represents the average retail value (in millions of dollars) per 1000 motor homes.A

05 12

0.05

A vtnt

0.1361t3 3.165t 2 23.02t 59.83.5806t3 82.577t 2 603.60t 1667.5

vt 0.1361t 3 3.165t 2 23.02t 59.8

nt 3.5806t 3 82.577t 2 603.60t 1667.5 (b)

(d) represents the rate of change of the average retail value per 1000 motor homes.A t

12

v(t)

05 12

350

120

5

n(t)

92. (a)

h r csc r rcsc 1

r h r csc

sin r

r h(b)

39602 3 79203 miradian

h 30 h6 h rcsc cot

93.

f x 3x12 3x

fx 6x12 f x 4x32 94.

f x 192x 4

fx 1 64x3

f x x 32x2


95.

f x 2x 13

fx x 11 x1x 12 1

x 12

f x xx 1 96.

f x 2x3

fx 1 1x2

f x x2 2x 1

x x 2

1x

97.

f x 3 sin x

f x 3 cos x

f x 3 sin x 98.

sec xsec2 x tan2 x

f x sec xsec2 x tan xsec x tan x

fx sec x tan x

f x sec x

99.

f x 2x

f x x2 100.

fx 2x2 2x2

f x 2 2x1 101.

f 4x 12

2x12 1x

fx 2x 102.

f 6x 0

f 5x 2

f 4x 2x 1

103.

One such function is f x x 22.

f 2 0

x

1

1

2

2

3

3 4

4

y 104. The graph of a differentiable function f such that and for all real numbers would, in general, looklike the graph below.

x

f

y

xf < 0f > 0

105.

0

22 4

f2 2g2 h2

fx 2gx hx

f x 2gx hx 106.

f2 h2 4

fx hx

f x 4 hx

107.

10

12 34

12

f2 h2g2 g2h2h22

fx hxgx gxhxhx2

f x gxhx

108.

14

34 12

f2 g2h2 h2g2

fx gxhx hxgx

f x gxhx


109.

It appears that is cubic; so would be quadratic and would be linear.f

f f

2

2

1

112x

f

y

f

f

110.

It appears that is quadratic; so would be linear and would be constant.

f ff

x2 3 4

1

2

1

3

2

ff

f

y

111.

123 1 2 3 4 5

345

1

2

3

4

x

y

f

f

112.

134

1

1

2

3

x

y

f

f

113.

1

2

3

4

1

4

y

x

f f

2 2

114.

1

2

y

x

f f

2

115.

The speed of the object is decreasing.

a3 6 msec2v3 27 msec

at 2t

vt 36 t 2, 0 t 6

116.

1500

2t 152

at 2t 15100 100t22t 152

vt 100t2t 15

(a)

(b)

(c) a20 1500220 152 0.5 ftsec2

a10 1500210 152 1.2 ftsec2

a5 150025 152 2.4 ftsec2

117.

at 16.50

vt 16.50t 66

st 8.25t 2 66t Average velocity on:

3, 4 is 132 123.754 3

8.25.

2, 3 is 123.75 993 2

24.75.

1, 2 is 99 57.752 1

41.25.

0, 1 is 57.75 01 0

57.75.

0 1 2 3 4

0 57.75 99 123.75 132

66 49.5 33 16.5 0

16.5 16.5 16.5 16.5 16.5at vt ftsec2vt st ftsecst fttsec


118. (a)

position function

velocity function

acceleration functiona

v

s

y

t11 4 5 6 7

8

4

12

16

s

v

a

(b) The particle speeds up (accelerates) when andslows down when

Answers will vary.

a < 0.a > 0

119.

Note: (read factorial)nn! nn 1 . . . 3 2 1

f nx nn 1n 2 . . . 21 n!

f x x n 120.

1nn!

x n1

f nx 1nnn 1n 2 . . . 21

x n1

f x 1x

121.

(a)

(b)

Note: (read n factorial)n! nn 1 . . . 3 2 1

n!

n 1!1! gn1xhx gnxhx

gxhnx n!1!n 1! gxh

n1x n!2!n 2! g xh

n2x . . .

nn 1n 2 . . . 21

n 1n 2 . . . 211 gn1xhx gnxhx

nn 1n 2 . . . 21

321n 3n 4 . . . 21 g xhn3x . . .

nn 1n 2 . . . 21

21n 2n 3 . . . 21 g xhn2xf nx gxhnx nn 1n 2

. . . 211n 1n 2 . . . 21 gxh

n1x

gxh4x 4gxh x 6gxh x 4g xhx g4xhx

g xhx g4xhx

f 4x gxh4x gxh x 3gxh x 3g xh x 3g xh x 3g xh x

gxh x 3gxh x 3g xhx gxhx

f x gxh x gxh x 2gxh x 2g xhx hxg x hxg x

gxh x 2gxhx hxg x

f x gxh x gxhx hxg x hxgx

fx gxhx hxgx

f x gxhx

122.

In general, x f xn x f nx nf n1x.

x f x x fx f x 2 f x x fx 3 f x

x f x x f x f x f x x f x 2 f x

x f x x fx f x


123.

When

When

When

When

For general n, fx x n1x cos x n sin x.

n 4: fx x3x cos x 4 sin x

n 3: fx x2x cos x 3 sin x

n 2: fx xx cos x 2 sin x

n 1: fx x cos x sin x

x n1x cos x n sin x

fx x n cos x nx n1 sin x

f x x n sin x 124.

When

When

When

When

For general n, fx x sin x n cos xx n1

.

n 4: fx x sin x 4 cos xx5

n 3: fx x sin x 3 cos xx 4

n 2: fx x sin x 2 cos xx3

n 1: fx x sin x cos xx2

x sin x n cos x

xn1

xn1x sin x n cos x

fx xn sin x nxn1 cos x

f x cos xx n

xn cos x

125.

x3y 2x2y x3 2x3 2x21x2 2 2 0

y 1x, y

1x2

, y 2x3

126.

y xy 2y 12 x12x 26x2 6 24x2 y 12

y 12x

y 6x2 6

y 2x3 6x 10

127.

y y 2 sin x 2 sin x 3 3

y 2 sin x

y 2 cos x

y 2 sin x 3 128.

y y 3 cos x sin x 3 cos x sin x 0

y 3 cos x sin x

y 3 sin x cos x

y 3 cos x sin x

129. False. If then

dydx

f xgx gx fx.

y f xgx, 130. True. y is a fourth-degree polynomial.

when n > 4.d n ydx n

0

131. True

0

f c0 gc0

hc f cgc gc fc

132. True 133. True 134. True.

If then at vt 0.vt c

135.

intercept at

on graph:

Slope 10 at

Subtracting the third equation from the second, Subtracting this equation from the first,Then, Finally,

f x 3x2 2x 1

3 2 c c 1.10 43 b b 2.3 a.3 b c.

10 4a b2, 7:

7 4a 2b c2, 7

0 a b c1, 0:x-

f x 2ax b

f x ax2 bx c


136.

f x 0 x 2b 4b2 12ac

6a

f x 3ax2 2bx c

f x ax3 bx2 cx d, a 0

(a) No horizontal tangents:

Example:

f x x3 x

a c 1, b 0:

4b2 12ac < 0

f x 0 (b) Exactly one horizontal tangent

Example:

f x x3 3x2 3x

a 1, b 3, c 3:

4b2 12ac 0

(c) Exactly two horizontal tangents

Example:

f x x3 x2b 1, a 1, c 0:

4b2 12ac > 0

137.

does not exist since the left and right derivativesare not equal.f 0

f x 2, 2, if x > 0if x < 0

fx 2x, 2x, if x 0if x < 0 2x f x xx x2, x2, if x 0if x < 0 138. (a)

(b)

fg f g False

fg 2f g f g

fg fg fg f g

fg fg fg

fg f g True

fg fg fg fg fg f g

Section 2.4 The Chain Rule

y f uu gxy f gx

2. y u12u x 1y 1

x 1

4. y 3 tan uu x2y 3 tanx2

6. y cos uu 3x2

y cos 3x2

1. y u4u 6x 5y 6x 54

3. y uu x2 1y x2 1

5. y u3u csc xy csc3 x

7.

y 32x 722 62x 72 y 2x 73 8.

y 154 x242x 30x4 x24 y 34 x25

9.

gx 124 9x39 1084 9x3 gx 34 9x4 10.

ft 23

9t 2139 639t 2

f t 9t 223

Section 2.4 The Chain Rule 135

11.

ft 12

1 t121 121 t

f t 1 t12 12.

gx 12

5 3x123 325 3x

gx 5 3x 5 3x12

13.

y 13

9x2 42318x 6x9x2 423

y 9x2 413 14.

gx 1,1, x > 1 x < 1 gx x2 2x 1 x 12 x 1

15.

y 2144 x2342x x

44 x23

y 24 x214 16.

f x 34

2 9x349 2742 9x34

f x 32 9x14 f x 3 42 9x

17.

y 1x 221 1x 22

y x 21 18.

2t 3

t2 3t 12

s t 1t 2 3t 122t 3

st t 2 3t 11

st 1t2 3t 1

19.

f t 2t 33 2t 33

f t t 32

20.

y 15t 34 15t 34

y 5t 33

y 5

t 33 21.

1

2x 232

dydx

12

x 232

y x 212 22.

t

t 2 232

g t 12

t 2 2322t

gt t 2 212

gt 1t 2 2

23.

2xx 233x 2

2xx 232x x 2

f x x24x 231 x 242x

f x x2x 24 24.

27x 324x 3

3x 929x 3x 9

f x x33x 923 3x 931

f x x3x 93

25.

1 2x2

1 x2

1 x212x2 1 x2

x21 x212 1 x212

y x121 x2122x 1 x2121 y x1 x2 x1 x212 26.

x3x2 32

216 x2

x3

216 x2 x16 x2

y 12

x21216 x2122x x16 x212 y

12

x216 x2


27.

1

x2 132

x2 132x2 x2 1

x2x2 132 x2 112

y x12x2 1322x x2 1121

y x

x2 1 xx2 112 28.

4 x 4

x 4 432

x 4 4 2x 4

x 4 432

y x 4 4121 x 12x 4 4124x 3

x 4 4

y x

x 4 4

29.

2x 52 10x x2

x2 23

g x 2 x 5x2 2x2 2 x 52x

x2 22

gx x 5x2 22

30.

2t 24t t 4

t 3 23 2t 34 t 3t 3 23

h t 2 t2

t 3 2t 3 22t t 23t 2

t 3 22

ht t2

t 3 22

31.

91 2v2

1 v4

f v 31 2v1 v 2

1 v2 1 2v1 v2

f v 1 2v1 v 3

32.

63x2 223x2 9x 2

2x 34

33x2 226x2 18x 4

2x 34

g x 33x2 2

2x 3 2

2x 36x 3x2 22

2x 32

gx 3x2 2

2x 3 3

33.

The zero of corresponds tothe point on the graph of where the tangent line is horizontal.

yy

y 1 3x2 4x32

2xx2 12

2

1 5

y

y

2 y x 1x2 1

34.

has no zeros.y

y 1

2xx 1326 6

1

y

y

7 y 2xx 1

35.

has no zeros.y

y x 1x2xx 1

2

5 4

y

y

4 y x 1x 36.

has no zeros.

2 10

2

g

g

6g

gx 12x 1

1

2x 1

gx x 1 x 1


37.

The zeros of correspond to the points on the graph of ywhere the tangent lines are horizontal.

y

x sin x cos x 1

x 2

dydx

x sin x cos x 1

x 2

3

5 5

y

y

3 y cos x 1

x38.

The zeros of correspond to the points on the graph of ywhere the tangent lines are horizontal.

y

dydx

2x tan 1x

sec2 1x

4 5

6

y

y

6 y x2 tan 1x

39. (a)

1 cycle in 0, 2

y0 1

y cos x

y sin x (b)

2 cycles in

The slope of at theorigin is a.

sin ax

0, 2

y0 2

y 2 cos 2x

y sin 2x40. (a)

3 cycles in 0, 2

y0 3

y 3 cos 3x

y sin 3x (b)

Half cycle in

The slope of at theorigin is a.

sin ax

0, 2

y0 12

y 12 cosx2

y sinx2

41.

dydx

3 sin 3x

y cos 3x 42.

dydx

cos x

y sin x 43.

gx 12 sec2 4x

gx 3 tan 4x 44.

hx 2x secx2 tanx2

hx secx2

45.

y cos 2x22 2x 2 2x cos 2x2

y sinx2 sin 2x 2 46.

41 2x sin1 2x2 y sin1 2x221 2x2

y cos1 2x2 cos1 2x2

47.

Alternate solution:

hx 12 cos 4x4 2 cos 4x

hx 12 sin 4x

2 cos 4x

2 cos2 2x 2 sin2 2x

h x sin 2x2 sin 2x cos 2x2 cos 2x

hx sin 2x cos 2x 48.

12 sec12sec212 tan212 g sec12 sec21212 tan12 sec12 tan1212 g sec12 tan12

49.

sin2 x 2 cos2 x

sin3 x

1 cos2 xsin3 x

fx sin2 xsin x cos x2 sin x cos x

sin4 x

f x cot xsin x

cos xsin2 x

50.

cos2 v sin2 v cos 2v

gv cos vcos v sin vsin v

gv cos vcsc v

cos v sin v

51.

8 sec2 x tan x

y 8 sec x sec x tan x

y 4 sec2 x 52.

5 sin 2 t

10 sin tcos t

gt 10 cos tsin t

gt 5 cos 2 t 5cos t2 53.

sin 2 cos 2 12 sin 4

f 214sin 2cos 22 f 14 sin2 2

14sin 22


54.

4 cot t 2 csc2 t 2

ht 4 cot t 2csc2 t 2

ht 2 cot2 t 2 55.

6 sec2 t 1 tan t 1 6 sin t 1cos3 t 1

ft 6 sec t 1 sec t 1 tan t 1

f t 3 sec2 t 1

56.

3 10 2x sinx2

dydx

3 5 sin 2x 22 2x

3x 5 cos 2x 2

y 3x 5 cosx 2 57.

1

2x 2x cos2x2

dydx

12

x12 14

cos4x28x

x 14

sin4x2

y x 14

sin2x 2

58.

13

cos x13

x23

cos xsin x23

y cos x1313 x23 13

sin x23 cos x

y sin x13 sin x13 59.

s 2 34

t 1

t 2 2t 8

s t 12

t 2 2t 8122t 2

st t 2 2t 812, 2, 4

60.

y2 12

9x2 4

53x3 4x45

y 15

3x3 4x459x2 4

y 3x3 4x15, 2, 2 61.

f 1 925

f x 3x3 423x2 9x2

x3 42

f x 3x3 4

3x3 41, 1, 35

62.

f 4 532

f x 2x2 3x32x 3 22x 3x2 3x3

f x 1x2 3x2 x2 3x2, 4, 116 63.

f 0 5

f t t 13 3t 21t 12 5

t 12

f t 3t 2t 1

, 0, 2

64.

f 2 5

f x 2x 31 x 122x 32 5

2x 32

f x x 12x 3

, 2, 3 65.

y0 0

6 sec32x tan2x

y 3 sec22x2 sec(2x tan2x

y 37 sec32x, 0, 36


66.

is undefined.y2

y 1x2

sin x

2cos x

y 1x

cos x, 2, 2

67. (a)

Tangent line:

y 5 95

x 3 9x 5y 2 0

f 3 95

f x 12

3x2 2126x 3x3x2 2

f x 3x2 2, 3, 5

68. (a)

Tangent line: y 2 139

x 2 13x 9y 8 0

f2 433

13

3 139

x2

3x2 5

13x2 5

fx 13

x12 x2 5122x 13

x2 512

f x 13

xx2 5, 2, 2

(b)

3

5 5

7

(3, 5)

(b)

9 9

6

6

69. (a)

Tangent line:

(b)

(1, 1)2

1

3

4

y 12x 11

y 1 12x 1

y 1 121 12

y 22x3 16x2 12x22x3 1

y 2x3 12, 1, 1 70. (a)

Tangent line:

(b) 6

5

1

2

(1, 4)

y 23

x 143

y 4 23

x 1

f 1 43813

23

f x 23

9 x2132x 4x39 x213

f x 9 x223, 1, 4

71. (a)

Tangent line:

y 2x 2x y 2 0

f 2

fx 2 cos 2x

f x sin 2x, , 0 (b)

2

0 2

2

( , 0)


72. (a)

Tangent line:

(b) 2

2

2

2

4 2

2( (,

y 32

2x

328

22

y 22

32

2 x

4

y4 3 sin34

322

y 3 sin 3x

y cos 3x, 4, 22 73. (a)

Tangent line:

(b)

4

4

4( (, 1

y 1 4x 4 4x y 1 0

f 4 212 4 fx 2 tan x sec2 x

f x tan2 x, 4, 1

74. (a)

Tangent line:

(b) 3

1

2

2

4( (, 2

y 12x 2 3

y 2 12x 4

y4 612 12 y 6 tan2 x sec2 x

y 2 tan3 x, 4, 2 75. (a)

(b)

(c) 5

4

2

2

12

32,( (

y 3x 3

y 32

3x 12

g12 3

g t 3tt2 3t 2

t 2 2t 132

gt 3t2

t 2 2t 1, 12,

32

76. (a)

(b)

(c) 12

60

0

(4, 8)

y 9x 28

y 8 9x 4

f 4 9

f x x 25x 22x

f x x 2 x2, 4, 8 77. (a)

(b)

(c)

4

1

2

0, 43( (

3

y 43

y 43

0x 0

s0 0

s t 21 t

3

2 t

31 t

st 4 2t1 t

3, 0, 43


78. (a)

(b)

(c) 35

18

2

(2, 10)

y 274

t 472

y 10 274

t 2

y 2 274

y 5t 2 8t 9

2t 2

y t 2 9t 2, 2, 10 79.

Tangent line

9

4

9

8

(3, 4)

y 34

x 254

;

y 4 34

x 3

f 3 34

f x x25 x2

f x 25 x2, 3, 4

80.

Tangent line

3

2

1

2(1, 1)

y 2x 1;

y 1 2x 1

f 1 2

f x 22 x232 for x > 0

f x x2 x2

, 1, 1 81.

Horizontal tangents at

Horizontal tangent at the points and

56 , 332 6, 332 , 32 , 0,

x

6,

32

, 56

sin x 12

x 6

, 56

sin x 1 x 32

sin x 12 sin x 1 0

2 sin2 x sin x 1 0

2 sin x 2 4 sin2 x 0

f x 2 sin x 2 cos 2x

f x 2 cos x sin 2x, 0 < x < 2

82.

Horizontal tangent at 1, 1

x 1

2x 132 0 x 1

x 1

2x 132

2x 1 x2x 132

f x 2x 112 x2x 112

2x 1

f x x2x 1

83.

125x2 1x2 1

f x 60x 4 72x2 12

12x5 24x3 12x

12xx 4 2x2 1

f x 6x2 122x

f x 2x2 13


84.

f x 2x 23 2x 23

f x x 22 1x 22

f x x 21 85.

2cos x2 2x2 sin x2

f x 2x2xsin x2 2 cos x2 f x 2x cos x 2 f x sin x 2

86.

2 2 sec2 x3 sec2 x 2

2 2 sec2 xsec2 x 2 tan2 x

2 2 sec4 x 4 2 sec2 x tan2 x

f x 2 sec2 xsec2 x 2 tan x2 sec2 x tan x

2 sec2 x tan x

f x 2 sec x sec x tan x

f x sec2 x

87.

h 1 24

h x 23x 13 63x 1

h x 19

33x 123 3x 12

hx 19

3x 13, 1, 649 88.

f 0 3128

f x 34

x 452 34x 452

f x 12

x 432

f x 1x 4

x 412, 0, 12

89.

f 0 0

4x2 cosx2 2 sinx2

f x 2x cosx22x 2 sinx2

f x sinx22x 2x sinx2

f x cosx2, 0, 1 90.

g 6 323 8 sec22t tan2t

g t 4 sec2t sec2t tan2t2

g t 2 sec22t

gt tan2t, 6, 3

91.

The zeros of correspond to the points where the graphof f has horizontal tangents.

f

3

3

2

1

22

2

3

x

f

y

f 92.

is decreasing on so must be negative there.is increasing on so must be positive there.f1, f

f, 1f

x1 2 33 2 1

1

2

3

1

3

2f

ff

f

y


93.


f

3

2

213x

f

y

f

94.


f

x41

4

23

4

3

2f

f

y

95.

g x f3x3 gx 3 f3x

gx f 3x 96.

g x f x22x gx 2x f x2

gx f x2

97. (a)

(c)

f 5 36 3232 129

43

f x hxg x gxh xhx2

f x gxhx

f 5 32 63 24

f x gxh x g xhx

f x gxhx (b)

Need to find

(d)

f 5 3326 162

f x 3gx 2g x

f x gx3f 5.g 3

f 5 g 32 2g 3

f x g hxh x f x ghx

98. (a)

(b)

(c)

Hence, you need to know

(d)

Hence, you need to know

s 2 f 0 13, etc.

fx 2.

sx f x 2 sx fx 2

r 1 3 f 3 34 12

r 0 3 f 0 313 1f 3x.

rx f 3x r x f 3x3 3 f 3x

hx 2 f x h x 2 f x

gx f x 2 g x f x0 1 2 3

4

4

8

12 1

42113s x

r x

8422343h x

4211323g x

4211323f x

12x

99. (a)

(b)

Since does not exist, is not defined.s 5g 6

s 5 g f 5 f 5 g 61

s x g f x f x

sx g f x, f 5 6, f 5 1, g6 does not exist.

112 12 f 4g 1

h 1 f g1g 1

hx f gx, g1 4, g 1 12, f 4 1, h x f gxg x


101. (a)

When , F 1.461.v 30

132,400

331 v 2

F 1132,400331 v21

F 132,400331 v1

100. (a)

12

f 51

h 3 f g3g 3

hx f gxg x

hx f gx (b)

2

12

g 82

s 9 g f 9 f 9

s x g f x f x

sx g f x

(b)

When .F 1.016v 30,

132,400331 v 2

F 1132,400331 v21

F 132,400331 v1

102.

When feet and feet per second.v 4 y 0.25t 8,

4 sin 12t 3 cos 12t

v y 13

12 sin 12t 14

12 cos 12t

y 13

cos 12t 14

sin 12t 103.

The maximum angular displacement is (since

When radians persecond.

ddt 1.6 sin 24 1.4489t 3,

ddt

0.28 sin 8t 1.6 sin 8t

1 cos 8t 1. 0.2

0.2 cos 8t

104.

(a) Amplitude:

Period:

(b) v y 1.755 sin t5 0.35 sin

t5

y 1.75 cos t5

10 210

5

y 1.75 cos t

A 3.52

1.75

y A cos t 105.

Since r is constant, we have and

4.224 102 0.04224.

dSdt

1.76 10521.2 102105

drdt 0

dSdt

C2R dRdt 2r drdt

S CR2 r 2

106. (a) Using a graphing utility, or trial and error, you obtaina model similar to

(b) 100

00 13

T t 65 21 sin t6 2.1.

(c)

(d) The temperature changes most rapidly in the spring(AprilJune) and fall (SeptemberNovember).

15

15

0 13

Tt 11 cos t6 2.1


107. (a)

(b)

The function is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue.dCdt

294.696t2 2317.44t 30.492

dCdt

604.9116t2 38.624t 0.5082

601.6372t3 19.3120t2 0.5082t 0.6162 1350

C 60x 1350

x 1.6372t3 19.3120t2 0.5082t 0.6162

108.

(a)

(b)

(c)

f 2k1x 1k1 2k1 cos x

f 2kx 1k2k sin x

f x 2 f x 2 sin x 2sin x 0

f 4 4 sin x

fx 3 cos x

f x 2 sin x

f x cos x

f x sin x 109. for all x.

(a) Yes, which shows that isperiodic as well.

(b) Yes, let so Since is periodic, so is g.f

g x 2 f 2x.gx f 2x,

ff x p f x,

f x p f x

110. (a)

Note that and

Also, Thus, r 1 0.g 1 0.

f 4 5 06 2

54

.g1 4

r 1 f g1g 1

r x f gxg x (b)

Note that and

Thus, s 4 12

54

58

.f 4 54

.

f 4 52

, g 52 6 46 2

12

s 4 g f 4 f 4

s x g f x f x

112. (a) If then

Thus, is even.f x

f x f x.

f x1 f x

ddx

f x ddx

f x

f x f x, (b) If then

Thus, is odd.f

f x f x.

f x1 f x

ddx

f x ddx

f x

f x f x,

113.

uuu2

uu

u, u 0

ddx

u ddx u 2 12

u2122uu

u u2 114.

gx 2 2x 32x 3, x 32

gx 2x 3

111. (a)

g x 2 sin x cos x 2 cos xsin x 0

gx sin2 x cos 2 x 1 g x 0 (b)

Taking derivatives of both sides,Equivalently, and

which are the same.g x 2 tan x sec2 x,f x 2 sec x sec x tan x

g x f x.

gx 1 f x

tan2 x 1 sec2 x


118. (a)

(b)

(c) is a better approximation than

(d) The accuracy worsens as you move away from x c 1.

P1.P2

P1

30

0

2

f

P2

P2x 12

2

4 x 12 f 1x 1 f 1 2

8x 12

2x 1 1

P1x f 1x 1 f 1

2x 1 1

f 1 2

821

2

4f x

2 sec2

x4

tan x4

4

f 1 4

2 2

f x 4

sec2 x4

f 1 1f x tan x4

119. (a)

28x 62

43x 6 2

P2x 12

56x 62

43x 6 2

P1x 43x 6 2

f 6 423 23 56

f 6 2 sec

3 tan

3 43

f 6 sec

3 2 4sec 2xtan2 2x sec3 2x

f x 22sec 2xtan 2x tan 2x 2sec 2xsec2 2x2

f x 2sec 2xtan 2x

f x sec2x (b)

(c) is a better approximation than

(d) The accuracy worsens as you move away from x 6.

P1.P2

00

0.78

f

P1

P2

6

120. False. If then

y 121 x121.y 1 x12, 121. False. If then

fx 2sin 2x2 cos 2x.f x sin2 2x, 122. True

115.

f x 2x x2 4x2 4, x 2 f x x2 4 116.

h x x sin x xx cos x, x 0

hx x cos x 117.

f x cos x sin xsin x, x k f x sin x

Section 2.5 Implicit Differentiation 147

123.

limx0

f xsin x 1 lim

x0 f xsin x sin xx

limx0

f x f 0x 0 a1 2a2 . . . nan f 0

f 0 a1 2a2 . . . nan

f x a1 cos x 2a2 cos 2x . . . nan cos nx

f x a1 sin x a2 sin 2x . . . an sin nx

124.

For Also,

We now use mathematical induction to verify that for Assume true for Then

kn1n 1!.

n 1kknn!

Pn11 n 1k Pn1

n.n 0.Pn1 knn!

P01 1.ddx

1xk 1

kxk1

xk 12 P1x

xk 12 P11 k.n 1,

Pn11 n 1kPn1

Pn1x xk 1n2ddx

d n

dxn1

xk 1 xk 1Pn x n 1kxk1Pnx

Pnx xk 1n1d n

dxn1

x k 1

xk 1Pn x n 1kxk1Pnx

xk 1n2

ddx

Pnxx k 1n1

x k 1n1Pnx Pn xn 1xk 1nkx k1x k 12n2

Section 2.5 Implicit Differentiation

4.

y x2

y2

3x2 3y2y 0

x3 y3 8 5.

y y 3x2

2y x

2y xy y 3x23x2 xy y 2yy 0

x3 xy y2 4

1.

y xy

2x 2yy 0

x2 y2 36 2.

y xy

2x 2yy 0

x2 y2 16 3.

yx

y x12

y12

12

x12 12

y12y 0

x12 y12 9


6.

y yy 2x

xx 2y

x2 2xyy y2 2xy

x2y 2xy y2 2yxy 0

x2y y2x 2 7.

y 1 3x2y3

3x3y2 1

3x3y2 1y 1 3x2y33x3y2y 3x2y3 y 1 0

x3y3 y x 0

8.

y 2xy y

4xy x

xy y 2xy 4xy y 0

x

2xyy

y

2xy 1 2y 0

12

xy12xy y 1 2y 0

xy12 x 2y 0 9.

y 6xy 3x2 2y2

4xy 3x2

4xy 3x2y 6xy 3x2 2y2 3x2 3x2y 6xy 4xyy 2y2 0

x3 3x2y 2xy2 12

10.

cot x cot y

y cos x cos ysin x sin y

2sin xsin yy cos ycos x 0

2 sin x cos y 1 11.

y cos x

4 sin 2y

cos x 4sin 2yy 0

sin x 2 cos 2y 1

12.

y cos xsin y

cos x sin yy 0

2sin x cos y cos x sin yy 0

sin x cos y2 2 13.

y cos x tan y 1

x sec2 y

cos x xsec2 yy 1 tan y1

sin x x1 tan y

14.

1

cot 2 y tan2 y

y 1

1 csc2 y

csc2 yy 1 y

cot y x y 15.

y y cosxy

1 x cosxy

y x cosxyy y cosxy

y xy y cosxy

y sinxy 16.

y2 cos1y cot

1y

y y2

sec1y tan1y

1 yy2

sec 1y tan

1y

x sec 1y

17. (a)

(c) Explicitly:

x

16 x2

x

16 x2

xy

dydx

12

16 x2122x

y 16 x2

y2 16 x2

x2 y2 16 (b)

(d) Implicitly:

y xy

2x 2yy 0

x

y1 x= 16 2

y2 = 16 2

2 626

2

6

6

y

x

Section 2.5 Implicit Differentiation 149

18. (a) (b)

(c) Explicitly:

x 2

y 3

x 2

3 4 x 22 3

x 2

4 x 22

x 2

4 x 22

dydx

12

4 x 22122x 2

y 34 x 22 y 32 4 x 22

x 22 y 32 4, Circle x1 2 3 4 51

2

3

4

5

y x= 3 + 4 ( 2) 2

y x= 3 4 ( 2) 2

yx2 4x 4 y2 6y 9 9 4 9

(d) Implicitly:

y x 2

y 3

2y 6y 2x 2

2x 2yy 4 6y 0

19. (a)

(c) Explicitly:

3x

416 x2

3x443y

9x16y

dydx

38

16 x2122x

y 3416 x2

y2 1

16144 9x2 9

1616 x2

16y2 144 9x2 (b)

(d) Implicitly:

y 9x16y

18x 32yy 0

y2

y1

x6 2 2 6

6

4

2

4

6x= 16 2

x= 16 2

3

3

4

4

y

20. (a)

(c) Explic

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