03 - CHAPTER 3 Differentiation Section 3.1 The Derivative...

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Unformatted text preview: CHAPTER 3 Differentiation Section 3.1 The Derivative and the Tangent Line Problem . . . . 123 Section 3.2 Basic Differentiation Rules and Rates of Change Section 3.3 Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . . . 148 Section 3.4 The Chain Rule . . . . . . . . . . . . . . . . . . . . 162 Section 3.5 Implicit Differentiation . . . . . . . . . . . . . . . . 178 Section 3.6 Derivatives of Inverse Functions . . . . . . . . . . . 193 Section 3.7 Related Rates . . . . . . . . . . . . . . . . . . . . . 204 Section 3.8 Newton’s Method . . . . . . . . . . . . . . . . . . . 216 Review Exercises . . 137 . . . . . . . . . . . . . . . . . . . . . . . . . . 224 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 CHAPTER Differentiation Section 3.1 3 The Derivative and the Tangent Line Problem 1. (a) At x1, y1 , slope 0. At x2, y2 , slope 5 2. 2 3. 2. (a) At x1, y1 , slope At x2, y2 , slope At x2, y2 , slope 3. (a), (b) y (b) At x1, y1 , slope 4 3. At x2, y2 , slope 5 2. (b) At x1, y1 , slope 5 4. 2. f )1) )x 1 f )1) 1) 4. (a) x 1 y f )4) 5 f4 4 f1 1 5 f4 4 f )4) 4 6 f3 3 5 )4, 5) f )4) f )1) Thus, 3 3 f )1) x f4 4 (c) y 3 4 5 6 f1 x 1 3 x 3 1 0.25 f4 4 f1 f4 > 1 4 f3 . 3 1 1 f4 4 f 1) f1 <f 1. 1 2 1x 2 x 5. f x 4.75 1 (b) The slope of the tangent line at 1, 2 equals f 1 . This slope is steeper than the slope of the line through 1, 2 and 4, 5 . Thus, 1 2 1 2 )1, 2) 1 2 3 5 4 2 2 5. 1 3 2x is a line. Slope 7. Slope at 1, 3 lim 2 g1 x x x→0 lim 1 1 2 x f0 3t lim 3 t→0 2 5 2 1 lim g2 2 3 2 x 5 lim x→0 10. Slope at 0 2, 7 1 x 4 4 x lim 4 x h 2 t 2 t h 2 3 2 7 t 4 4t t t→0 t→0 1 t lim lim 2 4 t→0 lim x x t→0 t t x x x→0 f0 t t→0 g2 x→0 x→0 2 t t t→0 lim lim x x→0 lim 3 x 3 2 1 is a line. Slope 8. Slope at 2, 1 4 x lim 2 lim 3 2x x x→0 9. Slope at 0, 0 g1 2 x x→0 lim 6. g x 2 4 t 4 t 4 123 124 11. Chapter 3 fx Differentiation 12. g x 3 fx fx lim x x x→0 3 lim fx 0 x lim x→0 x x 0 lim x→0 2 x hs x→0 3x 16. f x fx lim s→0 3 fx lim x x x→0 9 lim fx 12 x x x x→0 1 2 lim x→0 17. f x 2x2 fx lim fx x x 2x x fx 2 x x x→0 4x x 2 x 2 x x→0 lim fx 1 lim 4x x 2 2 x x→0 x 1 lim 4x 2x x→0 fx x x 2x2 x 1 4x 1 x fx x 2 x2 1 x 1 x2 2x x x→0 lim 1 x2 x→0 lim x x x→0 lim x 1 x x→0 18. f x 2x2 1 x 2x2 lim 1 2x 1 x→0 lim 9 1 2 x x 2 1 x2 x 2x x x x hs s s s→0 2 3 3 s 2 lim x→0 2x x 2x 1 5 2 3 s s hs s s s→0 1 x 2 9 lim lim 3x x→0 x x→0 x x 5x 2 s 3 3 2 lim lim 3 fx x→0 2 3 3 2 3s fx x x x→0 fx x 3x lim lim lim 15. h s fx fx gx x x→0 x→0 5x 5 5 lim 2 lim x x x→0 0 x→0 gx lim x lim 0 3x gx 3 x→0 14. f x fx 13. f x 5 5 5x S ection 3.1 19. f x fx x3 12 x fx lim x x x→0 x lim fx 3 x 12 x x3 x x→0 x3 3x 2 x 3x 2 x x x→0 3 12 x 3x 2 x 3x x 2 x→0 x 3x x x→0 x3 3 12 x 2 x 3x2 12 12 x2 fx lim x x x→0 x lim x fx 3 x x3 x2 x x3 lim 2 x x→0 3x 2 x 3x 2 x x 3 x2 x→0 2x x x x 3x 2 x x x→0 x→0 3 2x x x 3x x 2 x 2x 3x 2 x 2x 22. f x 1 fx x x x→0 lim 1 x x x x→0 lim x lim x 1x x xx x x x 1 x→0 x2 x x→0 xx lim 1 1 23. f x x fx lim 1 1 lim x→0 1 2x x4 2 2 x3 1 fx x x fx x x 1 x x→0 lim x→0 lim x→0 lim x→0 x x x x x x x 1 x 1 1 2x x xx x→0 1x 1 x fx 1 x 1 x x x x 1 1 x x 1 x 1 1 x 1 1 1 x 1 x 1 1 2x 1 2x x 1 x2 x2 x x lim 1 1x x x 1 x 1 fx x→0 x x→0 x→0 fx 1 x xx lim lim fx 1 x2 1 x→0 lim x2 2 1 lim x3 x 3x 2 lim x 2 x 3x 2 lim fx 12 x x lim 3x 2 21. f x x3 12 x x lim fx 12 x x lim 20. f x The Derivative and the Tangent Line Problem x x 2 2x2 x x x x 2x2 2x2 2 125 126 Chapter 3 Differentiation 4 24. f x fx x2 25. (a) f x x fx lim x x x→0 fx x x x x→0 lim x→0 lim x x x 4x xx 4 x x x x x x x x x x→0 fx lim y 4x 2 5 4x 8 y 2 xx 5 y x 4x 3. (b) 8 5 (b) 1 fx x x x→0 lim x 2 (−3, 4) 2x x x→0 lim x2 1 2x 2x x lim 2 x x→0 3 −1 2 x x x 2x 2 2x 2 At 3, 4 , the slope of the tangent line is m The equation of the tangent line is 4 4x 4x 2 3 2 4. 3 y 8. x3 27. (a) f x −6 1 x x→0 y 5 fx x (b) 10 (2, 8) fx lim fx x x x→0 lim x x→0 lim x3 x 3x 2 x fx −5 x3 3x x→0 x x x→0 3x x x 2 At 2, 8 , the slope of the tangent is m The equation of the tangent line is y 3 x lim 3x 2 8 12 x 2 y 12 x 16. 5 −4 2 3x 2 32 2 2x −2 2x 1 At 2, 5 , the slope of the tangent line is m The equation of the tangent line is x x x2 x −5 x2 1 x x (2, 5) 26. (a) f x fx 2 2x x lim 2x x 4 x x xx x x→0 x xx x x x→0 x 4x xxx x→0 fx x→0 lim 4x 4x lim x→0 xxx lim lim 4 4 lim fx 1 2 12. 22 4. S ection 3.1 x3 28. (a) f x fx 1 (b) fx lim x x x3 x→0 x lim x3 lim 4 fx (1, 2) −6 1 x x→0 3x 2 x 3 6 1 −4 x 3x x x→0 2 x 3 x3 1 1 x lim 3x 2 3x x→0 x 2 x 3x 2 At 1, 2 , the slope of the tangent line is m The equation of the tangent line is y The Derivative and the Tangent Line Problem 2 3x 3x 2 3. 1 y 31 1. 29. (a) f x x fx lim 3 (b) fx x x fx x x x x x→0 lim x→0 x lim x→0 x lim x→0 −1 x x x x x x x x 5 −1 x x x 1 x x (1, 1) 1 2x x At 1, 1 , the slope of the tangent line is m 1 21 1 . 2 The equation of the tangent line is 1 1 x 2 1 y y 1 x 2 1 . 2 30. (a) f x x 1 (b) 4 (5, 2) fx lim fx x x x x→0 x lim fx 1 x −2 x→0 x lim x→0 x lim x→0 x x x 1 x x x 1 x 1 1 1 1 At 5, 2 , the slope of the tangent line is m 1 25 1 . 4 1 The equation of the tangent line is y 2 y 1 x 4 1 x 4 5 3 . 4 x x 1 x x x x 1 1 2x 1 1 1 x x 1 1 10 −4 127 128 Chapter 3 31. (a) f x Differentiation 4 x x fx lim fx x x x→0 x (4, 5) fx − 12 4 x x lim xx 4x x2 x xx x x x x3 2x 2 x x2 x xx x x→0 lim −6 x xx x→0 lim 12 4 x x x x→0 lim 10 (b) x2 x x x→0 x x 2 xx 4 x3 x2 x 4x x 4 x x x x x lim x2 x xx x4 x x2 4 4 x2 x→0 1 x2 At 4, 5 , the slope of the tangent line is m 4 16 1 3 . 4 The equation of the tangent line is 5 3 x 4 4 y y 3 x 4 2. 1 32. (a) f x x fx lim fx x x x→0 lim 1 x x x x→0 fx 3x2 1 x x x x 1x 1 xx x x→0 x x 1 1 y 1 1 2 At 0, 1 , the slope of the tangent line is m 0 1 1 2 1. The equation of the tangent line is y (b) 3 (0, 1) −6 3 −3 x 1. 1 3x y 1x 1 x ± 1. Therefore, at the points 1, 1 and 1, 1 the tangent lines are parallel to 3x y 1 0. These lines have equations 1 1 lim 3 x 1 x→0 lim 3x2. Since the 33. From Exercise 27 we know that f x slope of the given line is 3, we have 1 3x 1 2 and y 1 3x 1 y 3x 2. S ection 3.1 34. Using the limit definition of derivative, f x 3x 2 The Derivative and the Tangent Line Problem 3x 2. Since the slope of the given line is 3, we have 3 1⇒x x2 ± 1. Therefore, at the points 1, 3 and 3 3x y y 1 and y 3x 1, 1 the tangent lines are parallel to 3x 1 3x y 3x 1 2x x 1 2, 1 1 2x 1 1 2x 1 x 2 y 1 1 x 2 1 x 2 1; x 1 32 1⇒x 2. 3 . 2 x⇒ f x we have At the point 2, 1 , the tangent line is parallel to x 2y 7 0. The equation of the tangent line is 1 2 y x 1 Therefore, at the point 1, 1 the tangent line is parallel to x 2y 6 0. The equation of this line is 1 1 2, 1 2 32 1 1. y 3 2. Since the slope of the given line is we have 1 2 x 0. These lines have equations 36. Using the limit definition of derivative, fx Since the slope of the given line is 4 4. 1 . 2x x fx y 1 35. Using the limit definition of derivative, 37. f x 129 1 1 ; decreasing slope as x → 2x x⇒fx 39. f x 1 x 2 2 1 x 2 2. x2 ⇒ f x 38. f x Matches (b). 1 y y 2x; 40. f does not exist at x Matches (d). 0. Matches (c). Matches (a). 41. g 5 g5 2 because the tangent line passes through 5, 2 . 2 5 0 9 2 4 42. h 43. The slope of the graph of f is 1 ⇒ f x 1. 44. The slope of the graph of f is 0 ⇒ f x 0. y 4 because the tangent line passes through h 1 2 1 1 6 3 4 1 2 4 1 2 45. The slope of the graph of f is negative for x < 4, positive for x > 4, and 0 at x 4. y y 4 2 3 1 2 −3 −2 −1 −2 2 f′ f′ x −1 f′ 4 1 2 3 −2 −1 1 −1 −2 x 2 −6 −4 −2 x −2 −4 −6 −8 2 4 6 1, 4 . 130 Chapter 3 Differentiation 46. The slope of the graph of f is for x < 4, 1 for x > 4, and undefined at x 4. 1 47. Answers will vary. Sample answer: y 48. Answers will vary. Sample answer: y x y 4 3 2 2 4 3 3 2 1 f′ 1 1 x −4 −3 −2 −1 −1 x 1 2 3 x −4 −3 −2 4 1 49. f x 3 5 6 3x and c 5 53. f 0 4 1 2 and f x fx 3x 2 fx x2 y y f 3 3 and f is odd, then f c fc 3. 57. Let x 0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x 4 2 x. The slope of the line through 2, 5 and x 0, y0 equals the derivative of f at x 0: 5 2 y0 x0 5 y0 4x0 4 2 x0 8 x0 0 2x x0 y 2x 2 1 x0 y (b) If f c 3 and f is even, then f 3 x 02 1, 3 5 2x 2 2x 9 x0 1 x 02 2x0 2 x 02 2x0 3 3 x0 0 0 ⇒ x0 1 y 3 6x 1 y 6x 9 y 10 6 (3, 9) 8 (2, 5) 6 4 (3, 3) 4 (1, 3) (−1, 1) 1 x 1 2 3 6 1 −8 −6 −4 −2 −2 −4 x 2 4 (1, −3) 6 3 2x 1 y y −2 3, Therefore, the points of tangency are 3, 9 and ( 1, 1 , and the corresponding slopes are 6 and 2. The equations of the tangent lines are: 7 2 3. x0 2x0 y 5 fc 2x0 y0 3 3 ⇒ x0 c 58. Let x 0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x 2 x. The slope of the line through 1, 3 and x 0, y0 equals the derivative of f at x 0: 3 y 1 −3 2x0 4x0 3 6 2x0 Therefore, the points of tangency are 1, 3 and 3, 3 , and the corresponding slopes are 2 and 2. The equations of the tangent lines are: 5 4 2 8x0 2 0 2 3 y0 1 x0 x0 4 2 −2 x 2x0 2 1 −1 2 −6 −4 −2 3 x −3 −2 4 f 56. (a) If f c f 1 6 −2 y 3 8 −3 0; f x > 0 if 2 x 2 9 x3 fx 10 −1 2 x and c 55. f 0 0; f 0 x0 y 1 5 52. f x 6 4 12 −3 −2 −1 x 2 and c 51. f x 2 2 4 −4 54. f 0 4, f 0 0; f x < 0 for x < 0, f x > 0 for x > 0 3, <x< 3 −3 x 3 and c 50. f x 2 −2 −4 2 −2 −3 1 x y y 2x 1 S ection 3.1 59. (a) g 0 The Derivative and the Tangent Line Problem 131 3 (b) g 3 0 8 3, (c) Because g 1 (d) Because g g is decreasing (falling) at x 7 3, 4 1. g is increasing (rising) at x 4. (e) Because g 4 and g 6 are both positive, g 6 is greater than g 4 , and g 6 g 4 > 0. (f) No, it is not possible. All you can say is that g is decreasing (falling) at x x2 60. (a) f x fx x3 (b) g x fx lim x x x→0 x2 x x lim x→0 lim x fx 2 2x 2. gx x→0 x 2 x x2 x 2x x y 2x 3x x x→0 2 x 3 x3 x 2 x 3x x x 2 3x 2 2 and the tangent line is 1 1 x 2x lim 3x 2 1, f 3x x x→0 At x x x 3x 2 lim x x→0 3x 2 x3 x→0 x lim 2 x x3 lim gx x3 x x lim x x→0 x x x→0 x2 x→0 lim gx lim 1 y or 2x At x 1. At x 0, f 0 0 and the tangent line is y 0. At x 1, f 1 2 and the tangent line is y 1, g y 2x 1 3 and the tangent line is 1 3x 1 or y 3x 2. At x 2 0, g 0 0 and the tangent line is y At x 1. 1, g 1 3 and the tangent line is y −3 1 3x 1 or y 3x 0. 2. 3 2 −3 −3 3 For this function, the slopes of the tangent lines are always distinct for different values of x. −2 For this function, the slopes of the tangent lines are sometimes the same. 61. f x 13 4x 2 32 4x . By the limit definition of the derivative we have f x −2 x 2 fx 62. f x 1 0.5 2 fx 1.5 27 32 1 4 1 32 3 27 16 3 4 0.5 1 1.5 2 0 1 32 1 4 27 32 2 0 3 16 0 3 16 3 4 27 16 2 3 −2 12 2x 3 By the limit definition of the derivative we have f x x. x 2 1.5 1 0.5 0 0.5 1 1.5 2 fx 2 1.125 0.5 0.125 0 0.125 0.5 1.125 2 fx 2 1.5 1 0.5 0 0.5 1 1.5 2 −2 2 −1 132 Chapter 3 Differentiation fx 0.01 0.01 fx 2x 63. g x 0.01 x 2 2x fx 64. g x 2 0.01 x 2 100 2x 0.01 0.01 3x 0.01 fx 0.01 3 x 100 8 f 3 g The graph of g x is approximately the graph of f x 2 2x. f −2 The graph of g x is approximately 3 . the graph of f x 2x g −1 8 4 −1 −1 65. f 2 24 3.99 2.1 f2 1 67. f x x 4, f 2.1 2.1 4 0.1 2 Exact: f 2 4 2 2.1 0 x3 4 68. f x 2x3 2 2, f 2.1 2.31525 2 2.1 2 f2 1 and f x 13 2 4 66. f 2 3.99 2.31525 32 x 4 3x and f x 5 Exact: f 2 3.1525 3 2 3 6 As x → , f is nearly horizontal 0. and thus f f −2 f′ 5 f′ −9 9 f −5 69. f x S x 4 x −6 x f2 3 x x 4 x S 0.5: S x 0.1: S f2 x x x x x 3 x x x 2 3 1 3 x x x 1 2 x 2 3 x 2x 1 3 2 19 x 10 f2 x2 22 2 1: S x 0.5: S x (b) As x 3 3 x 2 5 S 0.1 x x x x f2 2 19 x 10 3 f S1 −2 4 5 7 S 0.5 − 1 1 x x (a) 2 3 x 2 f2 x 2 2 22 S x x → 0, the line approaches the tangent line to f at 2, 3 . (b) As 70. f x 2 1: x (a) 2 0.1: S x x x 2 2 f2 52 x x x x 5 2 2 5 x 6 2 5 2 5 x 6 2 5 2 4 x 5 2x 22 9 10 x 3 x x 52 16 x 21 2 5 2 16 x 21 x 5 2 2 5 2 2 5 6 4 x 5 12 x 4 −6 6 S0.1 S0.5 41 42 x → 0, the line approaches the tangent line to f at 2, 5 . 2 S1 f −4 3 S ection 3.1 x2 g1 f2 2 1 x2 x, c g1 1 1, c fx lim x→ 2 fx x f1 1 x ,c x x 0 x3 lim xx x lim x→1 1 1 1 2x2 1 x2 x→ 2 x3 2x x→1 g0 0 x 2 lim x 2 x→2 4 lim x 1 lim x2 x x x→1 1 x→ 2 2 2 lim x 2 4 x→ 2 3 1 x2 x x1 x lim x →1 1 3 lim x 2 x→1 x 3 5 x 1 lim fx x x 6 lim fx x → . . 3 x→3 x . Does not exist. x 1 → x x 1 ,c x x lim x→0 x As x → 0 , f6 lim x As x → 0 , 77. f x 2x x2 0 gx x lim x→0 f3 x lim x →2 1 75. g x 76. f x x2 f2 2 x 2 x, c x→1 g0 3 2 2 lim lim f1 1 x x→1 x3 74. f x x2 1 2x2 x3 2 lim x→2 gx x lim x→1 73. f x f fx x xx 72. g x 2 lim f2 1, c x→2 71. f x The Derivative and the Tangent Line Problem x→6 f3 3 2 3, c lim x→3 1x x 13 3 lim 3 x x→3 3x 1 x lim x→3 3 1 3x 1 9 6 f6 6 lim 623 x6 x x→6 0 lim x→6 1 6 x 13 Does not exist. 78. g x g x 3 3 1 3, c 3 gx x lim x→ 3 g 3 313 x3 x lim x→ 3 3 0 lim x→ 3 1 3 x 23 Does not exist. 79. h x h x 5 5 ,c lim x→ 5 5 hx x h 5 x lim x→ 5 5 5 x 0 lim x→ 5 5 x x 5 5 Does not exist. 80. f x f4 x 4 ,c 4 lim fx x f4 4 x→4 Does not exist. lim x→4 x 4 x 0 4 lim x→4 x x 4 4 81. f x is differentiable everywhere except at x (Discontinuity) 1. 133 134 Chapter 3 Differentiation 82. f x is differentiable everywhere except at x (Sharp turns in the graph) ± 3. 83. f x is differentiable everywhere except at x (Sharp turn in the graph) 84. f x is differentiable everywhere except at x (Discontinuities) ± 2. 85. f x is differentiable on the interval 1, (At x 1 the tangent line is vertical.) 86. f x is differentiable everywhere except at x 0. (Discontinuity) 2x is differentiable x1 for all x 1. f is not defined at x 1. (Vertical asymptote) 88. f x 87. f x x 3 is differentiable for all x 3. There is a sharp corner at x 3. 89. f x all x at x 3. . x 2 5 is differentiable for 0. There is a sharp corner 0. 5 5 6 −6 −7 −6 2 6 6 −3 −1 −2 90. f is differentiable for all x 1. f is not continuous at x 1. 91. f x x 1 The derivative from the left is 3 lim x→1 −4 fx x f1 1 x lim x→1 1 x 0 1. 1 5 The derivative from the right is −3 lim x→1 fx x f1 1 lim x→1 x 1 x 0 1 1. The one-sided limits are not equal. Therefore, f is not differentiable at x 1. 92. f x 1 x2 The derivative from the left does not exist because lim x→1 fx x f1 1 x2 1 1 lim x→1 x 0 lim x→1 1 x x2 1 x2 x2 1 1 1 1 lim x→1 x x2 . (Vertical tangent) The limit from the right does not exist since f is undefined for x > 1. Therefore, f is not differentiable at x x x 93. f x 1 3, x ≤ 1 1 2, x > 1 The derivative from the left is lim x→1 fx x f1 1 lim x x→1 1 x 3 0 1 1 2 0. 1 lim x x→1 2 0 The derivative from the right is lim x→1 fx x f1 1 lim x x→1 lim x x→1 x 1 1 0. These one-sided limits are equal. Therefore, f is differentiable at x 1. f1 0 1. S ection 3.1 The Derivative and the Tangent Line Problem 135 x, x ≤ 1 x 2, x > 1 94. f x The derivative from the left is fx x lim x→1 f1 1 x x lim x→1 1 1 lim 1 1. x→1 The derivative from the right is fx x lim x→1 f1 1 x2 x lim x→1 1 1 lim x 1 x→1 2. These one-sided limits are not equal. Therefore, f is not differentiable at x 95. Note that f is continuous at x x2 4x fx 96. Note that f is continuous at x 2. 1, x ≤ 2 3, x > 2 x→2 fx x f2 2 1 2x fx x2 lim 1 x→2 2 x, x lim x 5 lim x→2 2 2 x→2 1, x < 2 x≥2 fx x f2 2 x→2 fx x f2 2 4x lim lim x→2 3 x→2 x lim x→2 4. The derivative from the right is lim 2. The derivative from the left is The derivative from the left is lim 1. 5 1 2 2 12 x 2 x2 1 . 2 The derivative from the right is lim 4 4. x→2 2 12x x lim x→2 The one-sided limits are equal. Therefore, f is differentiable at x 2, f 2 4. fx x f2 2 lim x→2 lim x→2 lim x→2 lim x→2 2x 2 x2 2x 2x 2 2 x 2x 2 4 2x 2 x 2x 2 2 2x 2 2 2x 2 1 . 2 The one-sided limits are equal. Therefore, f is differen1 tiable at x 2, f 2 2. 97. (a) The distance from 3, 1 to the line mx d Ax1 By1 2 A m3 4 y 0 is C B 3 2 11 m2 y 4 1 3m m2 3 1 2 . 1 x 1 5 (b) −4 4 −1 The function d is not differentiable at m 1. This corresponds to the line y x 4, which passes through the point 3, 1 . 2 3 4 136 Chapter 3 Differentiation x 2 and f x 98. (a) f x x 3 and g x (b) g x 2x y 3x 2 y 5 3 (c) The derivative is a polynomial of degree 1 less than the original function. If h x x n, then h x nx n 1. 4 f g′ 3 2 2 1 −4 −3 −2 −1 1 2 3 4 (d) If f x fx x −1 −2 f' g 1 x 1 2 −1 −3 x4, then lim fx x x x→0 lim x4 x x x→0 lim fx x4 x4 4x3 6x2 x 2 x x→0 lim 4x x 3 x 4 x4 x x 4x3 6x2 x x→0 4x x 2 x 3 x lim 4 x 3 6x2 x→0 x 4x x 2 x 3 4 x 3. x 4, then f x 4 x 3 which is consistent with the conjecture. However, this is not Hence, if f x a proof since you must verify the conjecture for all integer values of n, n ≥ 2. f2 99. False. The slope is lim x x x→0 f2 x 2 is continuous at x 2, but is not 100. False. y differentiable at x 2 (sharp turn in the graph). . 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 f x x , then the derivative from the left at x 0 is 1 and the derivative from the right at x 0 is 1. At x 0, the derivative does not exist. 102. True—see Theorem 3.1. x sin 1 x , 0, 103. f x x x 0 0 Using the Squeeze Theorem, we have x ≤ x sin 1 x ≤ x , x 0. Thus, lim x sin 1 x x→0 f is continuous at x 0. Using the alternative form of the derivative, we have lim x→0 fx x f0 0 lim x→0 x sin 1 x x0 0 lim sin x→0 x 2 sin 1 x , 0, x x 1 and 1), the function is not differentiable at x 0 0 Using the Squeeze Theorem again, we have x 2 ≤ x 2 sin 1 x ≤ x 2, x 0. Thus, lim x 2 sin 1 x x→0 and g is continuous at x 0. Using the alternative form of the derivative again, we have lim x→0 gx x g0 0 lim x→0 f 0 and 1 . x Since this limit does not exist (sin 1 x oscillates between gx 0 x 2 sin 1 x x0 Therefore, g is differentiable at x 0 lim x sin x→0 0, g 0 0. 1 x 0. 0 g0 0. S ection 3.2 104. Basic Differentiation Rules and Rates of Change 137 3 −3 3 −1 As you zoom in, the graph of y1 x 2 1 appears to be locally the graph of a horizontal line, whereas the graph of y2 x 1 always has a sharp corner at 0, 1 . y2 is not differentiable at 0, 1 . Section 3.2 1. (a) x1 y 7. y y x3 y 12 3x2 y1 8 0 fx 5 x1 x 1 x 5 5 1 5x4 45 2t 2 fx 3t 4t t3 st 2 2t 3t 2 15. s t 2 fx sin cos 6x2 8x 20. g t cos 23. y y 2 4x3 2x 12x 2 3x 17. f x 6x 5e x fx 3 21. y cos t 6 24. y y Differentiate 14. y 8 8 x9 9 3x 1 3 2x 22. y y 2 cos x 3x e 4 2 sin x Simplify 25. y 5 2x 2 y 5 x 2 2 y 5x 3 y 26. y 4 3x2 y 4 x 3 2 y 8 x 3 3 y t3 5 x3 8 3x3 2e t 3t 2 ht 1 sin x 2 3x e 4 3x 2 18. h t 1 cos x 2 x3 8 y 5e x x2 sin t y Rewrite x 8x gx x2 3 cos x Function 1 x8 10. g x 1 13. g x 3 sin x 1x e 2 1 y 1 sin 1x e 2 6x5 fx 2 4x2 y1 2 6 3 2x3 x 6. y x 1 x y x6 gx 2t y (b) 1 2 9. f x 1 6x5 16. f x 4 gt y 6 56 2t y 3 x1 x t2 12. y 6 y 1 x 6 gx 5 5. y 12 1 32 2x y1 0 6 x y 6 8. g x y 2. (a) 3 4. f x 11. f t 19. y y (b) 1 2 y1 y 2 1 2x y 3. y Basic Differentiation Rules and Rates of Change 5 2e t sin x cos x 138 Chapter 3 Differentiation Function 27. y 28. y 29. y 30. y 31. f x 3 2x Rewrite 3 2 y 5x x 25 3 3 x2 6x f1 35 30x fx ft x3 f 6x f5 ft 32. x 2, 5, 0 3x2 9 8x4 y 3 5 2 x 25 3 y 2 25x3 1 x 2 32 y 1 2x3 y 4x3 6 34. f x 4 y 6 fx Simplify y 2 3x 2, 1, 3 3 9 x 8 y 12 x y 4 3 x y x x 3 x 8 y Differentiate 12x2 3 12x2 y 3 ,3 5 1 2 33. f x 73 x, 5 3 5t2 fx 5 3 f0 1 2 0, 21 2 x 5 0 35. f 4 sin f 4 cos f0 75 3 , 5t 2 41 , 0, 0 gt 36. 2 gt 1 3 cos t, , 1 4t 3 3 sin t 0 37. f t ft f0 30 3t 3 e , 0, 4 4 3t e 4 30 3 e 4 4 x x fx 1 2x 42. h x hx 44. y y 2x2 2 3x 6x 36x g1 2 40. f x 1 38. g x gx 1 3 0 g 39. g t 4e t2 4 t3 gt 4e x, 1, 4e x 2t 12t 4e 41. f x x3 3x2 x2 4 x 3 4x 2 3 fx 2 x3 3x x 1 x2 5x 2 45x 2 1 2x 3 1 43. y y 2x2 1 x2 18x 2 x 15x3 1 8 x3 x3 x x2 1 x3 x x x1 3x2 45. f x fx 1 x 1 x 2 8 x3 6 12 3 2x 23 2 6x1 1 2x 3 2 x2 3 t2 4 2t 12 t4 S ection 3.2 3 46. f x 5 x 1 x 3 fx t2 48. f t 3 t1 3 52. g x 1 3x 2 47. h s 1 5x 4 3 5 1 t 3 s4 h (s 4 s 5 49. f x 2 3t1 23 1 3t2 3 13 2x 5 cos x 2 5 sin x 3x 4 5 sin x 3 s2 2 s 3 x fx 4 5s1 13 6x1 5 cos x 12 3x 51. f x 3 15 6x fx 3 5 2 2 3s1 5 2 5 cos x 3 5 sin x 3 5 sin x x 2e x 3 2x 2 x3 2e x 2e x 3ex gx 1 2x 53. (a) y x4 At 5 4 43 x y x1 5 cos x x 2 x 3 fx 3 45 3 13 2 50. f x 1 x 5 23 2 t 3 ft x1 x Basic Differentiation Rules and Rates of Change 3ex 54. (a) f x x 4x3 1 4 fx 1, 2 : y 4 Tangent line: y 1 3 1 2 5x y 5x 5 2 x3 3 x 2 3 2x7 4 74 3 2 At 1, 2 : f 1 34 2x 3 3 x 2 1 y (b) 2 y Tangent line: 3 x 2 7 2 5 (−1, 2) −2 3x 1 −2 (b) 2y 7 0 5 (1, 2) −2 7 −1 55. (a) g x x ex 56. (a) h t sin t 1t 2e gx 1 ex ht cos t 1t 2e At 0, 1 : g 0 1 (b) At 2 1 2x y Tangent line: y 1 2x , 1et : h 2 1 Tangent line: 0 1 2e y 1 1 y 8 (b) 1 2e 1 1 2e t t 20 (0, 1) −4 1 2e 4 −4 −1 6 −2 1 2e 1 2 e 139 140 Chapter 3 x4 57. y 8x2 4x3 y 2x 0⇒x 2 0, ± 2 x 1 cos x 14 , 2, 14 sin x, 0 ≤ x < 2 60. y 0 At x 3 Horizontal tangent: , At x 3 2 sin x 0 3 ⇒x 2 sin x . ,y 2 cos x, 0 ≤ x < 2 3x y 1⇒x cos x 2 cannot equal zero. x3 3 Therefore, there are no horizontal tangents. Horizontal tangents: 0, 2 , 2, y 2 x 2x y 4 4x x 59. y 1 x2 58. y 2 16x x2 4x y Differentiation 3 3 ,y 3 3 2 ,y 3 At x ex 4x 3 3 3 4 x 4 4x 9 4 3, k 3 x 4 3 4 Hence, k 66. k x x k 2x 1 Hence, k 4x 3 4x 4 Hence, x 9⇒ x2 2 and for x k x2 64. k 4 and 4x k x x2 2x Equate functions. Equate derivatives. 2x 2x For x 65. ln 4 4 ln 4 Hence, k 2 223 , 3 3 No horizontal tangents k 2x , 4e x cannot equal 0. y kx 63. x 2 3 3 4e x 0 x ln 4, 3 x ex 4 , . 1 62. y ex y 2 3 . 23 Horizontal tangents: 61. y or 3, k 9⇒x 7 Equate functions. Equate derivatives. 2 and k 4 8 ± 3. 10. Equate functions. Equate derivatives. 32 3 4 x2 x and 4 x 4 3 x 4 3⇒ 3 x 4 3 x 4 3⇒ 3 x 2 3⇒x Equate functions. Equate derivatives. 2 x and 2 x x x 4 ⇒ 2x x 4⇒x 4⇒k 4. 2⇒k 3. 7⇒k 3. 3 S ection 3.2 Basic Differentiation Rules and Rates of Change 67. (a) The slope appears to be steepest between A and B. 68. The graph of a function f such that f > 0 for all x and the rate of change the function is decreasing i.e. f < 0 would, in general, look like the graph at the right. (b) The average rate of change between A and B is greater than the instantaneous rate of change at B. (c) y 141 y x f BC A E D x 69. g x 6⇒gx fx fx y 71. 5f x ⇒ g x 70. g x 72. y 3 2 f −2 x 2 1 1 f 1 f 1 3 5f x 2 x −1 1 3 3 4 f′ 2 −3 −4 If f is linear then its derivative is a constant function. fx ax fx If f is quadratic, then its derivative is a linear function. a b fx x2 and y 2x ⇒ m y m and 2x1 2x1 2x2 x1 x2 6⇒m 2x 2x2 2ax b 6x c 5, respectively. 6 6 3 2 Since y1 y x2 bx fx 73. Let x1, y1 and x2, y2 be the points of tangency on y The derivatives of these functions are: ax2 x22 x1 and y2 6x2 y 5: y2 x2 m x22 y1 x1 6x2 5 x2 x1 x12 5 2x2 6 4 3 )2, 3) 2 x22 x22 6x2 x2 6x2 5 x2 x2 2 2x2 3 x22 5 3 12x2 2x22 2 x2 6x2 2 x2 9 2x2 14 4x22 4 1 ⇒ y2 0, x1 x 2 and y1 6 2x2 18x2 y x2 0 2 ⇒ y2 0 x 1 3, x1 3 )1, 0) 2 3 18 y 1 5 0 4 1 ⇒y 1 or 2 2 1 4 1 and y1 4x )2, 4) 3 x −1 Thus, the tangent line through 1, 0 and 2, 4 is 4 2 2 3 0 x2 x2 1 6 −1 6x2 2x22 )1, 1) −2 4. 1 Thus, the tangent line through 2, 3 and 1, 1 is y 1 3 2 1 x 1 1 ⇒y 2x 1. 142 Chapter 3 Differentiation x. m2 is the slope of the line tangent to y 74. m1 is the slope of the line tangent to y 1 ⇒ m1 1 and y 1 ⇒y x The points of intersection of y x and y 1 x. Since 1 x are y x x⇒y 1 ⇒ x2 x 1⇒x ± 1, m2 At x 75. f x 3x fx 3 1 m1, these tangent lines are perpendicular at the points of intersection. 76. f x 2 fx cos x Since cos x ≤ 1, f x 0 for all x and f does not have a horizontal tangent line. 77. fx fx 1 2x x, 1 x 2 4, 0 1 2x 12 1 . x2 ± 1. 1. Since m2 sin x 1 ⇒ m2 x2 x x x x 4, y 9x 2 x2 0 5 y x 10 x2y 10 2x x2 10 2 2x 2x 2x The point 4, 2 is on the graph of f. 02 x 44 y 2 4y 8 x x 4y 4x 10 x 5 ,y 2 2 x 4 5 4 0 Tangent line: 5 2 , 5, 0 x 2 x2 fx 2x 5x 2 2xx 4 4 fx 78. 2 xy 4 5x 3x 3 Since 5x 4 9x 2 ≥ 0, f x ≥ 5. Thus, f does not have a tangent line with a slope of 3. 0y 4x 4 x5 4 4 The point 54 2, 5 is on the graph of f. The slope of the tangent line is f 5 2 8 25 . 8x 1 80. f 4 20 8x 25y 40 1 16 3.64 0.77 3.33 8 x 25 25y 79 . f 1 4 5 y Tangent line: 1.24 −10 19 −1 0 5 2 20 S ection 3.2 Basic Differentiation Rules and Rates of Change 81. (a) One possible secant is between 3.9, 7.7019 and 4, 8 : y y 8 8 8 4 7.7019 x 3.9 2.981 x y Sx (b) f x 31 x 2 Tx 3x 4 (4, 8) 4 −2 8 3x 12 −2 3.924 3 2 2 ⇒f 4 2 20 4 2.981x 143 3 4 S x is an approximation of the tangent line T x . (c) As you move further away from 4, 8 , the accuracy of the approximation T gets worse. 20 f T −2 12 −2 (d) x 3 2 1 0.5 0.1 x 1 2.828 5.196 6.548 T4 x 1 2 5 6.5 0.1 0.5 7.702 8 8.302 9.546 11.180 14.697 18.520 7.7 f4 0 8 8.3 9.5 11 14 17 (b) f x 1 1.0221024 1.0073138 y Secant line: y 3.022 x 1 x 1 1 3x 1 3 1 1 3x 2 (c) The accuracy worsens as you move away from 1, 1 . 1 (Answers will vary.) 2 3x2 Tx 82. (a) Nearby point: 1.0073138, 1.0221024 1 2 2 (1, 1) −3 (1, 1) −3 3 f T 3 −2 −2 (d) x 3 2 1 0.5 0.1 0 0.1 0.5 1 2 3 fx 8 1 0 0.125 0.729 1 1.331 3.375 8 27 64 Tx 8 5 2 0.5 0.7 1 1.3 2.5 4 7 10 The accuracy decreases more rapidly than in Exercise 81 because y x2 and g x 2x, but f x 83. False. Let f x fx gx x2 4. Then gx. then dy dx 0. 85. False. If y 2, 87. True. If g x 3f x , then g x 2 is a constant. 3f x . x3 is less “linear” than y 84. True. If f x 86. True. If y gx x 88. False. If f x c, then f x 1 1 xn x3 2. gx x, then dy dx x n, then f x 0 1 nx g x. 1 n . 1 n 1 xn 1. 144 Chapter 3 Differentiation 1 , 1, 2 x 89. f x 90. f x fx 1 x2 fx 1 ⇒f 1 1 2 2, 1 , 32 1 4 f2 2 f1 1 12 2 91. g x x2 e x, gx 2x f 1 1 0, 4.718 1 e 1 e 2.718 st 1 2 e2 2 8 12 64 ft sec 1362 1362 4 4 16 t 2 32 t 118 ft sec 16 t 2 4.9t 2 vt 9.8 t v5 9.8 5 9.8 10 2t 22 t 2 8t 108 v0 t 1362 4 27 22 t 2 32 2 22 86 ft sec 1362 4 295.242 ft sec s0 71 m sec 22 m sec 4.9t 2 v0 t 4.9t 2 96. s t s0 120 120 220 22 0 v2 9.226 sec 120 t 120 22 t 0 t 1362 ⇒t 16 8 1362 4.9t 2 16 t 2 0 32 e2 17 220 112 height after falling 108 ft 2: v 2 16 t 2 8.305 v3 32 ft sec t2 v 10 12 e 2 12 st 48 ft sec 1: v 1 st 0, 2 32t 1346 When t 95. 1x e, 2 st 1298 When t (e) v h0 0 94. 1362 s1 1 (c) v t (d) 12 e: 2 32t s2 2 (b) 0.955 vt 16 t 2 vt 3 Average rate of change: h2 2 93. (a) s t 1 6 1 2 1 : 2 2, 8 1 12 1x e 2 Instantaneous rate of change: Average rate of change: g0 0 0.866 3x2 hx 1 e 3 2 3 f0 0 x3 92. h x 0, 1 Instantaneous rate of change: e: 2 ⇒f 3 6 1 2 ex 0, 1 : g 0 0 Average rate of change: Average rate of change: g1 1 3 sin x 0, 1 ⇒ f 0 1 ⇒f 2 1, 1 0, Instantaneous rate of change: Instantaneous rate of change: 1, cos x, s0 4.9t 2 4.9 6.8 s0 0 when t 2 226.6 m 6.8. 2.403 S ection 3.2 1 2t 97. From 0, 0 to 4, 2 , s t 1 m min 50 60 mph. 40 30 20 10 t v 2 4 6 8 10 Time (in minutes) 60 Velocity (in mph) 145 v 0 for 4 < t < 6. Finally, from 6, 2 to 4⇒ vt t 98. mi min. 30 mph for 0 < t < 4 60 Similarly, v t 10, 6 , st 1 2 Velocity (in mph) 1 2 vt ⇒ vt Basic Differentiation Rules and Rates of Change 50 (The velocity has been converted to miles per hour.) 40 30 20 10 t 2 4 6 8 10 Time (in minutes) 2 3 2 3 40 mph mi min 6 min s mi min Distance (in miles) v 99. 4 mi v 0 mph 0 mi min 2 min 0 mi min 0 mi v 60 mph 10 8 (10, 6) 6 (6, 4) 4 (8, 4) 2 1 mi min (0, 0) 1 mi min 2 min 2 mi t 2 4 6 8 10 Time (in minutes) 100. This graph corresponds with Exercise 97. s Distance (in miles) 10 8 (10, 6) 6 4 2 (0, 0) (4, 2) (6, 2) t 2 4 6 8 10 Time (in minutes) 101. (a) Using a graphing utility, R (c) T R B 0.00557v 2 0.417v 0.418v 0.02. 0.02 (b) Using a graphing utility, B (d) dT dv 0.01114v For v 0.418 40, T 40 0.0014v 0.04. 80 T (e) 0.00557v 2 B R 0.86. 120 0 0 For v For v 102. C 80, T 80 100, T 100 1.31. 1.53. (f ) For increasing speeds, the total stopping distance increases. x 10 15 20 25 30 35 40 15,000 1.55 x dC dx gallons of fuel used cost per gallon C 2325 1550 1163 930 775 664 581 233 103 58 37 26 19 15 23,250 x2 23,250 x dC dx The driver who gets 15 miles per gallon would benefit more. The rate of change at x in absolute value than that at x 35. 15 is larger 146 Chapter 3 s 3, 103. V Differentiation dV ds When s 3s 2 4 cm, s 2, 104. A dV ds 48 cm2 per cm change in s. When s dA ds 12 at 2 105. s t c and s t s t0 t0 Average velocity: dA ds 2s 4 m, 8 square meters per meter change in s. at t t s t0 t0 t 1 2 a t0 t 2 c t 1 2 a t0 t 2 c 2t 1 2 a t02 2t0 t t 2 1 2 a t02 2t0 t t 2 2t 2 at 0 t 2t at0 s t0 , 106. 1,008,000 Q C dC dQ C 351 C 350 When Q 108. dT dt 350, KT 5083.095 dC dQ 6.3Q 107. N fp (a) f 1.479 is the rate of change of gallons of gasoline sold when the price is $1.479 per gallon. 6.3 5085 $1.91 (b) f 1.479 is usually negative. As prices go up, sales go down. $1.93. bx c a0 1, 0 : 0 2 b0 c⇒c a1 0, 1 : 1 2 b1 1⇒b 1 2 ax a 1 2a 1 1 a a 1 1 x2 y At a, b , the equation of the tangent line is a 1 Thus, y a x 2 a 1 x 1. From the tangent line y x 1, we know that the derivative is 1 at the point 1, 0 . y 1 ,x>0 x 110. y Since the parabola passes through 0, 1 and 1, 0 , we have: 1 a y a 1 x a2 a or The y-intercept is 0, The area of the triangle is A y a Therefore, y 1 3 2x2 3x 1. x a2 1 bh 2 1 2 2a 2 a 2 . a 2 . a 1 1 y The x-intercept is 2a, 0 . 2 b t0 Ta ax2 109. y 1,008,000 Q2 instantaneous velocity at t 2 () 1 (a, b) = a, a 1 x 1 2 3 2. S ection 3.2 x3 111. y 9 Tangent lines through 1, 9: y 9 3x 2 9x 9 3x 3 3x 2 9x 0 x3 147 9x 3x 2 y Basic Differentiation Rules and Rates of Change 2x3 3x 2 x2 2x 3 3 2, 81 8 x 9x 1 9 3 2 0 or x The points of tangency are 0, 0 and . At 0, 0 , the slope is y 0 3 2, 9. At 81 8 , the slope is y 3 2 9 4. Tangent lines: y 0 y 9x 112. y y 9x 9x y 0 and y 81 8 9 4 9 4x y 0 9x 4y 27 3 2 x 27 4 0 x2 2x (a) Tangent lines through 0, a : y a 2x x x2 a 2x2 a x2 a x ± 0 The points of tangency are ± a, At a, a , the slope is y Tangent lines: y a 2 2 ax 2 a ax y a . At a and y a, a , the slope is y a. a ax y a 2 2 ax a 2 a. a a Restriction: a must be negative. (b) Tangent lines through a, 0 : y 0 2x x a x2 2x2 2 ax 0 x2 2ax xx 2a The points of tangency are 0, 0 and 2a, 4a 2 . At 0, 0 , the slope is y 0 Tangent lines: y 0 0x y 0 and y 0 Restriction: None, a can be any real number. 4a2 y 4a x 2a 4 ax 4a2 0. At 2a, 4a 2 , the slope is y 2a 4 a. 148 Chapter 3 Differentiation x≤2 x>2 a x 3, x 2 b, 113. f x f must be continuous at x 2 to be differentiable at x lim a x 3 lim f x x→2 lim x 2 lim f x x→2 b x→2 3ax 2, 2 x, fx 8a 8a x→2 4 2 8a 4 b b 2, the left derivative must equal the right derivative. 22 12a 4 a 1 3 b 8a f0 4 3 4 cos x, x<0 ax b, x ≥ 0 114. f x b f2 x 1 a, sin x is differentiable for all x n ,n 0. You can verify this by graphing f1 and f2 and observing the locations of the sharp turns. x<0 x>0 sin x, Hence, a sin x is differentiable for all x 115. f1 x an integer. 1⇒b cos 0 fx 0. Answer: a 116. Let f x fx b 4 x<2 x>2 For f to be differentiable at x 3a 2 2. 0, b 1 cos x. fx lim x x x→0 fx cos x cos x sin x sin x x lim cos x cos x x 1 0 sin x 1 lim x→0 x→0 Section 3.3 lim sin x x→0 Product and Quotient Rules and Higher-Order Derivatives x2 1 x2 gx x2 1 2x 2. f x 2 2x3 2x 2 4x3 ht t1 2t 4 2x 6x 2 4 t1 2 2x t t2 3 x2 2t 3 7t 2 4 3t 2 3 t2 2x 2x 2x3 6x 5 x3 fx 6x 5 3x 2 x3 18 x 3 15x 2 6x3 24 x 3 2x 3 sin x x sin x 1. g x 3. h t cos x 15x 2 12 4x 2 2 3 4 t2 1 t 3 4 23 4. g s gs s2 s4 s1 t2 4 3t 2 3 2 2s 2s3 4 2 5s 2 2s 1 2 2 s1 4 4 s2 2s1 2 2 26 12 4 1 s2 s 2 s2 12 S ection 3.3 Product and Quotient Rules and Higher-Order Derivatives x 3 cos x 5. f x fx x 3 6. g x sin x 3x 2 cos x 3x x sin x hx x 1 x2 11 x 2x x2 1 2 3 9. h x 8. g t 1 x2 fx x1 x 3 1 x3 x 1 3x 1 x2 12 gt 2t 23 x1 3 3x 2 1 1 8x 3 3x 2 3 x 3 1 gx s s fx x3 sin x 2 x x cos x f0 2 sin x ft x3 3x 3x 4x 3 12x3 10x 3x2 2x2 18x t3 fx 3x 5 3x2 x2 fx 3 x2 15 x2 3 41 2x 2 2 x 1 6 1 4 3 fx fx f 4 x 6x x 2 x 32 6x 3 4 fx 2 2 sin x 4 x 2x 1 2 5x2 2 2x 2 11 x 1 x x 1 11 2 x1 12 2 1 x cos x 1 2 2 2 1 1 x 4 x cos x x 1 2 1 4 2 x3 1 1 2 4 8 cos x x sin x 2 1 2 2 2 3 cos t 1 3x2 2x x x f2 17. 2 t sin t 0 16. f x x2 f1 2 1 t4 1 x3 2x 3x2 4 3 3 2x x s s 2 cos t 3t 2 sin t x x s 2 1 t3 2 14. f x 5 15 x2 x 1 2 1 1 2 12 cos t t3 f1 15. f x 4 2 2 3x 2x2 4 s 1s 2 11 s 12. f t x3 14 t 7 2t 1 s x2 2 13. f x 2t2 22 2 7 s sin x x2 x 2 cos x t2 7 2t 2t hs 2 1 sin x 2x x cos x s 10. h s 1 1 2x 2 7 3 3 x3 t2 2t x3 1 x 9x 2 3x 2 3 x 3 1 2 11. g x sin x 3 cos x x2 x cos x 2 x 7. f x x sin x gx 149 2 1 2x x2 x 2 1 150 Chapter 3 sin x x fx 18. Differentiation 19. f x x cos x fx e x sin x fx e x cos x e x sin x e x cos x sin x sin x 1 x2 x cos x sin x x2 f 6 f0 32 2 36 6 33 cos x ex 20. f x ex fx cos x e x sin x ex 1 2 sin x cos x ex 12 0 f0 1 1 1 18 2 3 3 6 2 Function 21. y 22. y x2 Rewrite y 3 4 12 x 3 2 x 3 y 2 x 3 y 2x 3 5x 2 Differentiate 52 x 4 3 4 y Simplify 10 x 4 2 3 2x y 2 3 5x 2 y 23. y 7 3x3 y 7 x 3 3 y 7x 4 y 7 x4 24. y 5 4x2 y 5 x 4 2 y 5 x 2 3 y 5 2x3 25. y 4x3 x y 4 x, x > 0 y 32 x 7 26. y 27. f x fx 2 3x2 5 7 3 2 2x2 4x x2 1 2 x 29. f x fx 1 x x 2x x2 2 x 3 1 6x 7 2x x2 2x 2 y x 6 x 7 y fx x3 3x x2 x2 2 1 1 3x2 x4 4x x2 x3 12 3 x2 2 12 12 2x x2 2 3x 2 2x 3 1 2 1 4 x1 1 2, 12 28. f x 1 1 2x y 5 7 x2 2x x2 x2 y 3 x 3 4 4x 1 x 32 4x x 30. f x 3 x2 x2 6x x 6x x3 9 32 3 2 12 fx x4 1 x4 2x3 2 x x 2x2 x x4 1 1 x x 1 x 1 2 2 2 1 x x 1 1 x x 1 4x3 1 S ection 3.3 31. f x fx 2x 5 2x1 x 5 x 2 12 x 2 32 x 32. f x 3 fx 12 5x 32 Product and Quotient Rules and Higher-Order Derivatives x1 5 2 x x x1 12 16 x 3 x1 3 1 x 2 3 5 x 6 2x 5 2x3 2 2x 5 2x x x x1 2 3 3 5 6x1 23 2 1 x 3 23 1 x2 3 6 Alternate solution: fx fx 33. h s s3 hs 5 35. f x fx 36. g x gx 2 2 s6 2 6s x2 2x xx 3x 2 x 2 x x 2x x2 4x x fx 9x2 4x 9x2 4 15x 4 16 1 2 x4 4x 4x 3x1 5 6x1 23 x 6 2x2 x2 3 1 x2 3 6 1 1 6x 4x 2 x 2 3x 8x 2x 2 x2 3 2 2x 3x 2x 2 2x 3 x2 x 3 2 3 2 1 2x x2 1 x 3x3 1 4x 3x 4 5 41x2 48x 3 x5 3 x2 2x 2 1 x 2x 1 2 2 1 5x 36x3 2x 2 3 2 x2 5x x2 x 1 3x x 1 2x x 2 1 x 12 3x3 4x3 x x2 2x 1 37. f x 9x 4 hx 2x 1 2x 3x 2 2 x2 2 1 3 1 x 2 3 5 x 6 34. h x 4 6s s 1x 3 x x2 2 12s 2 4s3 3 16x 33x 2 3x 3 20 32x 4x 1 x 4x 2 6x 4 3x3 4x 12x 3 4x x 51 3x 4 1 15x3 4x2 8x 2 20x 16x 20 38. f x x2 x x2 1 x2 x 1 fx 2x 1 x2 1 x2 x 1 2x 1 x4 2x5 x4 6x5 4x3 39. f x fx x2 x2 x2 x2 x3 2x2 3x3 x 3x2 1 x 2x5 x 2x x2 x2 1 2x2 x 2x5 x 2x3 x4 x2 1 2x2 2x x3 x2 40. f x c2 2x x2 2 x x2 x2 x 1 2x 2x3 1 x2 2x 1 c2 x2 2x x 1 c2 c2 4xc2 c2 x2 x2 22 c c2 2x fx c2 c2 c2 c2 x2 x2 x2 4xc2 x2 2x c2 2 x2 2 151 152 Chapter 3 Differentiation 41. f t t2 sin t ft t2 cos t 2t sin t t t cos t 2 sin t 42. f f 1 cos t t 43. f t 1 cos sin cos 1 t sin t t2 ft 1 sin cos t t sin t cos cos t t2 44. f x fx sin x x 45. f x ex tan x fx ex sec2 x cos x sin x x2 47. g t 4 gt 1 t 4 t t1 8 sec t 34 4 8 sec t 1 4t3 8 sec t tan t 4 y 51. y y tan x 3 sec x tan x 2 sec2 x 3 sec x tan x 2 tan2 x cos x sin2 x cos x 54. f x fx cos x hx y x cos x x sin x 2 cos x 1 sin x fx cos x x sin x 55. y 2 sin x cos x 2 sin x sin x y cos x cos x sin2 x 2 tan x 2 ex cos x 57. y ex sin x 2ex cos x x 2e x 2x sin x 2x cos x sin x 2e x x2 1 1 2e x 2e x 2x x2 1 2 59. g x 2e x x2 x2 2x 1 12 gx x 2e x 2 sin x 2x cos x 2 cos 2x 2ex cos x x2 2x tan x x xex x 2 sin x 2xex 2 ex 4x 4 xe x ex 4 2 x 2 4x e x 4x 2 16x3 2 y x2 sec2 x x x sec2 x cos x y 58. y x2 tan x 53. f x 1 2 cos2 x 56. h x 52. y sec x sec x x tan x x2 cos x cot2 10 csc s cot s x sec x tan x x2 y 1 sin x cos x csc2 x 1 s2 sec x x 50. y sec x csc2 x 10 csc s hs 3 sec x 2 csc x cot x 1 s cot x ex y 8 sec t tan t 3 1 sin x 2 cos x csc x x 48. h s 3 sec x tan x 2 49. y ex 46. y x x e x 2x 8x 3 1 2x 2 2x2 x 8x 2 2 x 16x 1 2 5 1 2 ex 4 x (form of answer may vary.) S ection 3.3 x2 60. f x x 3 x2 fx 2 62. f 1 x2 x 2x2 12 2 1 x5 2x3 x2 cos y 64. f x csc x tan x cot x fx sin sin cos (form of answer may vary) 2 1 2 csc x cot x 1 csc x 2 43 1 0 f1 1 (form of answer may vary) csc x cot x 1 csc x csc x cot x 1 csc x 2 22 3 1 22 6 sin csc x csc x 1 y 1 g (form of answer may vary.) cos 1 1 cos 2 1 1 1 y 63. 61. g 1 153 sin cos 1 f Product and Quotient Rules and Higher-Order Derivatives 0 65. ht ht sec t t sin x sin2 x sin 2x tan 1 x3 3x 1x 2, fx x3 3x 11 x 6x2 1 6x slope at 1, Tangent line: y 4 3 1, 3 2 3x2 68. (a) f x 3 fx 5 3 1x f2 1 ⇒y x 2 2 x x 1 , 1 x cos2 x sin x cos x 1 2 2 9 (b) (1, − 3) 2, 1 3 11 x x 1 1 3 4 −3 6 −4 2 x 1 2 1 3 2 x 9 10 − 10 11 2 slope at 2, Tangent line: y 10 − 10 cos cos x cos x 2 67. (a) f x 4x3 f 1 sin x cos 2x sin 1 2 (b) sin x cos x t2 sec f1 cos x fx sec t 1 sin x sin x sin x cos x t sec t tan t sec t t tan t t2 h fx 66. 2 ⇒y 2 x 9 1 9 154 Chapter 3 Differentiation fx fx f 2 4 ,1 slope at 4 f ,1 y 1 2x y 1 2x 2y 2 (b) 23 3 ,2 3 slope at 3 ,2 Tangent line: Tangent line: 4x sec x tan x 70. (a) sec2 x 4 sec x, fx tan x, fx 69. (a) y 4 6 3x 2 3y (b) 6 2 2 3x 23 3 0 6 0 4 (( − π ,1 4 − −2 −4 71. (a) f x 1 e x, 1, 0 x fx f1 1 ex x ex ex 72. (a) f x x ex e 0 ex 1 (b) ex 1 1 4 ex 1 4 (b) 3 x 16 y 3 (1, 0) ex x 3 x 42 2 y Tangent line: 3 −3 0, 3 16 f0 y , 4 ex x4 x fx Tangent line: y 4 3 x 16 0 1 4 1.5 −3 −2 2 − 0.5 8 73. f x x2 4 x2 fx f2 4 , 40 8 2x x2 4 2 16 2 42 1 2 2y x 1 1 x 2 y y 1 x 2 4 0 fx 74. 2, 1 x2 16 x 4 fx 2 f 27 x2 9 x2 , 2y x 3 2 1 x 2 y 2 3 2 90 27 2 x x2 9 2 54 3 9 92 3 y 2 3, 1 x 2 6 0 1 2 3 3 x2 54 x 9 2 S ection 3.3 75. fx x2 fx f 16 x , 16 x2 8 5 2, 16 4 20 2 12 x 12 x 25 12 x 25 fx 25y x2 1 1 2x x1 x2 x 2 0 or x 2x 12 xx x 2. 2 ex 8 gx 2 ex 8x e2x 2y x 2 1 x 1 x 1 x 1 x 2 2 1 0 when cos x 1 2 1 0, f 3 0 1 x 2 1 ⇒y 1 x 2 2 1 x 2 3 ⇒y 1 x 2 7 2 y f (x) = 6 x −2 2 4 6 −4 −6 1 3 . 4 sin x ⇒ x 3 23 , e 42 1 1 x x 1 2 x 4x 5 1x 1 f 1 1 x x 5x 4x 2 x sin x 4 2 1 10x 2 2x 1 2 4 1 2y + x = −1 y (− 1, 5) x x−1 f (x) = 6 y = −x + 4 2 (2, 2) y = − 4x + 1 1 1 x −4 2 x −2 2 ( −2 1 0 0⇒x 1, f 2 2; f 1 ,2 2 1 2 4, f 2 Two tangent lines: (3, 2) (−1, 0) −6 −4 −2 x+1 x−1 x 4x 1 2 y e x cos x 1 xx 1 1x x 2 y 2y + x = 7 e x sin x x, x x 1 be a Let x, y point of tangency on the graph of f. 5 1, 3; f 2 x x fx 2 3; Slope: 1 e x sin x, 0 ≤ x ≤ Horizontal tangent at 3 1 2 2x x2 0. fx ±2 x 0 when x e x cos x 4 1 x 1 2x x2 2x 2 2 x 1 fx 8x 1 2 2 0 1 x2 fx 82. f x 1 x 2 16 25 3. 6⇒y x 24 1 1 x fx 2 12 ex Horizontal tangent at 3, 8e x x 2 x 25 x2 80. f x 0 when x 81. f x 2 Horizontal tangent is at 0, 0 . ex gx 2 x 25 16 x2 fx Horizontal tangents are at 0, 0 and 2, 4 . 8x 4 5 2x 78. f x 0 when x 79. g x 2 25 y 0 4x2 62 24 x2 16 10 2 y 1 x fx 24 155 4 5 2, , 64 4x 2 x x2 6 2 16 25 16 x 6 x2 fx 2 x2 77. f x x2 f2 y 25y 16 x 2 16 2 256 x2 12 25 8 5 y 4x 76. f x 16 16 16 x 2 x x 2 16 2 256 2 Product and Quotient Rules and Higher-Order Derivatives y 1 4x 1 2 y 2 1x 2 ⇒y ⇒y 4x x 1 4 1 4 1 , 2 ) −1 156 Chapter 3 x 83. f x 25 x 5x x gx 6 2 3 3x 1 x 22 x gx Differentiation x 5x 22 4 2 6 4 2 2x x 2 fx 2 f xgx f xg x p1 f 1g1 f 1g 1 gx f x 87. Area At 1 3 q4 2 70 2t 31 t 2 3t1 1 1 t 2 2 6 1 2 1 x cos x sin x x2 sin x 2 sin x 2x 1 x cos x sin x x2 2x sin x 3x x x 2t3 f xgx 1 8 2 p4 t1 2 88. V 12 13 t 2 4 42 t 2 2 131 t 22 Vt 1 375,000 x C , x≥1 2 6 27 8 (a) C 200 90. 6 P dP dV 6x 375,000 x 2 2t 1 2 1 1 2 12 16 375,000 x2 2 t 3t 2 4t 1 2 12 k V k V2 $0 per unit (c) C 300 11 6 $1.83 per unit The rate of change of C is decreasing at a rate of $3.38 per unit when the order size is 200 units. The rate of change is not changing when the order size is 250 units. The rate of change of C is increasing at a rate of about $1.83 per unit when the order size is 300 units. 91. Pt Pt 500 1 500 500 P2 4t 50 92. t2 50 t2 4 4t 2t 50 t 2 2 200 50 4t 2 t2 2 2000 31.55 bacteria per hour F dF dd 50 50 t2 t2 2 3 4 t $3.375 per unit (b) C 250 5 f xg x 42 r 2h 12 6x2 375,000 x fx 4 gx q7 2 10 6t 1 2 cm sec 2t 89. C 5x f xg x gx f x (b) q x 1 3 t 1 t 2 2 14 2 32 At 3x 1 x2 86. (a) p x f xg x gx sin x f and g differ by a constant. 85. (a) p x (b) q x x cos x gx f and g differ by a constant. 3 x2 gx 2 x 2 x cos x 84. f x 2 41 3x x 2 Gm1m2 d2 Gm1m2 d Fd 2Gm1m2 d3 2 cubic inches sec S ection 3.3 93. (a) d sec x dx 1 d dx cos x d csc x dx d1 dx sin x 1 cos x 2 sin x 1 cos x sin x cos x 1 sin x sin x cos x cos x cos x sin x sec x tan x sin x 0 1 cos x sin x cos x sin x sin x 2 csc x cot x cos x sin x cot x d cot x dx d cos x dx sin x fx csc x, fx sin2 x cos2 x sin2 x sin x cos x cos x sin x 2 sin x 1 sin2 x csc2 x sec x gx 94. cos x 0 1 sin x csc x (c) 157 1 cos x sec x (b) Product and Quotient Rules and Higher-Order Derivatives gx 0, 2 csc x cot x ⇒ sec x tan x ⇒ sec x tan x csc x cot x sin3 x cos3 x 1 cos x 1 sin x 1⇒ 1 ⇒ tan3 x sin x cos x cos x sin x 1 ⇒ tan x 1 1 37 , 44 x 95. (a) n t 3.5806t 3 82.577t 2 vt 0.1361t 3 3.165t 2 603.60t 23.02t 1667.5 (b) 350 12 59.8 n(t) 0.1361t3 3.165t 2 23.02t 3.5806t3 82.577t 2 603.60t vt nt (c) A 59.8 1667.5 v(t) 5 5 12 0 12 0 0.05 (d) A t represents the rate of change of the average retail value per 1000 motor homes. 5 12 0 A represents the average retail value (in millions of dollars) per 1000 motor homes. r 96. (a) sin r r h r csc h r csc 97. f x 4x3 6x1 r csc r h 1 csc 3960 2 fx x 32 x2 1 64 x3 2 fx 3x 12 98. 3 x fx 192 x4 cot 6 fx 2 fx r h h 30 (b) h 3 7920 3 mi radian 158 Chapter 3 99. fx Differentiation x x 1 x fx x2 100. f x 11 x1 x1 2 x 1 1 fx 2 x 101. f x fx fx 3 sin x sec x tan x sec x sec2 x sec x sec2 x 103. g x gx gx ex x xe x 104. h t ht e t cos t ht et ex ex ex 2 x x3 105. f x x2 fx 2x ex x4 2x xe x ex 4 x f 5 x 2 f 6 x f 0 2 x 109. 1 sin t e t sin t e t cos t sin t e t cos t cos t sin t 2e t cos t 106. f x 2x tan2 x 2 2x 108. f tan x sec x tan x e t sin t x2 x2 xe x 1 x 2 sec x 3 cos x fx x 2 x3 3 3 sin x fx fx 102. f x 1 1 1 x2 1 2 fx 2x x 2x 2x 1 107. f x 2 x2 2 f 4 x 2x 1 2x 2 1 12 x 110. The graph of a differentiable function f such that f > 0 and f < 0 for all real numbers x would, in general, look like the graph below. y 4 3 2 1 y x f2 3 2 1 4 0 One such function is f x x 2 2. f x 111. f x 2g x hx 112. f x 4 hx fx 2g x hx fx 2g 2 h2 f2 h2 gx hx hx f2 113. f x 2 0 2 4 fx hxg x gxh x 2 hx 4 f2 h2g 2 g2h 2 2 h2 1 2 34 1 10 2 S ection 3.3 114. f x gxh x f2 3 2 f′ f f ′′ f 1 h2g 2 x 34 1 2 2 1 159 y f′ hxg x g2h 2 116. y 115. gxhx fx Product and Quotient Rules and Higher-Order Derivatives 1 −2 2 −1 x −1 2 3 4 −2 14 f″ It appears that f is quadratic; so f would be linear and f would be constant. It appears that f is cubic; so f would be quadratic and f would be linear. 117. 118. y f′ 4 3 f′ y 119. y f″ 4 3 2 1 f′ 1 x −3 −2 −1 1 12345 −1 f″ −3 −4 −5 −4 −3 x −1 −2 2 π 2 x f′ t 2, 0 ≤ t ≤ 6 36 at f″ 2π −4 121. v t 1 π 2 −3 −1 y 120. f″ 2 2t v3 27 m sec x π 6 m sec2 a3 The speed of the object is decreasing. 122. (a) (b) The particle speeds up (accelerates) when a > 0 and slows down when a < 0. y 16 12 4 −1 Answers will vary. s 8 v t 1 4 5 6 7 a s position function; v velocity function; a acceleration function fx 123. f xn n nn x Note: n! 124. nn 1n 2 ... 2 1 1 ...3 2 fx n! 1 (read “n factorial”) f n x 1 x 1 n nn 1n xn 1 2 ... 2 1 1 nn! xn 1 160 Chapter 3 125. f x Differentiation gxhx fx gxh x hxg x fx gxh x g xh x gxh x 2g x h x hxg x gxh x g xh x 2g x h x x 3g x h x 3g x h x g x g xh 3g x h 3g x h x (a) fx gxh 4 f x 4 gxh g x n g x hn x x 4g x h nn 1n h xg x 2g x h x x hxg x xhx 3g x h x x 6g x h x 2...2 1 g x hn 2...2 1 1n 1n nn 1 n 2 . . . 2 1 gn n 1 n 2...2 1 1 n! g x hn 126. 1! n 1! n Note: n! n! gn 1 !1! 1 . . .3 nn xf x xf x xf x fx xf x xf x 1 fx n 2x3 6x 6x 2 y y y y xh x gn x h x g x hn 2 2! ... x gn x h x xf n 1 1 ,y x 1 ,y x2 2f x x x3y 3f x 2x2y x3 2 x3 2 x3 2x 2 1 x2 2 2 0 12x y xy ... x 6 y x 10 y y 2! n 3 2 1 (read “n factorial”) 2f x nf y y 1 n! x xh x xf x x 128. xf xh x nn 1 n 2 . . . 2 1 g x hn 2 1 n 2 n 3...2 1 x 127. y fx n In general, x f x 129. 1 1 fx xf x 2 3g g4 x h x 4g x h x nn 1 n 2 . . . 2 1 g x hn 3 2 1 n 3 n 4...2 1 g x hn x h xg x g4 x h x xh x g x h4 x (b) f hxg x 12 2y 2 sin x x. 12 x 12x 2 6x 2 6 24x2 y 130. 3 sin x y 2 sin x 2 sin x 3 3 y 3 sin x cos x y 2 cos x 2 sin x 3 cos x 3 cos x sin x y 3 cos x sin x 3 cos x sin x 0 S ection 3.3 131. (a) f x ln x, f1 Product and Quotient Rules and Higher-Order Derivatives 0 (b) 3 P1 1 , x fx f1 1 , x2 P1 x f1x P2 x 1 f1x 2 −3 f1 1 f1 1 2 x f1x −3 1 1 P2 1 x P2 x 1 1, 1 2 x 1 0 1 P2 1 1, 1 x 2 f1 P1 1 1, P2 x 6 1 0, P1 x f P2 fx (c) P1 1 1 161 1 P2 1 1 The values of f, P1, P2 and their first derivatives agree at x 1. The values of the second derivatives of f and P2 agree at x 1. P2 x is a better approximation. (d) The accuracy worsens as you move away from x e x, 132. (a) f x f0 1. 1 (b) 5 f fx e x, f0 1 fx e x, f0 1 P1 x f0x P2 x 1 f0x 2 0 f0 0 2 P2 P1 −4 x f0x 1 0 4 −1 f0 12 x 2 x 1 (c) P2 is a better approximation. (d) The accuracy worsens as you move away from x 133. False. If y dy dx 134. True. y is a fourth-degree polynomial. dny 0 when n > 4. dx n f x g x , then f xg x 0. gx f x. 135. True hc f cg c fc 0 gc f c gc 0 0 136. True 138. True 137. True If v t 139. f x ax 2 bx fx 2ax c then a t b c x-intercept at 1, 0 : 0 2, 7 on graph: 7 4a Slope 10 at 2, 7 : 10 a b 2b 4a c c b Subtracting the third equation from the second, 3 b⇒b 2. Finally, 3 Then, 10 4 3 fx 3x 2 2x 1 b 2 c. Subtracting this equation from the first, 3 c⇒c 1. a. vt 0. 162 Chapter 3 Differentiation ax3 140. f x bx 2 fx 3ax 2 fx cx 0 ⇒x 2bx d, a 0 c 2b ± 4b2 6a 12ac (a) No horizontal tangents: f x 4b2 0 4b2 12ac < 0 Example: a c x3 fx 141. f x 1, b Example: a x3 x2, if x ≥ 0 x2, if x < 0 2, fx 12ac fx 4b2 0 1, b 3x 2 3, c 142. (a) 2x (b) fg fg fg 1. y x2 5. y e 9. y y 11. g x gx 13. f x fx 15. f t ft 2 7 2 9x x2 fu 5 y u4 1 y y 2x y 7 2 4 2. y fg fg True fg fg 2f g fg fg False f gx fg fg 3 9 108 4 9x 3 2 u3 6. y cos eu 8. y ln x 2x 39 4x x2 13 x2 2x3 y ft 1 1 21 t x y fu y u y 2 x2 u 3 tan u 12 3x 2 u 3x 2 y cos u 3 u ln x y u3 2 1 2 2x3 1 6x 2 x2 35 x2 9t 2 9t 3 4 7 15 5 12x2 2x3 1 5 23 7 16. g x 12 gx u 1 3 tan 14. f t 13 u x 4. y u 12. y 12 t y y 23 x2 t csc x 6 2x 9x 2 9 3 1 1 2 fg fg 10. y 12 4 1 y 3 34 9 x2 u 3 2x 6x u 7 gx u 1 2x 2x fg fg u 4 csc3 x 7. y 0: x2 fg u 5 3. y 1, c The Chain Rule f gx 6x fg fg 1, a x3 fx 3x f 0 does not exist since the left and right derivatives are not equal. y Example: b 3: if x > 0 2, if x < 0 Section 3.4 12ac > 0 fg 2x, if x ≥ 0 2x, if x < 0 fx 0: x xx (c) Exactly two horizontal tangents (b) Exactly one horizontal tangent 5 3x gx 1 5 2 3x 2x 13 9 3 5 12 x2 30x 5 3x 3 6 9t 4 7 12 3 25 3x S ection 3.4 17. y 9x2 12 9x 3 y 19. y y x ft 2 x 1 3 2x 3 1 12 x2 fx 2x x2 1 x x2 12 x 2 x x2 1 x x2 33. g x 2 x x2 2x x2 x2 1 12 1 y 1 32 32. y x2 2x 32 1 5 2 32 x2 x2 1 x2 12 t 2 5 2 1 12 y 1 x 3 3x 7 52 x2 x 10x 23 x 2 x 2 3 2t 3t t2 4 t 2t 2 3 t2 3x 7 2 9x 3x 7 2 12x 2 32 7 3 1 7 12 x 16 2 7 x2 121 x 16 22 1 x2 x x4 x4 1 x2 12 2x x 16 5 2x 2 x 16 x2 x2 32 x2 2 12 2 x 1 2 x4 x4 2 1 2 x4 x4 23 2 12 2 32 2 t2 34. h t 2 2t 3 7 x 4 2 2x 4 x4 2 3 2 12 1 2 32 2 x 3x 2 x2 34 12 2 x3 2 16 x 2 12 1 27 9x 44 12 x2 1 2 24 3 t 3x 30. y 9 1 4 3x 12 1 3 x 3x 2 2 16 1 x2 x2 gx 12 2x x 1 3t 3 t2 34 1 3 24 t 2 x2 12 x x2 8t 2 x1 x2 2 x 3x 1 1 2 x 3t 1 t2 28. f x 4 2x2 1 x2 31. y t2 gt 32 2 4 x2 1 y y 2x 2 1 9x st 2 3 32 2 x2 3 4 4 3 2 1 14 9x 26. g t x2 1 y 2 3 t 2 2x x x1 1 2 x 3 4x 2x x 29. y x2 4 12 x2 x fx 1 3 2 x2 4 x t3 2 2 ht 2 t3 t t3 2t2 163 x>1 1, x < 1 34 24. y 3 1 x 2 1 fx x 2x 2 3 2 dy dx 27. f x 34 2 x x 1, gx 23 4 22. s t 2t y 9x2 1 12 t 6x 18x 2x 20. f x x2 2 23. f t 23 14 1 4 4 y 25. x2 x2 18. g x 4 24 2 21. y 13 4 The Chain Rule 2 t4 4t t3 2 3 2 2t t2 3t2 3 2 t 2 2t3 4 t3 t3 2 3 4x 3 12 3 1 2 164 164 Chapter 3 1 1 35. f v fv 3 Differentiation 3 2v v 1 1 3x 2 2x 36. g x 2 2v v v 2 1 1 v 2v gx 2 3x 2 2x 1 3 2 2 1 5 3 3x 2 1 9 1 2v 1 v4 3 1 5 2 2x 6x 2 1 2 x x2 2 1 1 3x2 1 2 x x2 y −1 4x3 2 x y′ 7 1 5 y 1 y 2 1 1 4 2x 38. y y 12 2 15x 5 3x 2 52 4 3x 2 2x 37. y 30x 5 2x 6 3x 2 5 6x 2x y′ 2x x 32 1 −6 6 −2 The zero of y corresponds to the point on the graph of y where the tangent line is horizontal. 3t 2 t2 2t 39. g t gt y has no zeros. 24 g′ fx g −5 3 x 1x 2x x 1 −5 −2 t 1 st t 3 1 9 8t 2t y′ y has no zeros. 22 y 4 t2 5t2 42. y x 43. s t f′ −2 t 2 15 y′ 9 cos x x 6 − 15 3 t 44. g x s′ −3 6 gx s t x 1 1 2x x 1 1 2x 6 1 g 1 g′ −2 −2 3 1 46. y x sin x cos x 1 cos x 5 1 y′ x2 −3 x sin x 10 g has no zeros. −5 dy dx y −3 2 The zero of s t corresponds to the point on the graph of s t where the tangent line is horizontal. y 6 The zero of y corresponds to the point on the graph of y where the tangent line is horizontal. −3 45. f −3 4 1 y y 4 The zeros of f correspond to the points on the graph of f where the tangent lines are horizontal. −2 The zeros of g correspond to the points on the graph of g where the tangent lines are horizontal. x x2 x2 x 2 5x 2 2x 40. f x 1 3t t2 3t 2 t2 2t 1 3 2 41. y −1 x2 The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. y dy dx x2 tan 2x tan 1 x 1 x 6 y sec2 1 x The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. −4 5 y′ −6 S ection 3.4 47. (a) y cos x y y0 y 1 y sin 3x 3 3 cycles in 0, 2 sin 2x y 165 3 cos 3x y0 1 cycle in 0, 2 (b) y 48. (a) sin x The Chain Rule y (b) sin x 2 2 cos 2x y0 y 2 2 cycles in 0, 2 y0 The slope of sin ax at the origin is a. 1 x cos 2 2 1 2 Half cycle in 0, 2 The slope of sin ax at the origin is a. e3x 49. (a) y 50. (a) y 3e3x y (b) y 3. y (b) y 3x 3e y At 0, 1 , y y ln x3 2 y 3 2x 12 sin 2 4 2. y ln x1 2 1 ln x 2 11 2x 54. y 1 2x 2. 56. y dy dx 1 sin 2 4 sin 57. g x x cos 2 2 2 3 tan 4x 58. h x 12 sec2 4x gx cos hx x 60. g t 5 cos3 t gt 15 cos t 15 cos t sin 2 t sin 3 61. 1 sin 2x 4 t 1 sin 4x2 4 dy dx 1 x 2 12 1 2x 2 5 cos 2 x 3x y 3x 5 cos 4x 2 3 5 sin 4x2 8x 3 40x sin 2x 2 63. 3x2 sec x3 tan x 3 x y t 2 sec x3 x 5 cos 1 sin 4 2 sin 2 cos 2 62. y 1 . 2 At 1, 0 , y 1 2 sin 2 4 f 3. 2x 2 ln x At 1, 0 , y 3 . 2 3 ln x At 1, 0 , y 2 x y 3 sin 3x 59. f 2e ln x2 53. y cos 3x dy dx 2. 2x At 0, 1 , y 3 ln x 2 At 1, 0 , y 55. 3. 31 2x 52. y e ln x3 3 x y At 0, 1 , y 3x e 51. y 2e2x y At 0, 1 , y e2x y sin cos x dy dx cos cos x sin x sin x cos cos x 2 1 cos 4x2 8x 4 2x cos 2x 2 166 166 Chapter 3 y sin x1 3 cos x1 64. y 3 sin x 1 x 3 1 cos x1 3 x2 3 66. y Differentiation 13 1 sin x 3 23 3 cos x sin x 2 dy dx 23 67. y dy dx x2 2xe 2e2x fx cos x 3 x2 e e2x 65. f x e x x2e y 68. ex 2x dy dx x2e x xe 69. g t gt 71. y dy dx t e et t 3e 70. g t et 2 et e 2 x2 72. y 2x dy dx 1 1 y dy dx 76. y dy dx ex e ex 1 y dy dx 83. y dy dx x2ex 2xex dy dx e 2 ex x 2 ex 2ex ex 2x 2 ex x2 77. f x ex ex x ex ex x 1 e x e x 2 x 2 e x 1 x xex e x 79. e3 x y ex sin x ln ex 81. g x x gx 1 4 4 ln x ln x2 ex cos x 3 1 x 4 ln x x ln x sin x 2 ln x 2 x2ex 2 x e sin x 1 x ln x hx y dy dx x ln x x 1 x ln x cos x ex 2ex cos x 82. h x 2 x 84. 3 2x 2x cos x dy dx e3 ln x ex x2 ln x fx 1 ln x 2 dy dx ex 2 cos x 80. t3 ex 2ex e2x x 3 t2 2 ex e ex e x y x x t2 ex 75. xex fx y ex ex 1 2 6e 6 t 3e3 ex ln 1 2 78. f x 3 6t 2xe x e 3 t2 x 2 3t 73. ex 2 ex e e ex ex ln 1 74. 3 t2 t ln 1 ln e x e gt 3 x 1 ln x ln 2x2 4x 2x2 3 3 e ex x e 1 x S ection 3.4 85. y dy dx ln x x2 1 x fx x ln 1 x 93. x 1 2t ln t x x ln 1 1 1 1 2x 1 x2 x y xx 1 1 1 1 x2 x2 Note that Hence, 2 dy dx x2 4 4x 4 4 4x x2 4 2x x 4 1 12 2 2x x2 x2 4x x2 95. y dy dx ln sin x cos x sin x 4 4 1 x2 x2 x2 1 1 x2 2 1 x 3 1 x 3 ln t 1 t2 ln 3 x x 2 2 1 1 3x 2 ln t t2 1 ln x 3 2 4 1 x ln x 2 3 x2 4 1 ln 2 4 1 x2 4 4 x2 x2 4 x2 4 x3 4 1 4x x2 x x2 4 x2 x2 4 x2 4 x3 x2 1 x2 12 4 x2 4 x3 x2 x 2 2 4 1 1 x2 2x2 1 42 x2 x2 x2 2 x x2 4 4 1 ln x 4 1 4x 4 4 . 1 4x 1 4x x2 4 . x3 96. y cot x 1 3 x 1 1 1 x2 4 1 2 x 1 x2 x 1 2 ln 4 1 x2 1 x2 2 1 x ln x 1 1 1 x2 1 x2 x2 92. y ln 2x 3 t1 t y x x 4 2x2 x dy dx x2 1 x2 1 1 x2 1 ln x x2 ln x 9 ln t t ht 1 1 1 x2 2x2 1 9 x x2 2x x 1 2 2x fx 2 ln t t3 1 ln x 2 x 1 x2 x2 y ln 90. h t t4 dy dx 94. 88. f x 1 1 ln x2 2 9 11 2x 2 x2 9 y 1 x2 x x2 1 2x 2 1 1 ln x2 ln x 1 t2 1 t gt dy dx 2x2 x x2 ln x2 86. y 1 ln t t2 89. g t y x2 1 ln x2 2 ln x 1 2x 2 x2 1 87. f x 91. 1 The Chain Rule y ln csc x 1 csc x csc x cot x cot x 4 2 167 168 168 97. Chapter 3 y Differentiation cos x cos x 1 ln 98. ln cos x dy dx 99. y ln cos x sin x cos x sin x cos x 1 ln dy dx 1 tan x ln 2 cos x 1 sin x 3 cos x 1 sin x sin x 101. f x 2 x2 1 fx 6 x2 1 2 12x x 4 2x 72x2 12 104. f x fx sec2 12x 2 sec x sec x tan 2 2 sec4 x 2 sec2 x sec2 2 sec2 2 sec2 x 3 sec2 x 2 2x e x 2 sec2 2 tan 2 sec2 x tan2 2 tan2 x x fx 3 2x 3e 3x 7 6x e 7 6x 3e 3 6x t2 12 t 2 3 x 2 2x 2x sin x2 2 cos x2 3 x tan x x x 2e 2t 1 2t 3x x 3x gx 6e 3x ex ln x ex x 1 2x 1 4x3 gx 3x 5e 2t t t2 fx 2 3x st 2x cos x 2 2 3x st fx 2 106. g x 3 3 4 x x x 4 fx s2 2x 1 2 2 2 sin x 2 x 2 107. fx x tan sec2 fx x 103. f x 1 2 105. sin x cos x 1 sin2 x x 2 sec2 fx 1 2 sin x cos x 2 1 sin2 x sin2 x 12 x2 1 1 ln 1 2 sin2 x 1 2 1 24x3 5x2 x fx 2x2 60x 4 sec x 2 102. f x 12x5 ln 1 y dy dx sin x 3 fx sec x sec x tan x sec x tan x cos x sin x 2 tan x sec x tan x sec2 x sec x tan x sin x cos x 1 100. sin x ln sec x dy dx 1 1 sin x 2 sin x ln y ex ln x xex 2 1 2, 8 12 108. 2, 4 2t 2 y y 3x3 13 3x 5 4x 15 4x 9x2 4 5 3x3 4x 4 8 y2 1 2 ex x ex 2x 1 x2 1 4x x 8 ex x2 , 9x2 ex ln x 2, 2 45 5 ex ln x 4 2 cos x2 2x2 sin x2 S ection 3.4 3 fx 109. x3 3 x3 fx 3t t 111. f t , 3 5 1, 9x2 3x2 0, 13 x3 4 21 t 5 1 x 2x 2x f2 sec3 2x , 37 1 x y0 0 y 115. (a) f x 3x2 2, fx 12 3x 2 2 3x 3x2 3, 5 12 6x 5 2 2x 3 x2 3x 3 3 2, 3 31 2x x 12 5 2 3 2 2x 3 2 2 , sin x 2 cos x x2 3 x2 2 1 3 5 4 33 f2 1 3 3 x2 12 x 3 5 12 5 13 9 Tangent line: 9 x 5 (b) 1 , 3 2x 1 x x2 5 , 2, 2 3 112 x x 5 1 2 2x 32 fx Tangent line: y 1 16 4, 2 is undefined. 116. (a) f x 9 5 f3 1 x2 y 6 sec3 2x tan 2x 3x cos x, 3 sec2 2x 2 sec(2x tan 2x y , 5 114. y 0, 36 3 2 3x 169 5 32 fx 2 x2 2 3x 2 x2 112. f x 5 y x2 fx 2 2 3t 12 t f0 1 110. f x f4 4 2 , 1 t ft 113. 2 1 4 9 25 1 f 3 x3 4 The Chain Rule 3 ⇒ 9x 5y 2 7 0 y 2 (b) 13 x 9 2 ⇒ 13x 9y 8 0 6 (3, 5) −5 −9 5 9 −3 117. (a) f x −6 sin 2x, fx 2 cos 2x f y 118. (a) ,0 cos 3x, 2 y y Tangent line: y (b) ⇒ 2x 2x y 2 0 4 (π , 0) , 3 sin Tangent line: y 3 4 32 2 2 2 y 2 (b) 2 2 3 sin 3x 2 0 4 2 −2 −π 2 ( π4 , − 22 ( π 2 −2 32 x 2 32 x 2 4 32 8 2 2 170 170 Chapter 3 2 tan3 x, y 119. (a) 12 tan2 x, fx sec2 x 61 2 4 fx 120. (a) ,2 4 6 tan2 x y y Differentiation 2 tan x sec2 x f 4 21 2 4 ,1 4 Tangent line: Tangent line: y 2 12 x y 12x 4 y (b) 2 64. 3 4x (b) ( π , 2( 4 −π 2 1 − dy dx ln 1 x 2 1, 1 1 2x 2x 122. (a) y 0, 4 y 1 12 x 2, 2e1 2 2e1 x y1 1 x 0, x2 4xe1 2x 4 Tangent line: y 2 dy dx 1 . 2 (b) 4 1 x 2 1 x 2 2 4x 4x 1 6 7 0 y Tangent line: y (b) 0 1, 2 y 2x When x 1 −4 x2 4 y ( π , 1( 4 π 2 y ⇒ 4x 4 −1 121. (a) 4 3 4 −4 4 −1 8 (0, 4) −4 4 −4 4x 123. f x ln 4 4x fx 126. y y 124. g x x 6 127. g t 2 ln 6 6 2x x gx 2x x6 5 2x 2x ln 6 6 gt 2x ln 5 5 t ln 2 129. h h 2 2 sin 2 2t 2 t ln 2 cos ln 2 2 sin cos 2 32t t 2 t ln 2 g 5 2 5 22 t 2 ln 3 32t t2 32t 32t 2t ln 3 t2 ft 130. g cos 2 ln 5 5x 128. f t t 2 ln 2 2 t 2 tt 2 5x y dy dx x t 22 t t2t 1 125. 1 sin 2 cos 2 1 2 ln 5 5 2 sin 2 S ection 3.4 y log3 x dy dx 131. 1 x ln 3 132. y dy dx log10 2x log10 2 1 x ln 10 0 log10 x The Chain Rule 133. f x log2 1 x ln 10 x2 x 1 2 log2 x 2 x ln 2 fx 171 log2 x 1 1 1 ln 2 x x2 ln 2 x x 1 134. h x log3 xx 2 1 135. log3 x 1 x ln 3 hx 1 log3 x 2 1 2 2x 1 3x ln 3 2x x y log10 x2 10 log4 t t 137. g t 1 t 3 2 log2 log10 x x 1 ln 5 x2 t 1 10 1 t 2 ln 4 x2 1 ln 10 x x2 1 x t3 2 5 1 t 2 ln 2 ln t ln t 1 1 1 ln t 1 2 ln 2 3 12 t ln t 2 10 ln t ln 4 t 10 t 1 t ln t ln 4 t2 gt 1 x ln 10 1 1 t3 2 2 ln 2 t1 ft x2 1 1 1 2x ln 10 x2 1 138. f t 2x 1 ln 5 1 2 1 2x 1 ln 10 x2 1 2 1 log5 x2 2 1 0 x log10 x2 dy dx x2 log5 1 11 ln 3 x 136. dy dx log3 2 1 1 ln 3 x 1 y 139. y f′ 3 2 1 1 x 2 2 3 f 2 3 The zeros of f correspond to the points where the graph of f has horizontal tangents. 140. f y 141. y 3 142. y 4 3 f′ f f 2 2 3 2 1 −3 f′ −2 −1 x −1 1 2 3 x 3 1 2 −3 f is decreasing on ,1 so f must be negative there. f is increasing on 1, so f must be positive there. f′ The zeros of f correspond to the points where the graph of f has horizontal tangents. x −1 −2 −2 f 4 f′ −3 −4 The zeros of f correspond to the points where the graph of f has horizontal tangents. 172 Chapter 3 143. g x Differentiation f 3x 3 ⇒ g x gx 145. (a) f x (b) f x fx gxh x f5 3 g xhx 2 63 (d) f x gxh x 36 3 3 sin2 x gx 2 12 9 2 2 1⇒g x cos 2 x 2 sin x cos x g h 5 to find f 5 . gx 2 cos x sin x 0 0 3 2 3gx f5 4 3 (b) tan2 x 2 3 gx 6 1 162 sec2 x 1 gx 2g 3 3 fx 2 hx 146. (a) g x g3 Need g 3 hxg x f5 g hx h x f5 24 2x f x 2 ghx fx gx hx fx f x 2 2x ⇒ g x gx 3 f 3x gxhx (c) f x f x2 144. g x f 3x fx Taking derivatives of both sides, g x f x. Equivalently, f x 2 sec x sec x tan x and g x 2 tan x sec2 x, which are the same. gt 147. (a) t2 13 , 22 , 148. (a) f x x fx 3 (b) y 3 2 3x y 3x x 2, x2 f4 1 2 (b) y 1 3t t 2 3t 2 t 2 2t 1 3 2 gt g 3t 2 2t 2 5x 2x (c) 2 9 8 9x 4 y 1 2 4, 8 9x 28 12 3 (4, 8) (c) 5 0 (1 , 3 ( 22 −2 6 0 4 −2 4 st 149. (a) 2t 1 3 21 3 st s0 (c) t , 0, 2 31 4 3 5t y t 0x t 4 3 t 8t (b) y (c) 9 27 4 10 27 t 4 2 27 t 4 47 2 3 −2 (0, ( 4 3 2, 2 y 0 3 5 (2, − 10) 4 −1 9 2 2t y2 4 3 −2 t2 y 150. (a) 0 y (b) y t − 18 2, 10 S ection 3.4 fx 151. x2, 25 x fx fx x2 25 3 4 f3 f1 y x2 2 x fx 2 x2 2 3 x 4 y 3 x 4 25 , 4 1, 1 for x > 0 2 1 3 , 32 2x 1 y 4 y 152. 3, 4 The Chain Rule 2x 1, Tangent line 3 Tangent line 8 (1, 1) −2 (3, 4) −9 2 −1 9 −4 fx 153. sin 2x, 0 < x < 2 2 cos x fx sin x sin x 1 1 2 sin x 2 1 2x fx 0 2x 2x 1⇒x 1 ⇒x 2 x 2x 3 2 x 2x 5 , 66 Horizontal tangent at the points 6 , hx 1 3x 9 1 3, 1 3 3x 9 1 hx 2 3x h1 2 1, 3 64 9 156. 3x 1 fx fx 2 1 x 1 13 2 1 13 cos x 2 , 6 3x fx 1 3 x 4 fx sin x 2 2x fx 4x 2 cos x2 2x 2 sin tan 2t , gt 1 2, 0, 32 4 sec 2t 3 52 4x 2 sin x 2 6 , 52 6 32 3 3 sec 2t tan 2t 2 8 sec2 2t tan 2t 0 g 4 2 sec2 2t gt x2 2x cos 4 gt 2x sin x 2 x2 4 4 3 128 158. 0, 1 x 4 1 x 2 f0 f0 x 32 0⇒x 2 12 1 33 3 , 0 , and , 2 2 24 fx 1 1 x 2x 1 33 2 hx 157. 12 1 Horizontal tangent at 1, 1 35 ,, 626 Horizontal tangents at x 1 13 1 2x 0 sin x 155. 4 sin2 x 0 sin x 5 , 6 x 2x fx 2 cos 2x 2 sin x 2 sin2 x 2 sin x 154. 1 2 173 174 Chapter 3 159. (a) f Differentiation 132,400 331 f 1 v 1 132,400 331 160. y v 2 1 v 132,400 331 v 2 When v (b) f 1 3 y 1.461. 3 cos 12t 8, y When t 1 12 cos 12t 4 12 sin 12t 4 sin 12t 30, f 132,400 331 f 1 sin 12t 4 1 cos 12t 3 0.25 feet and v 4 feet per second. 1 v 1 132,400 331 v 2 1 132,400 331 v 2 When v 161. 30, f 1.016. 0.2 cos 8t 162. y The maximum angular displacement is 1 ≤ cos 8t ≤ 1 . d dt 0.2 When t second. 8 sin 8t 0.2 (since A cos t y 1.6 sin 8t 3, d dt 2 10 y (b) v y 1.75 S C R2 dS dt C 2R dR dt 10 4.224 2 10 2 10 5 0.04224. 64.18 22.15 sin t (b) 100 1. 6 0 22.15 cos 11.60 cos t 6 t 6 20 0 − 20 t 5 0 and 164. (a) Using a graphing utility, or by trial and error, you obtain a model of the form (c) T t sin t 5 dr dt 2r 105 2 1.2 1.76 Tt sin t 5 r2 Since r is constant, we have dr dt dS dt 5 1.75 cos 5 0.35 163. 1.75 1.75 cos t Period: 10 ⇒ 1.4489 radians per 1.6 sin 24 3.5 2 (a) Amplitude: A 13 13 0 1 1 6 (d) The temperature changes most rapidly when t 4.1 (April) and t 10.1 (October). The temperature changes most slowly T t 0 when t 1.1 (January) and t 7.1 (July). S ection 3.4 4 3 dr r, 3 dt 165. V dv dt 4 r2 2 768 166. (a) g x 3 3 cubic inches/second 2⇒g x fx fx 2f x ⇒ h x (b) h x (c) r x 3x ⇒ r x f r0 r 3f 0 1 (d) s x f Hence, you need to know f x 2 s 1 3, f0 3 3f 0 1 2 3 4 1 3 1 2 4 gx 3x 3x . 2 ⇒s x 1 2 3 4 2 3 1 3 1 2 4 hx 3x 3 2 8 4 3 2 3 2 4 8 fx 1 3 3 3f 3 fx x 2f x Hence, you need to know f 167. (a) 175 dr dt 48 The Chain Rule rx 1 4 sx 12 fx 12 1 3 1 1 2 4 2 2. etc. 168. V t 350 20,000 (a) 3 4 t V 20,000 16,000 0 100 0 (b) 12,000 Tp 34.96 p 3.955 p 8,000 4,000 in.2 T 10 4.75 deg lb T 70 0.97 deg lb in.2 t 2 4 8 6 3 4 V2 (b) 20,000 dV dt 20,000 ln 10 2 $11,250 3 4 3 4 t When t P 1.05 (a) C 10 dV dt 4315.23 When t 169. C t 1: 4: dV dt 1820.49 t 24.95 1.05 10 (b) dC dt P ln 1.05 1.05 $40.64 t dC (c) dt ln 1.05 P 1.05 t ln 1.05 C t When t dC 1: dt 0.051P When t 8: dC dt 0.072P The constant of proportionality is ln 1.05. 176 Chapter 3 170. f x Differentiation sin x 171. f x (a) f x cos x f f 4 4 sin x 3 x cos x f 2x , so g x (b) Yes, let g x Since f is periodic, so is g . 2 x f 2k x 1 k 2k 2k 1 x 1 k (c) f 2 sin x sin x 0 sin x 1 2k 1 cos x 173. (a) If f 172. (a) r x f gx g x r1 f g1 g 1 4 and f 4 5 6 0. Thus, r 1 Also, g 1 (b) s x 5 . 4 f x 1 x fx fx f x. Thus, f x is even. g f4 f 4 (b) If f 5 ,g 2 Note that f 4 d dx x f g fx f x s4 0 2 f x , then x d f dx 0. Note that g 1 5 2 6 6 4 2 15 24 5 . Thus, s 4 4 f4 2 f 2x . sin x 2f (b) f x f x for all x. f x , which shows that f is (a) Yes, f x p periodic as well. 2 fx p x f x , then d f dx f 5 . 8 x d fx dx x 1 fx f 1 and 2 x f x. Thus, f is odd. 174. u2 u d u dx 175. g x d dx uu u2 176. f x x2 fx 2x 179. (a) f x fx fx 12 u 2 u2 u , u u 12 u 2uu 177. h x x2 x2 4 2 4 , 4 f1 sec2 x 4 f1x P2 x f1 tan 1 x 24 x 1 1 3 , 3 3 2 x 178. f x x cos x, x x x sin x 1 fx 0 (b) sin x cos x sin x ,x sin x k 2 P1 x 4 sec2 2x 2x x cos x hx x 4 P1 x 8 ±2 x 2 3 0 4 tan gx 2x 1 2 2 2 f1 2 f1x x 1 P2 2 f x 44 f1 4 2 x 1 1 8 21 0 4 3 0 (c) P2 is a better approximation than P1. (d) The accuracy worsens as you move away from x c 1. 1 1 f1 S ection 3.4 180. (a) f x sec 2x (b) 2 sec 2x tan 2x fx 2 2 sec 2x tan 2x tan 2x 4 sec 2x tan2 2x f f 6 6 6 sec 2 sec 3 4 3x P2 x 1 56 x 2 fx e fx tan 0 (d) The accuracy worsens as you move away from x 6. 43 56 2 6 2 4 3x 6 2 4 3x 2 6 2 6 x2 2, f0 x2 2, f0 fx x2e x2 2 e P1 x 1 0x 0 P1 x x2 2 e x2 2 x2 1, 2 P 1 0 f0 f 3 1 P2 0x 0 1 x 2 P1 0 0 2 1 x2 2 1 P1 0 1 1, 0 (c) P2 is a better approximation than P1. , P2 0 1 (d) P1 and P2 become less accurate as you move further from x 1. 0 P2 x x, P2 0 P2 x 182. (a) (b) 1 −3 0, P2 x 0.78 0 sec3 2x 3 6 xe f (c) P2 is a better approximation than P1. 23 423 28 x P1 2 sec 2x sec2 2x 2 2 3 P1 x 181. (a) 6 P 2 fx f The Chain Rule 1, P2 0 fx x ln x, f1 0 fx 1 f1 fx P1 x f1 1 1 1 , x −2 (b) 3 f ln x, P 1 P 2 f1 1 P1 1 −2 4 0 −1 P2 x f1x f1 x P1 x 1 P2 x x, 1 x 1 1 f1x 2 1 x 2 1 2, 1, 1 (c) P2 is a better approximation than P1. 2 (d) P1 and P2 become less accurate as you move further from x 1. x 1 x, P2 1 0 P1 1 1, P2 x f1x 1 1 P2 1 1 P2 1 1 177 178 Chapter 3 Differentiation 183. False. If y 1 1 x1 y 21 1 2, x 2 sin2 2x, then 184. False. If f x fx 2 sin 2x 2 cos 2x . then 1. fx 186. a1 sin x fx a1 ... 2a2 a1 cos x f0 a1 nan fx x xk Pn x 1 xk n f0 0 1 n 1P n x 1 xk xk 1 Pn x xk xk 1 n 1 dn 1 dx n x k 1 xk 1 n 2 n d dn 1 dx dx n x k 1 1 x Pn 1 1 n For n 1, nan lim x→0 Pn x n 1 2n 1 kx k n 1 Pn x Pn 1 xk kx k d 1 dx x k 1 1 xk n 1k 2x 1 1P n ⇒ xk 1 1 2 P1 x xk 1 n n 36 1 Pn x 2 ⇒ P1 1 1. k nn! for n ≥ 0. Assume true for n. Then x2 y2 2 y1 1 y 2 2 9 1 2y xy 81 2yy 0 y 2x x y x x3 xy x k. Also, P0 1 2. 3x2 x y y2 12 12 y3 3y2y 8 0 x2 y2 y y x x2y 6. 4 2 xy y 2yy 0 2y xy y 3x2 y 2y 3x2 y x3 4. 0 y 3x2 1P n y x1 5. 1 kx k n 0 2yy 12 x 2 1 !. y 1 x 2 1 Implicit Differentiation y2 3. fx ≤1 sin x k nn! Section 3.5 x2 1 n kx k lim x→0 1 k Pn 1 k 1. sin x x 1 k Pn 1 n 1 fx sin x 2 We now use mathematical induction to verify that Pn 1 Pn nan cos nx f0 lim d dx ... an sin nx ... 2a2 cos 2x 2a2 x→0 187. ... a2 sin 2x 185. True x 2xy 2 y2x y 2yxy x2 2xy y y 3 0 y2 2xy y y 2x x x 2y S ection 3.5 xe y xe y dy dx 10x ey 3y 3 dy xe y dx x3y3 9. 3x3y2y dy dx 0 3 10 0 10 ey dy dx 7. 10 xe y ey 3 y 3x2y3 x y x y exy 2x y2 2y dy dx xy 0 yexy 1 xy 2 3x2y3 12 x 1 3x2y3 3x3y2 1 2 xy xy xy x y 0 2y 0 1 2y 0 4 xy y 0 2 xy y 2 xy 2 xy 4 xy y x3 2x2y 3xy2 38 4xy 6xyy 3y 2 0 11. 3x 2 2x 2 y 2x 3y xy y sin x 13. cos x 2 cos 2y 4 sin 2y y cos x y x1 17. x2 19. 2x 3 dy y dx 2y cot x cot y x cos y cos cot y csc2 y y y 18. x sec y cos xy 0 20. x 2 yy sin yy 2x 3y 2y 1 x 2xy 3 2y 2 y 1 y 1 1 csc2 y 1 cot 2 y tan2 y 1 y ln xy ln x 2y x y2 sec 1 y tan 1 y y 10 2x sin y y 1 1 sec tan y2 y y 1 y cos xy 1 x cos xy dy 3 dx y cos y 1 dy dx x x cos y cos xy dy dx sin tan y 1 y2 3 ln y x 2 sin 1 cos x cos y sin x sin y y 16. xy y 0 14. sin xy x cos xy y cos y cos x cos x 4 sin 2y tan y x sec2 y y y 4xy 2x 3y tan y y sin y y y x 1 3y 2 0 x sec2 y y cos x 2 sin x 3x 2 4xy 2 sin x cos y 3y2 1 y 15. sin x 12. 3x2 2x 2y 1 y y 2y 2x yexy xexy 2y 12 179 10 dy dx 10. 1 y dy dx x2 dy xy xe dx 0 1y exy 8. 0 1 3x3y2 Implicit Differentiation y2 cos 5x 30 ln y 5x 30 1 dy y dx 5 1 1 cot y y 0 1 dy y dx 1 x 5 dy dx y x 5y y 5xy x 2 0 0 cos sin x y 180 180 Chapter 3 21. (a) x2 y2 Differentiation 16 y2 16 y (b) ± x2 16 x2 12 16 22. (a) x2 ± 2 x x x 16 − x 2 2yy 0 x y y 6y 2 y 3 2 4 3 2 4 2 y2 = − (d) Implicitly: 2x y y2 4 6 x −6 y 4x 2 −2 2x x 16 x2 x 16 − x 2 2 −6 1 16 2 ± y1 = 6 (c) Explicitly: dy dx y x2 9 9 4 9 (b) y x (Circle) 1 2 3 4 −1 x 3± y 2 4 5 2 y = −3 + 4 − (x − 2) 2 −2 x 2 2 −3 −4 (c) Explicitly: −5 dy dx 1 ±4 2 x x 2 2 x 2x 4 x 2y 3 4 4 6y x y 2 2 3 24. (a) 4y 2 9 16 16 9x2 y2 x2 x2 16 y x2 y1 = 3 4 4 x2 ± x2 1 4 (b) y 6 6y 2 9x2 1 144 16 (b) 4 2 2 4 144 ± 4 − (x − 2)2 2 3 y y 2yy 2 3± y2 y = −3 − 2 2 x 23. (a) 16y2 2x (d) Implicitly: x x 12 2 4 ± 2 4 4 x2 2 4 6 16 − x2 −9 9 2 −6 −2 x 2 6 −6 −4 −6 (c) Explicitly: 3 y2 = − 4 dy dx (c) Explicitly: 16 − x2 ± 3 16 8 x2 32yy y dy dx 2x ±x 2 x2 4 (d) Implicitly: 3x 4 16 (d) Implicitly: 18x 12 x2 0 9x 16y 3x 44 3y 9x 16y 4y 2 x2 8yy 2x y x 4y 4 x 4y 0 2x x y 2 2 3 S ection 3.5 xy 25. y1 xy 3x2 0 y xy 4, At 1, 1 : y 3 x3 y3 3 3 x 2 xy2 y2 3xy x2 2 x 3 y 3x2 3y2y y2 y2 13 2 y 3 1 3y 9 2x x2 x2 yy xx 3 x y 13 3 13 y3 2xy 0 2xy 2y 0 2x y 31. tan x 1 2y x y sec2 x y 1 3x2 2y 3y2 y 3x2 1 y sec2 x y sec2 x y tan2 x y xy 1 2x tan2 1 x cos y cos y y :y 1 23 3exy 0 cos y x sin y 0 3exy 33. 1 1 cot y x 3 y x 1 At 0, 0 : y At 2, 18x 9 2y 0 x2 y sin y 9 2x 2 5 x2 x 9 2x 2y y 32. x2 1 2 At 8, 1 : y 2xy 3y2 At 1, 1 : y 3 y 2xy y 1, 1 : y x3 x2 29. y 30. y 0 y y At 3 2 0 3xy 3x2 2xy 2yy 0 2xyy 3x y 3x2 2y 9 9 x2 0 3 y 2 x2 x2 y2 27. 181 At 3, 0 : y is undefined. x2y 2 0 1 4 1: y x x 2y 2yy y x 28. x3 y2 y y At x3 26. 4 Implicit Differentiation xy x 0, y 1 0 3exyxy y cot y x At 3, 0 : y 1 9 3, 0 1 3yexy 1 3yexy 3xexy 1 sin2 x y 182 182 Chapter 3 y2 34. Differentiation ln x, x2 1 x 2yy x2 35. e, 1 4y 4y y 2x At e, 1 : 0 y 1 2xy y 8 x2 2x 8 x2 4 x2 4 1 2e y 2xy 4 x2 32 64 At 2, 1 : y 16x 4 2 1 2 Or, you could just solve for y: y x y2 3x2 y 4 x3 1 4 36. 3x2 2y 4 y2 x 2yy At 2, 2 : y x2 37. 2 x2 4x3 y2 x y2 2x 4x2yy 2 4x2y 4x2y y 8x 4y3y 4x2y 8xy 4x2y 4y3y 4y x2y 2 8xy y3 x2 4 2xy 2xy x2y y At 1, 1 : y x3 38. 3x2 6xy 0 6xy 3y2y y3 6y 0 2 y 3y 6x 48 , :y 33 x 40. 2x 1 2 1 y 2 2y 2y 2y 2 20, 32 40 3, 4 2y y2 x2 2x 2x x 1 1 2 2 Tangent line: y 4 2x 3 y 2x 10 4, 0 4 2 y 2 Tangent line: y 0 1x y 4 5 1 x 41. xy 1, y 4 4 0 y 1, 1 y x At 1, 1 : y Tangent line: y At 3, 4 : y 3, At 4, 0 : y xy y 4x y 3x2 6x 0 y 2 2y 3x 16 9 83 2y 2 2y 6y 3y2 16 3 64 9 y 0 2 6y y At 39. 4 2yy 4xy2 4x2yy y2 8 x2 1 1 1x y x 1 2 4x3 x3 x3 y3 4xy2 xy2 xy2 x2 S ection 3.5 7x 2 42. 14x 13y 2 6 3 xy 6 3 xy 16 6 3y 0, 26yy 6 3 y 14x 26y 6 3 x 6 3 14 3 26 6 3 3 3, 1 : y Tangent line: y 1 x2y 2 43. x 22yy 3x 9 x2 2xy2 4 4y 2 18x 0, 8 yy 24 48 3 Tangent line: y 3 x2 45. 6 x2 y2 y2 2x 2 960 480y 880y y 11y 8 3 x2 2 x 8 1 x 2 4, 2 5 y2 x2 46. 2x y2 8 yy 4 y 2xy 2 2y 4 2 4y 4 2 1 3 Tangent line: y 0 2 11 x (b) x2 a2 y2 b2 y0 y0 y b2 2x Since 1 y 2x 2y 8 1 3x x 1 2 3 2yy b2 b2x0 x a2y0 x0 , x02 a2 x0 x a2 y02 b2 1⇒ 1 y 4 1 x 2 1⇒y 2x 4, b2x a2y 0⇒y tangent line at x0, y0 x02 a2 1, you have 4 1x 2 1 3 2x a2 1⇒ y02 b2 2 2 1 30 11 4x y Note: From part (a), 1, 1 4x y x 0 Tangent line: y 2x 2 4y3y y 2 11 1, 2 At 1, 2 : y y4 6y 30 2x 2, At 1, 1 : y 1, y2 y 2x 2 400y 2 13 3 4y y 47. (a) 1 x 2 2 11 Tangent line: y 1 160 y y x 13 Tangent line: y 2yy 800 13 x y y2 , 100 8 0 4 100 2x 4y 8, 1 1 2 2yy x 2 48 1 3y 5, At 8, 1 : y At 4, 2 : 6 16 3 y 3 6 100 x 2 2yy y2 2 y 3 13 3 8y 1 23 3 x 6 y 2xy2 4 12 16 3 3 x 6 23 2 x 3 0 18x 2x2y x2 44. 4, 2 3 18 4 2 2 16 2 3 4, 2 3 : y 3 3 y At 83 8 3x y 3, 1 0 y At Implicit Differentiation tangent line. y0 y b2 x0 x a2 1. 183 184 184 Chapter 3 x2 6 48. (a) Differentiation y2 8 x 3 3, 2 (b) x2 a2 y2 b2 1⇒ y 1, y0 x0 b2 x y0a2 y y 4 0 y y 4 x 3 yy0 b2 y02 b2 x0 x a2 4x 3y Since x02 a2 y02 b2 2x 4, y At 3, 43 32 2: y Tangent line: y 2 tan y 2x 2y 8 1⇒ 1 x 2 y 4 1⇒y sec2 y sin y cos2 y, tan2 y 2 <y< x2 2x y2 x2 1 sin2 y x2 1 , sin y cos2 y 0 x y y y y y 1 xy y2 y2 x2 y3 x xy y2 x2y2 52. 2x2yy 2x 2xy2 2 0 xy2 1 0 y x2 y 2 4xy2 x 4xy2 4x2y 4 1 2xy2 1 1 xy2 xy y2 0 x2yy y2 0 x2yy 2 xy2 x2y 2xyy x2 y 4xyy 4 x2yy 36 y3 3 x2yy 2xyy 1. 1 y2 0 x 2y 4 0 2 22 x2y 4 x 4y 3y x 4y 3y y 2x2y 4 2x2y 4 2xy2 1 2xy2 1 x 4y3 0<y< 1 sin y 36 2yy yy0 b2 1 y 2 1 1 x0 x a2 x y y 51. 1, you have sin2 y 1 y x02 a2 tangent line. cos y 50. 1 sec2 y tangent line at x0, y0 4 1 y x0 , xb2 ya2 0⇒y 3 x y sec2 y 2x 3x 6 Note: From part (a), 2yy b2 2 y 49. 2x a2 cos2 y cos2 y 1 1 , 1<x<1 1 1 x2 x2 S ection 3.5 x2 53. 2x y2 16 54. 1 xy x xy x 1 y x 1 1 x 2yy 0 y x y yy 0 y y 1 yy 1 yy 2 y x y y2 2 y3y y 56. y2 3x2 2y x3 y2 2x 3 x2 16 y3 3 y 3y 4x2 57. x 1 x 2 12 1 y 2 y 4 1 2y 2 2y 2y y2 2 y x x2 1 3 At 9, 1 : y x2 At 2, 1 1 1 (2, ) 5 5 11 x2 x 12 1 2x2 x2 1 2 1 2x −1 2x −1 1 2x x2 2y x2 1 2 5 :y 5 1 4 2 5 54 y Tangent line: 10 5y x 5 10 5y 1 10 5 4 1 2 5 5 1 x 10 5 10 x 8 0 2 2 14 −1 x 4 y3 −1 y y x2 y 3x 4y (9, 1) y x 2yy 6y 9 3y 1 1 x 3 y y2 3y 2x 4x2 0 Tangent line: 58. 3y 2 4x2 x2 y2 xy xy 3y 2x 2x 3y y 2 y y 3x2 2y 0 4 y y 1 3y 2x 4x 2yy 3x2 0 x 185 x3 2yy 0 0 y2 55. y y Implicit Differentiation 1 x 3 12 0 9 4 186 186 Chapter 3 x2 59. 2x y2 Differentiation 25 2yy 0 x y y At 4, 3 : 6 Tangent line: y Normal line: y 4 x 3 3 3 3 x 4 (4, 3) 4 ⇒ 4x 4 ⇒ 3x 3y 4y 25 0 −9 0 −6 3, 4 : At 6 Tangent line: y 4 60. x2 y2 (−3, 4) 3 x 4 4 Normal line: y 3 ⇒ 3x 4 x 3 4y 3 ⇒ 4x 3y 25 0 −9 0 −6 x 4 y (0, 3) At 0, 3 : −6 Tangent line: y 6 3 Normal line: x 0 At 2, −4 5: 4 Tangent line: y 5 2 x 5 Normal line: y 5 5 x 2 x2 2x 9 9 y 61. 9 y2 2yy 5y 9 2⇒ 2y 0 5x r2 slope of tangent line slope of normal line Let x0, y0 be a point on the circle. If x0 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes through the origin. If x0 0, then the equation of the normal line is y0 y0 x x0 y y −6 y0 x x0 6 −4 y2 4x 2yy x y 0 62. 0 y y x (2, 5 ) 2 ⇒ 2x 4 y 2 y 1 at 1, 2 Equation of normal line at 1, 2 is y 2 1x 1, y 3 x. The centers of the circles must be on the normal line and at a distance of 4 units from 1, 2 . Therefore, 1 2 3 x 2x 2 2 16 1 2 16 x x 1 ± 2 2. x0 which passes through the origin. Centers of the circles: 1 1 2 2, 2 2 2 2 2, 2 Equations: x 22 2 22 2 x 1 1 y y 2 2 and 2 2 22 2 16 22 2 16 S ection 3.5 63. 25x2 16y2 200x 50x 160y 32yy 400 200 0 160y 0 200 160 y Horizontal tangents occur when x 16y2 25 16 200 4 160y 4, 0 , 400 200x y2 8x 4y 6 (− 8, 5) 400 0 10 0⇒y (0, 5) 4 0, 10 (− 4, 0) x −10 − 8 − 6 − 4 400 25x x 64. 4x2 10 2 −2 5: 800 8 Vertical tangents: 0, 5 , y (− 4, 10) 4, 10 Vertical tangents occur when y 25x2 50x 32y 4: yy Horizontal tangents: Implicit Differentiation 0 0⇒x 0, 8 8, 5 4 y 0 (1, 0) 8x 2yy 8 4y 0 −1 1 x 2 3 4 −1 8 8x 2y 4 y 4 y Horizontal tangents occur when x 41 2 y2 81 4y Horizontal tangents: 1, 0 , 1, 2 2 8x y 65. ln y 1 dy y dx dy dx x x2 1 x y 4 8x 2x2 x x2 4 0 4x x 2 0⇒x 0, 2 y ln y 1 1 dy y dx 1 1 1 0, 66. x x2 0⇒y 2 1 1 ln x2 2 ln x 4 2: 2 2 , 2, 2x2 x2 (1, − 4) −5 4 4x2 Vertical tangents: 0, 4 −4 yy Vertical tangents occur when y 4x2 −3 0 4y (2, − 2) (0, − 2) 1: 4 y2 4x 2 x 1x 2x 3 1 ln x 2 1 ln x 2 1 1 2x 1 1 2 1 1 dy dx 3x2 2 1 x 2 3x x 1x 12x 2y ln x 1 2 x 3 12x 11 2x 3 11 3x2 12x 11 x 1x 2x 3 3 187 188 188 Chapter 3 y 67. ln y x2 3x 2 x 12 dy dx 1 2 3 2 3x 2 3x2 y 2x 3x 2 ln x x 15x 2x 1 8 1 x x2 2 2 1 ln x 2 2 1 y y 2 y ln y 68. 1 ln 3x 2 2 ln x 2 x 1 dy y dx Differentiation 1 1 2x 2 1 2 y ln x2 2 2x x2 x x2 2 2x 2 1 2x 2 x2 2 3x3 15x2 8x 2 x 1 3 3x 2 y 69. ln y 1 dy y dx xx x 70. 1 3 ln x 2 ln x 1 x dy dx 32 1 3 1 2x 1 y2 2x ln y 1 1 dy y dx 1 1 2x 1 3 1 x 4x x x2 2 2x 1 x 13 1 x 1x 1x ln x 1 y ln y 1 dy y dx 1 x2 ln x 2y 1 x2 ln x x 2x 1x 1x ln y x 1 ln x y x x 1 x x x x 1 ln x ln x 1 dy dx ln x x x 2 y 1 dy y dx 2 1 2 dy dx ln x 2 y xx ln y 2 1 2 2 2 2 1 1 2x x2 2 1 ln x x 1 x 1 x ln x 1 ln x x 2 x 1 x 1 x ln x 1x 1 ln 1 x x 1 1 x1 x y 1 xx 1 1 x x ln 1 ln x x 1 x2 1 x 1x 1 x ln x 1 ln x 2 2 6x2 12 1 x2 4 6 x2 2 1x 1x 2x x 6 x2 2 12x 2 xx 74. 2 dy dx 2 2 x y 4 1 1 1 4 y 1 dy y dx ln x 1 x x 2 1 x2 x x2 ln y 2 x2 ln x 1 2 1 2 2 ln x x y 1 dy y dx x 2 y 2 1 x x 72. x 21 xx dy dx 73. x2 ln x 1 x 71. 2 2 1 1 2x2 x x x dy dx 1 x 4x2 y 2 1 ln x 2 1 y 1 x 2 S ection 3.5 75. Find the points of intersection by letting y2 2x2 4x and 6 x 3x 4x in the equation 2x2 1 y2 4x 6. y 2 = 4x 2x 2 + y 2 = 6 4 Parabola: 2yy 2yy 6 2 y 2x y y (1, 2) −6 4 y 0 189 0 The curves intersect at 1, ± 2 . Ellipse: Implicit Differentiation (1, − 2) −4 At 1, 2 , the slopes are: y y 1 At 1, 1 2 , the slopes are: y y 1 1 Tangents are perpendicular. 76. Find the points of intersection by letting y2 2x2 3x3 5 Intersect when x 3x3 and 2x2 x3 in the equation 2x2 5 0 5. 2 2x2 + 3y2 = 5 1. (1, 1) Points of intersection: 1, ± 1 y2 3y2 x3: −2 4 (1, − 1) 2x2 3y2 5: 6yy 0 −2 2yy 3x2 y 3x2 2y 4x y y 2= x 3 2x 3y At 1, 1 , the slopes are: y At 1, 3 2 2 3 y 1 , the slopes are: 3 2 y 2 3 y Tangents are perpendicular. 77. y x and x sin y 78. Rewriting each equation and differentiating: Point of intersection: 0, 0 y x x: x3 sin y: y 1 sec y y 1 y cos y y y 1 y 3y 1 x3 x 3y 1 3 29 y x2 y 3 13 3x 1 x2 At 0, 0 , the slopes are: y 1 Tangents are perpendicular. For each value of x, the derivatives are negative reciprocals of each other. Thus, the tangent lines are orthogonal at both points of intersection. x (3y − 29) = 3 15 4 x 3 = 3y − 3 x = sin y −6 6 (0, 0) −15 −4 x+y=0 12 −3 29 190 190 Chapter 3 xy x2 C y 79. Differentiation 0 xy y2 2x K 2yy C=4 x y −3 y x y 2 0 y 2 −3 3 C=1 3 K = −1 K=2 −2 −2 At any point of intersection x, y , the product of the yx xy 1. The curves are orthogonal. slopes is x2 80. 2x y2 2yy C2 y 0 Kx y 2 K −3 x y y 2 K=1 K = −1 C=1 −3 3 3 C=2 −2 −2 At the point of intersection x, y , the product of the slopes is xy K x Kx K 1. The curves are orthogonal. 81. 2y2 3x 4 82. x2 0 12x3 12x3 4y 83. cos y (a) dy dt 3x3 (b) 2x dy dt 3y2 2x 3y2 3y2 2x 6xy dx dt 84. 4 sin x cos y cos x (a) 4 sin x 0 dx dt dy dt 6xy 3y2 3y2 dx dt dy dt 3 cos x (b) 4 sin x 0 dx dt sin y y 4 cos x cos y y 0 sin y dy dt 6xy 3y2 dy dt 1 3 cos x sin y cos x 3 3y 2 2x y sin y 3y2 y dx dt 1 3 0 y 0 x 3y2y 3x3 y dx dt 3 sin 6xyy 6xy 12x3 sin y y (b) 10 3y2 (a) 2x y (b) 4y y3 12x3 y dy dt 0 4yy (a) 4yy 3xy2 sin y dy dt 4 cos x dx dt 0 cos x cos y sin x sin y dx cos y dt cos x cos y dx dt 0 sin x sin y dy dt 85. A function is in explicit form if y is written as a function of x: y f x . For example, y x3. An implicit equation is not in the form y f x . For example, x 2 y 2 5. 86. Given an implicit equation, first differentiate both sides with respect to x. Collect all terms involving y on the left, and all other terms to the right. Factor out y on the left side. Finally, divide both sides by the left-hand factor that does not contain y . 87. 88. Highest wind speed near L 18 00 1671 B 1994 A 18 Use starting point B. 00 S ection 3.5 89. (a) x 4 4 4x2 16x2 x4 4x2 4y2 y2 14 x 4 10 − 10 y2 − 10 4x2 3⇒9 y (b) 10 14 x 4 4x 2 ± y 14 x 4 10 − 10 36 y3 y4 y2 x4 16x2 16x2 36 10 y1 0 − 10 x2 8± Note that x2 1 8±2 7 28 7, 1 7, 1 To find the slope, 2yy For x 7, y 7x 1 y1 1 3 For x 1 y2 1 3 7 7 For x y3 1 3 7x 1 7 1 3 7 7 1 3 7 7 7 87 7x 23 . 7 23 8 7. 1 3 3 7 7x 23 8 7. 7 7x 87 1 7 3 7 77 7 , and the line is 1 1 7x 7 , and the line is 1 7x 7 1 3 3 7 1 3 7, y 1 3 3 7 , and the line is 1 7x x 8 x2 . 23 7 , and the line is 7 7 7, y 1 y4 1 3 7, y 7 7 1 3 3 23 . (c) Equating y3 and y4: 1 3 7 7x 7 7x 7 3 7 7x 7x 1 7 7 7x 1 7x 7 77 7x 7 7 7 16 7 x If x 1 3 87 , then y 7 16 ± 7 2. Hence, there are four values of x: 7 x3 ⇒ y 1 3 1 1± 7, 1 8x For x Implicit Differentiation 7 7x 14x 87 7 5 and the lines intersect at 87 ,5 . 7 x4 256 2 144 8± 28 191 192 192 Chapter 3 90. Differentiation x y c dy 2 y dx 1 1 2x 91. yq 0 qy q px p y0 y0 x0 x 1 p q x0 x0 y0, 0 xp yq p q y Tangent line at x0, y0 : y y-intercept: 0, y0 xp 1y y x dy dx x-intercept: x0 x p q; p, q integers and q > 0 y xp 1 p x xp Thus, if y p q 1 x n, n y4 1 x p 1y yq pp x q q q 1 nx n x2 p q, then y 1 . x0 y0 Sum of intercepts: x0 x0 y0 y0 x0 y0 x0 2 x0 y0 x0 c x2 92. y2 2x 3 ⇒y 4 1, x2 1± 0 4 36 ⇒ 4y 2 Points on ellipse: 3:y 3:y Tangent lines: y y 1, ± 3 2 36 9x 2. Hence, 3 9x 4y 9 4 32 3 3 2 3 2 3 x 2 3 x 2 1 1 2 4x2 If you graph these four equations, you will see that these are horizontal tangents at 0, ± 1 , but not at 0, 0 . 3, 4 4y 2 4y 2 0 0 4, 0 y x 4 3 2 y2 9x 4y 9x 4y At 1, 0⇒x 4y3 0 y 3 2 2x 2y y2 x2 4 2yy 9 4y3 y Note: y 4 4 and At 1, 2y ±3 y2 9 2x Horizontal tangents at 0, 1 and 0, Points: 3, But, 9x 2 2yy y 25 x 9x x y2 2x 4 x 3 25 25 2 x 9 2x 4 93. 4y3y x y 16 2 x 9 94. c 0 y x2 2 2 3 4 25, slope 2yy y0 y0 3 x 2 4 4 3 x 2 23 23 9x 2 36x 4y 2 36 9x 2 ⇒ x 1. S ection 3.6 Derivatives of Inverse Functions y2 95. x 1 2yy y 1 , 2y slope of tangent line y 2, y on the parabola. Consider the slope of the normal line joining x0, 0 and x, y y y2 2y y2 0 x0 1 2 x0 x0 1 2 (a) If x0 1 4, then y 2 (b) If x0 1 2, then y2 0⇒y 1, then y2 1 2 y2 (c) If x0 1 4 1 2 1 4, which is impossible. Thus, the only normal line is the x-axis y 0. Same as part (a). x and there are three normal lines: 11 1 , and the line joining x0, 0 and , , 2 2 2 The x-axis, the line joining x0, 0 and If two normals are perpendicular, then their slopes are 2y y y2 1 0 ⇒y x0 1 2 x2 32 y2 8 2x 32 x 3 4 and y x 1 4 x0 1 2 2x 4 4x 2 24x 36 32 96x 112 0 17x Slope of normal line is 2. y 2 2x y 2x (b) Second point: 4 6 4 −6 6 −4 Section 3.6 Derivatives of Inverse Functions fx f 1 x3 2x fx 1. 3x2 2 2 1 ff 1 2 1, f 1 1 f1 3 . 4 x2 32 x2 x 4y 4 42 1 ⇒ x0 2 3 4. (c) 0⇒y At 4, 2 : y 1⇒ 17x 2 1 2yy 8 1 2 1 and 1. Thus, 12 14 x0 and The perpendicular normal lines are y 96. (a) 0. 2 a 1 31 2 2 1 5 6 2 8 28 x 4 28 , 17 46 17 1 0⇒x 4, 28 17 193 194 Chapter 3 Differentiation fx 15 x 27 2x3 , f fx 1 5x4 27 6x2 2. 1 f 1 11 1 ff fx sin x, f fx 1 2 ff 1 4. 54 11 27 3 5 3 4 6 3 2 a 1 17 a 1 12 1 f 6 cos 1 32 6 23 3 2 sin 2x 1 1 1 ff x3 fx 3x2 1 a 1 f0 6 1 2 sin 0 1 which is undefined. 0 a 6. fx 1 ff fx 6 11 , 28 x3, fx 1 f2 1 32 2 4 22 1 3 x, 3 3 x 1 1 8 1 11 , 82 f 4x, 1, 1 128 1 1 3 x x 4 , 1, 1 1 4 1 4 x 1 4 3 x fx 1 2x f5 x x2 f 1 x 2x f 1 1 2 4, 5, 1 fx 10. 1 2 1 1 x2 fx f 3 1 f8 2 4 1 f a 4 f 1 x 1 fx 3 4 2 4 f1 1 f 2 f ff fx 8. 4, f 8 1 2 3x2 7. 9. 1 2x 1 f 1 13 x fx 4 x2 1 6 f 1 1 4 ,f2 x fx f 1 2 cos 2x, f 0 1 f f 1 1 fx 5. 1 11 6 fx f 243 cos x 3. f 1 27 3 fx 4 1 f 4, 1, 5 f 1 4 1 x2 , x ≥ 0, 1, 2 x2 8x ,f 1 2 12 4 x x x 2 x x2 , 2, 1 ,f 4 x x 1 2 1 2 4 1 14 4 S ection 3.6 fx 11. (a) arcsin 2x 1 2 4 f y 2 2 2 2x 1 2 1 y (b) f 2 4 2 2x 1 1 2 1 x 2 4 (b) 1.25 ( ( 3 −2 1 fx y x4 1 f0 14. (a) fx 2x 2x 1 f 0x 2 1 2 0 , tangent line (b) −1.5 , 1, 4 π 2, 4 ) −4 1.5 2 4 ( (( 2 2 2 x 2 2 π 0, 2 1 2 x 2 4 y (b) 1 x2 x 2 y y arcsec x fx x4 0 2 tangent line ( arccos x 2 fx , ) − 1.0 13. (a) 4 −3 0.8 2π , 44 1 2 2 − 1, − π 4 − 0.6 1 1 x 2 y tangent line x2 1 1 y 1, 4 arctan x fx 4x2 22 12 2 2x 4 fx 12. (a) 1 fx Derivatives of Inverse Functions 5 0 −2 15. x y3 1 3y2 7y2 dy dx dy dx 3y2 At 4, 1 , dy dx 1 14y dy dx 1 3 14 1 . 11 x3 Then f x Hence, dy dx 7x2 2 3x 1 11 2. 14x and f 1 1 . 11 2 y2 11. 3 1 y2 3 2y dy dx 3 4y At 0, 4 , Alternate Solution: Let f x 2 ln y2 dy dx 14y x 1 16. 2 dy dx 16 3 16 13 . 16 tangent line 195 196 Chapter 3 Differentiation x arctan x x ey arctan x 17. ey 1 1 x2 At 1, ln 1 2 : 4 dy dx 1 1 dy 4 dx 4 dy dx 2 arctan 2x 3 arcsin xy 18. xy 2 dy dx x 1 1 y 342 1 2 3 21 y 32 At 1 ,1 : 2 1 2 2 3 1 4x2 y 1 2 3 2 3 3 y 19. f x 2 arcsin x 2 x 1 fx 21. g x 1 fx 25. g x gx arctan 24. f x a a2 fx x2 1 1 9x2 1 x 1 1 x x 1 2x hx 2x arctan x 1 2 x1 x x2 arctan x ex arcsin x x2 1 1 1 2 arccos x gx x2 1 x2 ex 1 1 ex arcsin x x2 2 x2 arccos x 1 x2 arccot 6x 30. f x 6 36x2 fx 1 arctan 1 9x2 arcsin 3x 1 9x2 1 x 1 x 3 9x2 3x 28. g x arcsin 3x 1 gx arcsec 3x 26. h x 9x2 x2 1 1 x 2t 1 t4 x2 arccos x x1 x hx 4 arcsin 3x x x3 ft fx x a x2 29. h x 3 1a x2 a2 1 x2 arcsin t 2 22. f x 31 2 1 x2 4 3x 27. g x 2 2x 2 x 2 3 arccos gx 23. f x 20. f t 1 12 arcsec 2x 2x 2 4x2 1 x 1 4x2 1 23 3 6 S ection 3.6 31. h t sin arccos t 1 1 2 ht t 2 t2 1 12 32. f x 2t Derivatives of Inverse Functions arcsin x fx arccos x 197 2 0 t t2 1 33. y y x arccos x x2 1 34. y x arccos x 1 1 2 x2 1 x2 12 2x 2 ln t2 2 y t2 arccos x 4 y 11 x ln 22 x 1 1 dy dx 1 1 4x 1 x y 1 x4 2 x2 x2 1 2 40. y dy dx x 12 x2 1 ln x 1 2 arccsc x x2 2 1 1 x2 1 t2 1 2 4 t t t2 1 4 x t2 1 t2 1 gt 2 39. 2 y dy dx 1 ln 1 4 x arctan 2x 2x 4x2 25 arcsin 5 2 4 t tan arcsin t t2 1 32 4x2 41. y 1 8x 4 1 4x2 arctan 2x arctan 2x y 1 x 8 arcsin x 5 1 1 x2 25 25 x2 2 43. y x2 x 25 25 25 x2 25 x2 x2 x 1 25 2 x2 25 x2 x2 12 2x 2x2 25 x2 y 1 x4 x 16 x2 2 16 x2 1 1 16 2 2 arctan x x2 x 16 4 x2 2 16 x2 x2 16 x2 x x2 1 x2 x 2x 1 x2 2 1 x2 x2 1 x2 16 2 2 x2 1 1 22 x arcsin x x2 x 16 2 1 2 1 x 4 x arcsin x x2 1 x2 1 x arcsin x 16 y 4t t2 4 1 arctan x 2 8 16 42. y 2 x4 0 1 t2 4 37. g t 4 x2 12 t2 1 1 1 x 2 2x 4 arcsec x fx 2t t 2 x2 4 38. f (x 1 12 2 4 1 4 arcsin x2 1 ln x 4 arctan x 1 11 x4 22 36. y arctan 4t t2 35. 4 x2 2 1 x2 2 x2 x2 12 2x 198 Chapter 3 44. y y Differentiation arctan x 2 1 1 21 1 x2 2 x2 12 x 2 2 2 x2 45. y 4 2 4 2x 4 2 y x x2 4 At 2 2x2 8 x x2 4 2 1 , 23 2 arcsin x, x2 1 1 2 Tangent line: y 14 4 . 3 3 1 , ,y 23 4 x 3 1 2 y y 46. y y 1 arccos x, 2 23 , 28 1 y 2 21 x 23 , ,y 28 At 1 2 12 3 8 Tangent line: y y At 4 16x 2 24 y y At 1 1 1 , ,y 24 3 2 Tangent line: y for x > 0 1 4 y 2 4 1 . 4 4 1 x 4 4 1 x 4 4x arccos x 4x 1 1, 1 x 1 2 1 2 4 1, 2 4 arccos x 2 1 4 3 3 3 3 6 2 2 x x 4 2 2 4x 2 4x 2 1 2 3 2 . 1 2. y 2 4 2 2x 1 34 y x2 Tangent line: y 2 2. 3 arcsin x x2 ,y 2 4 y 1 , 24 3x arcsin x, 3x 2 2 2x 4 4 23 3 3 4 1 1 x2 4 2 1 2 3 3 At 1, 2 , y 1 2 , ,y 44 43 x 3 2, 1 2 49. y 1 x , 2 Tangent line: y 3 8 1 x 16x 2 1 Tangent line: y 50. y 2 2 2 x 2 arctan At 2, 2 , 44 arcsec 4x , 4x 2 . 2 2 x 2 y 48. y 47. y 4 x 3 1 4 S ection 3.6 51. f x arccos x 1 fx 3 2, f 32 y Tangent lines: y 5 6 1 fx x2 1 1 x 2 4 1 x 2 3 2, f ⇒y 3 2 2x y 53. 3 2 2x arctan x, g x Tangent line: 6. When x 6 y 3 . 2 ± 2 when x x2 1 When x 52. g x Derivatives of Inverse Functions 2x ⇒y 2x 32 6 5 6 5 6. 3 3 1 2 ,g 1 1 1 2 4 1 2 arcsin x, a y 1.5 1 fx 0.5 x2 1 P1 1.0 x fx x x2 1 P1 x P2 x 54. 1 f 2 f fx fx fx 32 − 1.0 f 1 f 2 1 2 1 2 f arctan x, a x 1 2 6 23 x 3 x 1 2 1 f 2 1 2 6 23 x 3 1 2 23 x 9 −4 x −2 2 2 P ( x) 2 P2 x f1 f1x arctan x, a 1 x 2 1 1 f1x 2 1 1 4 2 4 1 x 2 1 1 x 4 1 2 0 y 1.5 0 1.0 1 1 f π 4 2x x2 1 2 P1 (x) x2 1 1 2 π 2 1 P1 = P 2 f 0.5 , x2 f0 1 x − 1.0 1 2x ,f0 x2 2 P1 x f0 f 0x x P2 x f0 f 0x 1 f 0 x2 2 fx 2 1 f1x fx 1 2 x − 1.5 y f1 f0 1 2 1 P1 x 55. f x 0.5 1.0 1.5 P2 0.5 1.0 1.5 − 1.0 0 − 1.5 x 199 200 56. Chapter 3 fx Differentiation arccos x, a 0 y f3 f0 2 2 1 fx x x x2 1 , f0 , f0 f 0x P2 x f0 f 0x 2x 1 1 f 0 x2 2 y y2 1 y2 arctan y 2x 2 1 y2 2 2 , : 2 2 2 y 16 2 2 , 2 2 x2 Not continuous at x 1 1 1 x At 1, 0 : y 2 1 1 1 2 y2 y y 4 , 1, 0 2yy 0⇒y y 0 1x y x 1 1 1 2 0 2 arctan x 2 2 1x 2n y x Tangent line: y 1 f 60. 1 61. f is not one-to-one because many different x-values yield the same y-value. Example: f 0 1 2 1 y y ⇒y 1 2 4 1 , 1 2 2 Tangent line: y y 2 0 y y 1 x 1 4 x 8 1 At xy 0, 0 2 2 y2 1 y y, 8 2 x 8 arcsin y 1 2 Tangent line: y 2 2 1 x2 xy arcsin x arctan y x 1 y2 4 4 1 1 1 arctan xy 1 2x 2 ,1 :y arcsin x 58. ,1 y y y 59. 4 At 0, 0 : 0 x Tangent line: y x 2 1, 1 4 0 1 x 1 y y At 1 x 2 x arctan y arctan y x −1 −1 32 f0 x2 1 −2 P1 x 57. P 1 = P2 1 1 fx 62. f is not one-to-one because different x-values yield the same y-value. Example: f 3 , where n is an integer. f Not continuous at ± 2. 4 3 3 5 S ection 3.6 63. Theorem 3.17: Let f be a function that is differentiable on an interval I. If f has an inverse function g, then g is differentiable at any x for which f g x 0. Moreover, 1 , f gx f gx gx (b) 0. d dt x 5 15 x5 1 x 3 66. (a) cot arccot x 3 arccot dx dt 2 201 64. The derivatives are algebraic. See Theorem 3.18. x 5 65. (a) cot Derivatives of Inverse Functions 5 dx 25 dt x2 If dx dt 400 and x 10, If dx dt 400 and x 3, d dt (b) 16 rad hr. d dt d dt 3 dx 9 dt x2 If x If x 58.824 rad hr. 10, 3, d dt d dt 11.001 rad hr. 66.667 rad hr. A lower altitude results in a greater rate of change of . ht 16t 2 16t 2 256 67. (a) 0 when t (b) tan 16 arctan 500 d dt 4 sec t2 θ 15,625 1000t 16 16 arccos 500 d dt 16 d ds 750 s ds dt 16 t2 2 2 When t 1, d dt 0.0520 rad sec. When t 2, d dt 0.1116 rad sec. h 300 69. tan y 300 dh dt 5 ft/sec 200 h arctan 300 d dt 1 1 300 h2 3002 1500 3002 h2 70. (a) 100 θ dh dt h x 300 5 3002 h2 3 rad/sec when h 200 100 200 300 100 (b) 15 y 0.361 0.002t y 20 110 0 0.123 million yr y 60 0 θ 750 1 ds 750 s s2 7502 dt 8t 125 4 125 t2 1 s h 16t2 256 500 h 500 750 s 68. cos 256 0.219 million yr 79.564 1 t2 1 750 s 2 750 ds s2 dt 202 Chapter 3 71. (a) Let y Differentiation arctan u. Then tan y u 1 + u2 u dy sec2 y dx u u sec2 y dy dx (b) Let y y 1 u 1 u2 . arcsec u. Then sec y u dy sec y tan y dx u u u2 − 1 y u sec y tan y dy dx 1 u u2 u 1 . Note: The absolute value sign in the formula for the derivative of arcsec u is necessary because the inverse secant function has a positive slope at every value in its domain. (c) Let y arccos u. Then cos y u dy sin y dx u 1 − u2 y u sin y dy dx (d) Let y 1 u u . 1 u2 arccot u. Then cot y u dy dx u csc2 y 1 + u2 1 y u csc2 y dy dx (e) Let y u 1 u 2. u arccsc u. Then csc y u dy csc y cot y dx u u 1 y dy dx u csc y cot y u u u2 1 u2 − 1 . Note: The absolute value sign in the formula for the derivative of arccsc u is necessary because the inverse cosecant function has a negative slope at every value in its domain. 72. f x kx sin x fx k cos x ≥ 0 for k ≥ 1 fx k cos x ≤ 0 for k ≤ Therefore, f x kx 1 sin x is strictly monotonic and has an inverse for k ≤ 74. False 73. True d arctan x dx 1 or k ≥ 1. 2 1 1 x2 > 0 for all x arcsin2 0 arccos2 0 0 2 1 S ection 3.6 75. Let x arctan x2 1 1 < x < 1. , x 1 sin arcsin x x2 1 x2 1 Hence, cos x x x Then tan x2 1 arctan 76. Let x tan Derivatives of Inverse Functions , as indicated in the figure x and 2 which gives arccos x Thus, arcsin x x arctan x2 1 1 < x < 1. for x < 1. , arccos x 2 x arctan 2 1 x2 π− θ 2 1 x 1 θ x 1 − x2 θ 1 − x2 77. f x 0≤x< sec x, 2 , ≤x< 1 or 3 2 y 4 2 −π 2 x π 2 −2 −4 (a) y x≤ x≥1 (b) 0≤y< 2 or y x sec y 1 3 2 ≤y< arcsec x sec y tan y y arcsec x, 1 sec y tan y 1 sec2 y ± sec2 y y 3π 2 tan2 tan y π 2 2 sin arcsin x Domain: On 0 ≤ y < x −6 −4 −2 78. y1 4 6 x y 1, 1 3 2 Range: y2 1, 1 , −2 −1 , y2 x −1 −2 −3 Range: y1 1 arcsin sin x Domain: y 22 1 2 3 2 and y 1 x x2 1 1 ≤y< 3 , tan y ≥ 0. 2 . 203 204 Chapter 3 79. f x Differentiation x arcsin 2 12 fx 1 x 22 2 0≤x≤4 , 14x 2 2 1 1 4 x2 21 x 2 arcsin 2 1 4x 4 x2 2 1 2x1 1 x4 1 x2 4 2x x2 4 2x Since the derivative is zero, we conclude that the function is constant. (By letting x 2.) you can see that the constant is 80. 0 0 in f x , 2 y2 y1 −3 3 −2 Since ex > 0 for all x, yz Section 3.7 y Related Rates x dy dt arctan e x > 0 for all x. 1 1. dx dt 2. dy dt (a) When x dy dt 1 3 24 dx dt dy dt y dy 6 dt 3 and dx dt 62 43 (b) When x 2, dx dt 20. 4 dx dt 1 4x dy dt 25 and dy dt xy dx dt 6 (a) When x 3 . 4 2 25 2 3. 3, 3x 4x dx dt 4 and dx dt (b) When x x dy dt dx 2 x dt 2x 2 x2 y 0 2x dx dt 12. 1 and dy dt 1 41 6 y2 5 . 2 5 2y dy dt 0 dy dt y dx x dt dy dt x dx y dt dx dt x dy y dt dx dt y dy x dt (a) When x dy dt 8, y 1 2, and dx dt 12 10 8 (b) When x dx dt 5, 25 x2 4. 2, 1, y 1 4 6 10, 5 . 8 4, and dy dt 3 . 2 (a) When x dy dt 6, 3, y 3 8 4 (b) When x dx dt 8, 6. 4, y 3 4 4, and dx dt 2 3, and dy dt 3 . 2 2, S ection 3.7 5. y x2 dx dt 2x 6. 2 dy dt 1 dx dt dx dt 1, 2 dy dt 7. y 4 cm sec. 0 cm sec. dy dt 1, 2, 2 22 25 0, 0 cm sec. dy dt 8. 2 22 2 25 y 2 dy dt (a) When x cos x dx dt 3, (a) When x dy dt 2 2 2 8 cm sec. (b) When x dy dt cos 6, 2 1 2 2 4 cm sec. dy dt 0, 2 2 cos 4, 2 cm sec. dy dt dy dx negative ⇒ positive dt dt 10. (a) (b) dy dx positive ⇒ negative dt dt (b) 11. Yes, y changes at a constant rate: dy dt dy dx is a multiple of . dt dt a dx . dt cos 2 4 (c) When x 9. (a) No, the rate 3 cm sec. 4, (c) When x dy dt 2 6 (b) When x dy dt 8 cm sec. 25 sin x dx dt dx dt 8 cm sec. 25 2, 4 cm sec. tan x sec2 x dx dt 2 (c) When x 21 2 dy dt 2x x2 1 (b) When x 20 2 dx dt 2 dy dt 0, (c) When x x2 (a) When x 12 (b) When x dy dt 1 1 dy dt (a) When x dy dt y Related Rates 2 cm sec. 3, 3 2 1 cm sec. dx dy negative ⇒ negative dt dt dy dx positive ⇒ positive dt dt 12. Answers will vary. 205 206 13. Chapter 3 x2 D dx dt 14. 14 x 2 D x2 dx dt x2 x2 2 1 x4 3x2 1 12 1 y2 x2 sin2 x 12 x 2 A 3x2 12 4x3 6x 2x3 3x dx 4 2 x 3x 1 dt dx dt 4x3 6x x4 3x2 1 sin2 x 2 dD dt 15. y2 2 dD dt Differentiation r2 2x 2 sin x cos x dx dt x sin x cos x dx x2 sin2 x dt 2 2 sin x cos x x2 sin2 x dr dt 3 dA dt 2r dr dt A r2 dA dt 16. 2r If (a) When r 6, dA dt 2 63 (b) When r 24 3 dA dr is constant, is not constant. dt dt 24, dA dt dr dt 17. (a) sin cos 2 dr dA depends on r and . dt dt 36 cm2 min. 144 cm2 min. 1 2b ⇒b s 2 h ⇒h s 2 1 bh 2 A s cos 2s sin 18. 2 dr dt s cos s2 2 sin cos 2 2 2 2 s2 sin 2 2 dV dt 2 1 2s sin 2 2 43 r 3 V 4 r2 (a) When r When r dr dt 6, dV dt 24, dV dt 4 6 4 2 288 in.3 min. 2 24 2 2 4608 in.3 min. θ s (b) If s h b (b) dA dt s2 cos 2 d d where dt dt When dA , 6 dt s2 2 When dA 3 dt s2 1 22 (c) If , 1 rad min 2 3 2 3s 2 . 8 1 2 1 2 s2 . 8 dA d is constant, is proportional to cos . dt dt dr dv is constant, is proportional to r2. dt dt S ection 3.7 19. 1 h 108 h2 36 h 3 dV dh dh 36 h2 dt dt dt dV 3 and h 2, When dt 13 h, 0<h<6 3 3 dh dt 36 dh dt 3 36 3 32 4 3 dV dt 3x 2 dx dt (a) When x 1, dV dt 0.0298 m min. 31 (b) When x s 6x2 dx dt 3 ds dt 12x 22. dr dt ds dt (a) When x (b) When x 23. 1, 10, V 12 rh 3 36 cm2 sec. 12 1 3 ds dt 1 3 12 10 3 92 hh 4 dh 9 2 dh h ⇒ 4 dt dt When h 24. 24, 3h 10 dV dt V r 15, dh dt 12 rh 3 9 1 25 3 h 3 144 By similar triangles, 4 dV dt 9 h2 4 10 15 r 5 dV dt dh 25 2 dh ⇒ h 144 dt dt 25 h3 3 144 h ⇒r 12 5 5 h. 12 10 dV dt 8 ft min. 405 2 When h r 12 h 8, dh dt 144 dV 25 h 2 dt 144 10 25 64 9 ft min. 10 dV dt 6, h dV dt 900 cm3 sec. 3 r3 dr dt (b) When r 33 h 4 2 12 r 3r 3 (a) When r 360 cm2 sec. since 2r 3 r2 9 cm3 sec. 3 2 dV dt dx dt 3 10 12 rh 3 V 2 10, dV dt 21. 207 x3 20. dh ⇒ dt 4 V dx dt V Related Rates dV dt 3 6 3 2 2 24 2 216 2 in.3 min. 3456 in.3 min. 208 Chapter 3 25. Differentiation 12 1 6 3 1 1 2 12 6 2 (a) Total volume of pool 1 166 2 Volume of 1 m. of water (b) Since for 0 ≤ h ≤ 2, b V dV dt 26. V (a) 1 bh 6 2 36h 12h When h (b) If dh dt 2x dx dt 2, 2, then y2 2y 1 144h h 12 ft 3 ft dh dt 1 2 12 1 dV dt 12 2 1 ft min. 6 3 8 h ft 3 ft 9 ft3 min. 252 dy dt 1 m min. 144 1 144 1 1 dV 12h dt dV dt 3 and h 8 x2 27. 1 and b=6 18h2 3 6h h dh dh ⇒ dt dt 12 6h, you have 6h2 since b 6bh 0 25 y dy dt (a) When x x dx dt 7, y 24, y 24, 15, y When x 2x dx since y dt 576 y When x (b) A dA dt 400 7, dy dt dy dt 20, 2. 27 24 dy dt 7 ft sec. 12 2 15 20 2 24 7 x 3 ft sec. 2 48 ft sec. 7 1 xy 2 1 dy x 2 dt y dx dt From part (a), we have x Thus, 2 12.5% dh 1 ⇒ 4 dt dh dt 1 bh 12 2 dV dt 3bh 18 m3 (see similar triangle diagram) h=1 18 100% 144 % pool filled 144 m3 1 6 12 dA dt 1 7 2 —CONTINUED— 7 12 7, y 24 2 24, dx dt 527 24 2, and dy dt 21.96 ft2 sec. 7 . 12 S ection 3.7 27. ––CONTINUED–– (c) x y tan d dt 1 y d dt sec2 dx dt x y2 cos2 1 y dy dt dx dt θ 25 y x y2 dy dt x Using x dx 24, dt 7, y d dt x2 y2 dy dt 1 2 24 7 and cos 12 7 24 7 12 24 , 25 1 rad sec. 12 25 0 28. 2x 2 24 25 we have dy 2, dt dx dt 2y 2 5 y dx dt When x dy dt 0.15y dy since x dt 0.15. x 2.5, dx dt 18.75, y 29. When y 18.75 0.15 2.5 122 6, x x2 s y x 62 0.26 m sec. 6 3, and 2 s 12 − y 12. 12 y y x 108 36 ( x, y ) 12 x2 2x dx dt 2 12 x dx dt Also, x2 2x dx dt Thus, x dx dt dx x dt x s2 1 dy dt 2s 12 dy dt s y y y2 2y 2 12 y ds dt ds dt 122. dy dt 0⇒ y s dx dt dy dt y x dx y dt 12 12x y dy dt x dx dt s ds . dt ds ⇒ dt sy 12x ds dt x dx y dt 12 6 12 6 3 63 6 0.2 3 15 1 53 3 15 m sec (horizontal) 1 m sec vertical 5 Related Rates 209 210 Chapter 3 Differentiation 30. Let L be the length of the rope. (a) L2 dL 2L dt dx dt 144 x2 dx 2x dt L dL x dt When L (b) If 4L dL since x dt dL dt 4 ft sec. L2 144 169 4 13 5 dx dt 144 5 4, and L 4 20 ft sec 13 13 ft 12 ft 52 5 13: x dx L dt 5 13 13, 4 ft/sec x dx dt 10.4 ft sec. As L → 0, dL increases. dt Speed of the boat increases as it approaches the dock. s2 31. (a) x2 y2 450 dy dt 600 ds 2s dt dx 2x dt ds dt x2 2x 150 450 1 hr 3 y2 x 200 100 Distance (in miles) 0 2s 200, s 200 250 250 and 600 750 mph. 20 min 33. ds dt since dy dt 0 10, x 10 53 2s 100 480 3 240 25 75 902 30 dx dt s ds x dt When s s2 x s2 dx dt dx dt s 100 y dy dt 150 and y 250 750 dx dt 200 s ds dt 32. dy 2y dt x dx dt When x (b) t y Distance (in miles) dx dt 5 3, 28 ds dt 2x When x 160 3 dx ds ⇒ dt dt x s s 902 ds dt 30 30 10 x 302 28 2nd s 30 ft 3rd dx dt 30, 277.13 mph. y 5 mi x2 x 1st s x 90 ft Home 30 10 28 10 8.85 ft sec. S ection 3.7 34. s2 902 x 28 ds dt x s 2nd 30 ft 60 dx dt x2 When x x 3rd 90 ft dx dt Home 60, 902 s 602 ds dt 35. (a) 15 6 y y x 56 13 ⇒ 15y 15.53 ft sec. 15x 6y 5 x 3 dx dt 5 3 15 5 dy dt (b) 30 13 60 28 30 13 y 6 dx dt dy x dt 5 5 3 dy dt dx dt 20 6 20x x 25 ft sec 3 25 3 5 y 10 ft sec 3 6y 20x y 20y y 14y 36. (a) y 10 x 7 x 20 6 x y dx dt dy x dt 5 dy dt (b) 1st s 10 dx 7 dt dy dt dx dt 10 7 5 50 7 50 ft sec 7 5 50 7 35 7 15 ft sec 7 Related Rates 211 212 Chapter 3 Differentiation 1 t sin , x 2 2 6 37. x t (a) Period: 2 (b) When x y2 1 ,y 2 12 0, 1 2 2 dx 2x dt 1 1 4 2 t 1 cos 26 6 y2 1, dy 0⇒ dt 12 cos t 6 dx dt 14 cos 6 15 4 12 1 1 3 24 15 12 2 5 5 120 Thus, 5 120 2y 4 r2 1 R1 cos dy dt 3 5 dR1 dt x dx . y dt cos 1 R2 dr . dt 95 . 125 0.5058 m sec 1.5 dR dt 1 R12 R dR1 dt dR2 dt 1 R22 50 and R2 75, 30 dR dt 30 2 1 50 2 1 1 75 2 0.6 ohms sec. pV1.3 1.3 pV 0.3 dV dt V 0.3 1.3p dV dt V1.3 V dp dt dp dt dV 1.3p dt k 42. rg tan 32r tan 0 v2 v2, r is a constant. dp V dt Likewise, d dt 2v dv dt 32r sec2 0 6 1 When R1 41. t t 1 R2 dR2 dt dr ⇒k dt 0⇒ t ⇒ sin 95 125 1 R 40. dr . dt 4 r2 1 dy dt 15 and 4 2 9 25 5 Therefore, k 4 r2 3 5 3 10 15 4 Speed dV dt 3 sin 5 dy dt 5 m sec. 120 39. Since the evaporation rate is proportional to the surface area, dV dt k 4 r 2 . However, since V 4 3 r 3, we have 1 4 1 y2 x2 4 m. 5 2 4 5 dx dt 2x Speed 0, 3 10 x dx . y dt 3 5 1 3 ,y 10 (c) When x Thus, dy dt 3 ,y 5 (b) When x 15 and t 4 1 2 seconds 1 dy 2y dt y2 Lowest point: dx dt x2 3 m. 2 t, x2 2 (a) Period: 3 2 1 ,y 4 3 sin 5 38. x t 12 seconds 6 Lowest point: (c) When x 1 dv dt 16r d sec2 v dt d dt v dv cos2 . 16r dt 1.5 1 ⇒t 2 1 6 S ection 3.7 43. y 30 tan y dy dt d dt 1 dy 30 dt d dt sec2 3 m sec 1 cos2 30 When y 2 2. Thus, 1 rad sec. 20 10 x dx dt 1 ft sec x 10 x2 d dt d dt 10 dx dt θ 10 dx sec x 2 dt 10 252 25 252 102 1 4347 e369,444 400,000,000 45. H x 30 4 and cos 11 3 30 2 sin cos θ dy dt 30, d dt 44. y 50 t (a) t 65 ⇒ H 80 ⇒ H 2 25 21 2 21 525 0.017 rad sec 19,793 99.79% t 10 1 25 5 21 60.20% (b) H H At t 75 and t d dt sec2 d dt dx dt 2, H 2 t 4.7%. x 50 Police tan 46. 369,444 50 50t 19,793 30 2 60 rad min rad sec θ 50 ft 1 dx 50 dt 50 sec2 d dt (a) When 30 , dx dt 200 ft sec. 3 (b) When 60 , dx dt 200 ft sec. (c) When 70 , dx dt 427.43 ft sec. x Related Rates 213 214 47. Chapter 3 d dt Differentiation 10 rev sec 2 rad rev d dt sin θ x d dt 20 sin 20 sin 20 400 sin (c) 2000 4 0 dx 400 sin dt dx is least when dt x y x y dx dt dy dt dy dt 1 y dx dt sin 22 240 y ,y x x ⇒x 50 cm sec. 50 tan 50 sec2 d dt 2 50 sec2 d dt 89.9056 mi hr L 600 mi hr d dt 5 x2 d dt cos2 y=5 θ x dx dt x2 L2 5 dx x2 dt 52 L2 1 dx 5 dt 5 dx x2 dt sin2 (a) When 30 , d dt 120 4 (b) When 60 , d dt 120 (c) When 75 , d dt 120 sin2 75 1 5 30 rad hr 3 4 600 120 sin2 1 rad min. 2 90 rad hr 3 rad min. 2 111.96 rad hr 1 cos2 , 25 4 ≤ ≤ 4 (b) y changes slowly when x 0 or x L. y changes more rapidly when x is near the middle of the interval. 5 dx dt sec2 200 3 cm sec. dx dt 22˚ 50. (a) dy dt 3 dx dt means that y changes three times as fast as x changes. tan 200 400 sin 60 d dt 51. n. x x y2 0 400 sin 30 49. tan y 2 n. dx dt dx 60 , dt For 1⇒ is greatest when sin 30 , (d) For − 2000 48. sin 22 x 1 dx 20 dt dx dt (b) 20 x 20 cos (a) 20 rad sec 1.87 rad min. S ection 3.7 52. 4.9t2 yt dy dt 20 y 9.8t y1 4.9 y1 9.8 20 15.1 20 20x When y y 20 x By similar triangles, x 240 15.1, 20x 20 240 x 15.1 x 20 53. x2 1, y2 240 . 4.9 dx dt dy dt dx dt y x dx dt At t 240 xy dx dt dy y dt 20 240 4.9 20 15.1 97.96 m sec. 9.8 d 2y dt 2 25; acceleration of the top of the ladder 2y x dy dt 0 y dy dt 0 dx dt dx dt First derivative: 2x dx dt dx dt Second derivative: x d 2x dt 2 d 2y dt 2 y dy dt dy dt 0 d 2y dt 2 When x 7, y d 2y dt 2 54. L2 x xy. 15.1 x 240 x (0, 0) 12 x 20x y 12 1 24 144 dy dt 24, 70 7 12 2 dL dt 2x L dL dt x Second derivative: L d 2L dt 2 d 2x dt 2 13, x 5, 1 13 0 5 dx dt dx dt 2 dy dt 1 24 625 144 4 49 144 1 24 2 0. 0.1808 ft sec2 d 2x dt 2 dx dt dx dt dL dt dL dt x d 2x dt 2 When L d 2x dt 2 dx d 2x is constant, 2 dt dt 2 x2; acceleration of the boat First derivative: 2L x 2 (see Exercise 27). Since 7 dx , and 12 dt 2 1 y 1 x 10.4, and 4 2 d 2x dt 2 L dL dt 10.4 dx dt d 2L dt 2 dx dt dL dt 2 dx dt 2 4 (see Exercise 30). Since 2 1 16 5 108.16 1 5 dL d 2L is constant, 2 dt dt 92.16 0. 18.432 ft sec2 Related Rates 215 216 Chapter 3 Differentiation 55. (a) m s 0.3754s3 18.780s2 dm dt 1.1262s2 37.560s (b) If t fx ds dt 0.75, then s 17.8 and dm dt 1.1154 million year. xn 1 Section 3.8 x2 313.23 1707.8 n 10 and 1. f x ds dt 313.23s Newton’s Method 2x 3 2. f x fx 4x x1 1.7000 0.1100 3.4000 0.0324 1.7324 1.7324 0.0012 3.4648 0.0003 1.7321 xn f xn 1 2x2 f xn f xn n 1.7 f xn 2 x1 f xn 1 1 3 1 2 3. f x cos x 1.25 0.125 f xn f xn xn 1 4 4 5.0 0.025 sin x fx 5 4 f xn f xn f xn f xn f xn xn 5 4 1.225 f xn f xn f xn f xn f xn f xn 3.0000 0.1411 0.9900 0.1425 3.1425 3.1425 0.0009 1.0000 0.0009 3.1416 xn f xn f xn f xn f xn 1 0.1000 0.1003 1.0101 0.0993 0.0007 2 0.0007 0.0007 1.0000 0.0007 0.0000 n xn f xn f xn f xn f xn 1 0.5000 0.3750 1.7500 0.2143 0.7143 2 0.7143 0.0788 2.5307 0.0311 0.6832 3 4. f x fx x1 5. f x fx 0.6832 0.0021 2.4003 0.0009 0.6823 n xn f xn f xn f xn f xn 1 0.5000 0.4688 1.3125 0.3571 0.8571 2 0.8571 0.3196 3.6983 0.0864 0.7707 3 0.7707 0.0426 2.7641 0.0154 0.7553 4 0.7553 0.0011 2.6272 0.0004 0.7549 1 n 3 tan x sec2 x 0.1 x3 x 3x2 1 1 Approximation of the zero of f is 0.682. 6. f x fx xn 2 x1 n x5 5x4 x 1 1 Approximation of the zero of f is 0.755. xn xn xn xn f xn f xn f xn f xn f xn f xn S ection 3.8 7. f x 3x 1 x f xn f xn f xn 1.2000 0.1416 2.3541 0.0602 1.1398 2 1.1398 0.0181 3.0118 0.0060 1.1458 3 1.1458 0.0003 2.9284 0.0001 1.1459 xn 1 Approximation of the zero of f is 1.146. Similarly, the other zero is approximately 7.854. 8. f x fx 3 2x x 1 1 2x 1 1 x fx x1 e 1 f xn f xn f xn f xn 5 0.1010 0.5918 0.1707 4.8293 2 4.8293 0.0005 0.5858 .00085 4.8284 x e x x 10. f x x 3 fx 1 x1 xn 1 0.5 ln x 2.0 4 2.204569 3 0.567143 2.0 2 0.567143 xn 1 0.566311 3 n 2.207939 4 2 xn 1 x 0.5 n f xn f xn xn 1 1 Approximation of the zero of f is 4.8284. 9. f x xn n 1 f xn f xn f xn n fx Newton’s Method 2.207940 Approximate zero: 0.567 Approximate zero: 2.208 11. f x fx x3 3 n 3x2 xn f xn f xn f xn f xn xn f xn f xn 12. f x fx 3 1.5000 0.3750 6.7500 0.0556 1.4444 1.4444 0.0134 6.2589 0.0021 1.4423 3 1.442. 1 2 Approximation of the zero of f is 1.4423 0.0003 6.2407 0.0001 1.4422 2x3 6x2 Approximation of the zero of f is 1.1447. n xn f xn 1 f xn f xn f xn xn f xn f xn 1 1 6 0.1667 1.1667 2 1.1667 0.1759 8.1667 0.0215 1.1451 3 1.1451 0.0032 7.8679 0.0004 1.1447 217 218 Chapter 3 Differentiation x3 3.9x2 4.79x fx 3x2 7.8x 4.79 n xn 1 13. f x 1.881 f xn f xn f xn f xn 0.5000 0.3360 1.6400 0.2049 0.7049 2 0.7049 0.0921 0.7824 0.1177 0.8226 3 0.8226 0.0231 0.4037 0.0573 0.8799 4 0.8799 0.0045 0.2495 0.0181 0.8980 5 0.8980 0.0004 0.2048 0.0020 0.9000 6 0.9000 0.0000 0.2000 0.0000 0.9000 xn f xn f xn Approximation of the zero of f is 0.900. n xn f xn 1 1.1 0.0000 f xn f xn f xn 0.1600 0.0000 f xn f xn xn 1.1000 Approximation of the zero of f is 1.100. n xn f xn f xn f xn f xn 1 1.9 0.0000 0.8000 0.0000 f xn f xn xn 1.9000 Approximation of the zero of f is 1.900. 14. f x 14 2x 3x fx 2x3 3 n xn 3 f xn f xn f xn f xn xn f xn f xn 1 1 0.5 5 0.1 0.9 2 0.9 0.0281 4.458 0.0063 0.8937 3 0.8937 0.0001 4.4276 Approximation of the zero of f is n xn f xn 2 1 2.0769 3 2.0720 f xn f xn 13 2 0.0725 0.0003 0.8937 0.8937. f xn 1 0.0000 xn 0.0769 f xn f xn 2.0769 14.9175 0.0049 2.0720 14.7910 0.0000 2.0720 Approximation of the zero of f is 2.0720. 15. f x x sin x 1 fx 1 cos x 1 Approximation of the zero of f is n 0.489. xn f xn f xn f xn f xn xn f xn f xn 1 0.5000 0.0206 1.8776 0.0110 0.4890 2 0.4890 0.0000 1.8723 0.0000 0.4890 S ection 3.8 16. f x x3 Newton’s Method cos x 219 f xn f xn 17. h x hx fx gx 1 2x 2 fx hx 1 3 x 1 x2 1 2x x2 1 0.0334 0.8666 0.8666 0.0034 3.0151 0.0011 0.8655 0.8655 0.0001 3.0087 0.0000 0.8655 xn h xn h xn h xn h xn 0.6000 0.0552 1.7669 0.0313 0.5687 2 0.5687 0.0001 1.7661 0.0000 0.5687 n xn h xn h xn h xn h xn 2.9000 0.0063 0.9345 0.0067 2.8933 2.8933 0.0000 0.9341 0.0000 2.8933 n 4 gx 3.2133 xn h xn h xn h xn h xn 1 4.5000 0.1373 21.5048 0.0064 4.4936 2 4.4936 0.0039 20.2271 0.0002 4.4934 n xn h xn h xn h xn h xn 1 0.8000 0.0567 2.3174 0.0245 0.8245 2 0.8245 0.0009 2.3832 0.0004 0.8241 4 Point of intersection of the graphs of f and g occurs when x 0.569. 18. h x 0.1074 2 x 0.9000 1 1 f xn f xn 1 2x f xn n Approximation of the zero of f is 0.866. f xn 1 sin x xn 3 3x n 2 fx 2 2 xn h xn h xn xn h xn h xn xn Point of intersection of the graphs of f and g occurs when x 2.893. 19. h x hx fx gx sec2 1 x tan x x Point of intersection of the graphs of f and g occurs when x 4.493. 20. h x hx fx gx 2x x2 cos x sin x One point of intersection of the graphs of f and g occurs x2 and g x cos x are when x 0.824. Since f x both symmetric with respect to the y-axis, the other point of intersection occurs when x 0.824. 21. h x hx x1 ln x 1 x 22. h x x hx 1 0.5 2 1x e 2 x2 2 2 h xn h xn xn 2x Two points of intersection n xn n 1 0.5 1 2 0.564382 3 0.567139 4 0.567143 Approximate intersection at x ex h xn h xn xn xn xn 1 1 1 2 1.231898 2 0.770312 3 1.205595 3 0.742528 4 0.567 n 1.205250 4 0.742115 Approximate intersections at x 1.205 and x 0.742 220 Chapter 3 Differentiation arctan x hx arccos x n xn 1 23. h x 1 1 0.5 x2 1 2 1 x 2 25. (a) f x x2 fx xn 1 xn2 a 2xn 1 5: xn nxn xi f xi f xi xi xi n a nxi n 1 1 1 xi n nxi n 1 n a xn 5 . x1 xn 2 (b) 0.489027 1 xi 0.489048 a, a > 0 26. (a) f x 1 x 2n 1 xn 0.5 3 fx a, a > 0 1 x 2n (b) x xn 2 x 1 2 1 n 1 1 Approximate intersection at x 0.489 f xn f xn xn 0.786151 arcsin x hx 2x xn 0.786251 4 Approximate intersection at x 0.786 0.798537 3 24. h x 4 6: xi a 3xi 4 6 , x1 4xi 3 1 1.5 n 1 2 3 4 i 1 2 3 4 xn 2 2.25 2.2361 2.2361 xi 1.5 1.5694 1.5651 1.5651 4 6 1.565 3 15: xi For example, given x1 1 2 2 x2 5 2, 5 2 9 4 2.25. 2.236 7: xn 2.5 i 1 x 2n 1 2xi 3 15 , x1 3xi 2 1 7 , x1 xn 3 2 3 4 xi 2 1 2.5 2.4667 2.4662 2.4662 1 2 3 4 2 2.75 2.6477 2.6458 6x2 6x 1 y 6x2 12x 6 fx x1 1 12x2 12x 3 y 12x2 24x 12 fx 3 2 2.646 2x3 4x3 2.6458 7 2.466 5 xn 15 x1 n 27. y fx fx 28. y 0; therefore, the method fails. f x2 n xn f xn 1 1 0 0; therefore, the method fails. f xn 1 fx n xn f xn f xn f xn f xn 1 3 2 3 2 3 1 2 1 2 1 1 0 — — xn f xn f xn S ection 3.8 x3 29. y 6x2 3x2 y 10x 12x 6 10 fx 30. f x x1 1 x3 2 x4 2 sin 2x 2 x2 2 cos x 221 cos 2x fx fx 2 sin x Newton’s Method 1 and so on; fails to converge x1 3 2 n Fails because f x1 0. xn 1 3 2 3 1 f x1 . f x1 x1 −1 c 2 b x −2 Continue this process until xn final approximation of c. xn 1 32. Newton’s Method could fail if f c 33. Let g x x a3 −1 f x2 . f x2 x2 f (x) x1 x2 Calculate a third estimate by x3 0 y 31. Answers will vary. See page 191. Newton’s Method uses tangent lines to approximate c such that f c 0. First, estimate an initial x1 close to c (see graph). Then determine x2 by x2 f xn f xn fx x cos x is within the desired accuracy. Let xn 1 be the 0, or if the initial value x1 is far from c. x 1 fx 0.4597 1.8415 0.2496 0.7504 0.7504 0.0190 1.6819 0.0113 0.7391 0.7391 0.0000 1.6736 0.0000 0.7391 1. The fixed point is approximately 0.74. 34. Let g x 1.0000 3 sin x x cot x x 1. g xn g xn g xn xn 2 gx g xn g xn g xn n x g xn g xn g xn g xn 1 10 n 1 Approximate fixed point: 1.12 x3 fx (a) 3x2 3x2 37. f x 6x 3 −4 —CONTINUED— 0.0087 0.8603 0.8603 0.0001 2.7403 0.0000 0.8603 x 0.3579 1.118326 x2 1 f x1 f x1 1.333 x2 0.5 2 (c) x1 xn 0.564382 3 1 x1 n ln x Approximate fixed point: 0.57 Continuing, the zero is 1.347. 5 −2 0.8516 2.7668 1.118238 (b) x1 4 0.1484 0.0240 1.0 2 x 2.4123 0.8516 xn 3 ex 35. g x 1.0000 36. g x The fixed point is approximately 0.86. g xn xn xn 3 gx n 2 csc2 g xn xn 0.567139 1 4 x1 f x1 f x1 2.405 Continuing, the zero is 2.532. 222 Chapter 3 Differentiation 37. —CONTINUED— (d) y 3x (e) If the initial guess x1 is not “close to” the desired zero of the function, the x-intercept of the tangent line may approximate another zero of the function. The x-intercepts correspond to the values resulting from the first iteration of Newton’s Method. 4 y f 3 x 2 1 4 y 38. f x 1.313x sin x, f x (a) 5 3.156 cos x (d) 2 y (1.8, 0.974) 2 − π 2 −2 (b) x1 x2 (c) x1 x2 39. f x 1 −1 1.8 f x1 f x1 (3.143, 0) The x-intercepts correspond to the values resulting from the first iteration of Newton’s Method. 6.086 3 f x1 f x1 x1 1 x a (e) If the initial guess x1 is not “close to” the desired zero of the function, the x-intercept of the tangent line may approximate another zero of the function. 3.143 40. (a) xn 0 xn xn 2 1 3xn i 1 xn a 1 xn2 xn xn2a xn2 xn 1 xn xn2a 2xn a xn 2 1 2 3 4 xi 1 x2 xn x π −2 x1 fx xn (3, 0.141) (6.086, 0) 1 0.3000 0.3300 0.3333 0.3333 1 3 0.333 (b) xn axn xn 2 1 11xn i 1 2 3 4 xi 0.1000 0.0900 0.0909 0.0909 1 11 41. 76x3 2,500,000 3 76x Let f x fx 4830x2 76x3 228x2 2,820,000 4830x2 4830x2 0.091 320,000 0 2,820,000 n xn 1 40.0000 44000.0000 21600.0000 2 37.9630 17157.6209 38131.4039 3 38.4130 780.0914 4 38.4355 2.6308 9660x. From the graph, choose x1 40. The zero occurs when x 38.4356 which corresponds to $384,356. f xn f xn f xn f xn 2.0370 xn f xn f xn 37.9630 0.4500 38.4130 34642.2263 0.0225 38.4355 34465.3435 0.0001 38.4356 S ection 3.8 0.808x3 42. 170 17.974x2 Let f x 0.808x3 fx 2 2.424x 35.948x From the graph, choose x1 n xn 1 1.0000 2 3 110.843, 1 ≤ x ≤ 5 71.248x 17.974x2 71.248x 59.157 71.248. 1 and x1 f xn 3.5. Apply Newton’s Method. f xn f xn f xn 5.0750 37.7240 1.1345 0.2805 1.1429 0.0006 f xn f xn xn 0.1345 1.1345 33.5849 0.0084 1.1429 33.3293 0.0000 1.1429 f xn n 1 2 3.6762 24.8760 0.1878 3.6878 0.3286 3.6878 3 f xn f xn 28.3550 0.0116 3.6762 0.0009 3.5000 f xn 4.6725 xn Newton’s Method 28.1450 0.0000 3.6762 xn f xn f xn The zeros occur when x 1.1429 and x 3.6762. These approximately correspond to engine speeds of 1143 rev min and 3676 rev min. x2 1 x 1 . x 1 is a discontinuity. It is not a zero of f x . 43. False. Let f x p x q x is given in reduced form. This statement would be true if f x 44. True 45. True 47. f x 13 4x fx 32 4x Let x1 3 4x 3x2 46. True 2 y 3 4 6x 60 40 12. 20 x −10 − 5 n xn f xn f xn f xn f xn 1 12.0000 7.0000 36.7500 0.1905 11.8095 0.2151 34.4912 0.0062 11.8033 0.0015 34.4186 0.0000 20 11.8033 3 15 11.8095 2 5 11.8033 Approximation: x 48. f x Domain: x Let x1 2 2, 2 2 and x fx f xn f xn 11.803 x2 sin x 4 xn y 2 are both zeros. x2 cos x 4 2 1 x 4 x2 sin x x 2 −2 1 1. −2 n 1 f xn xn 1.0000 f xn f xn f xn 0.2444 1.7962 0.1361 xn f xn f xn 1.1361 2 1.1361 0.0090 1.6498 0.0055 1.1416 3 1.1416 0.0000 1.6422 0.0000 1.1416 Zeros: x ± 2, x 1.142 −3 2 223 224 Chapter 3 49. f x Differentiation sin x fx cos x Let x0, y1 x0, Thus, x0 0 0 y0 x0 cos x0 be a point on the graph of f. If (x0, y0 is a point of tangency, then sin x0 sin x0 . x0 y0 x0 tan x0 x0 4.4934. Slope cos x0 0.217 You can verify this answer by graphing y1 sin x and the tangent line y2 0.217x. 2 −1 5 −2 2x3 50. f x 20x2 12x 24 (a) There is one real root between 10 and 11. (b) Using Newton’s Method and x1 n 1 2 xn 2 0.3529 2, the first few approximations are very poor. ... 0.8547 14 15 16 17 ... 3 10.8270 10.6723 10.6679 10.6679 10.6679 Zero (c) Using x1 10: n 1 2 3 4 5 xn 10 10.7660 10.6696 10.6679 10.6679 10.6679 Zero Using x1 100, x11 10.6679. (d) The convergence is faster if you select a starting value x1 close to the actual zero. Review Exercises for Chapter 3 1. f x fx x2 lim 2x 3 fx x x x→0 lim x x fx 2 2x x x→0 lim x→0 2x 2x x 3 3 x x2 2x x x 2 x→0 lim x2 3 2x x x x 2 2x 2x lim 2x x→0 x2 x 2x 3 2 2x 2 R eview Exercises for Chapter 3 2. f x x x 1 1 fx fx lim x x x→0 x lim fx x x2 lim lim 1x xx 1 x→0 x x xx 3. f x x fx lim 2x x 1x xx x x 1 1 x x x x x 1x 1x 1 x x xx x→0 lim x x x→0 x→0 1 x lim x→0 fx x x x x x 1 x x x x→0 x x lim x→0 x 1 x x fx x x x→0 2 1x x x x x x 2 1 2 2 x x x x fx lim x x 2x 2x xx x→0 x0 x x→0 xx 2x x 1. x 2 x2 xx 6. f is differentiable for all x 8. f x 2 (a) Continuous at x xx 2 x→0 5. f is differentiable for all x 2x xx 2x lim 1 2 x x x lim x x lim x fx 2 x lim 4 x x x→0 1 x 1 x x 1 fx x→0 lim 1 4. f x x→0 lim 1 1 x2 x x 1x 1 1 1 lim 7. f x 225 2 x2 1 4x 4x 2, x2, 3. if x < if x ≥ 2 2 (a) Nonremovable discontinuity at x (b) Not differentiable at x in the graph. 2 because of the sharp turn (b) Not differentiable at x discontinuous there. 2 2 because the function is y y 7 6 5 4 3 2 5 4 1 x −1 123456 −5 −4 −2 −3 1, g x 1 2 −2 9. Using the limit definition, you obtain g x At x −2 −1 −1 1 4 3 1 6 3 2. 4 3x 1 6. 10. Using the limit definition, you obtain h x At x 2, h 2 3 8 4 2 67 8. 3 8 4x. 226 Chapter 3 Differentiation 3x 2. 11. (a) Using the limit definition, f x At x 1, f y 2 3. The tangent line is: 3x y (b) 1 12. (a) Using the limit definition, f x At x 1 3x y 2x y 2x (b) 2 4 −6 6 −4 g2 2 x2 x lim x3 x x2 lim x x→2 2 2 x 2 lim x→2 8 16. y f′ x 25. h x 19. f x 12 fx 0 8t 5 3x 12 27. g t 2 t 3 4 t 3 x 23 6x1 2 3x1 x 1 13 1 9 3 x 4 3t3 x3 3x2 6x 3x x gs 2 4 x 9 1 x 2 29. f 3 4 9x3 f 2 2 2 x 12 12t 3 ht 2 x1 3t 4 21. h t 24. g s fx 2 x 9 0 12x11 26. f x 1 x2 hx x12 3x2 3 25 f′ gx 3 28. h x 3 1 13 x 20. g x fx 2 gt x 2x y π 2 8x7 40t 4 33 x π 2 x8 23. f x hx 3 x f −1 1 6x 2 17. y 1 − ft x y 1 22. f t 1 3 1 f 2 −1 x x→2 lim x x→2 lim x2 f2 2 1 4 2 x→2 fx x 4 2 x lim lim x→2 2 x2 x→2 14. f 2 1 x lim 15. −4 gx x lim x→2 x→2 y 2 (0, 2) (− 1, − 2) 18. y 0 0 −4 13. g 2 2 . 12 2. The tangent line is: 0, f 0 2 1 x 4s4 5s2 16s3 10s 12 1 x 2 3 sin 3 cos 32 x1 2x3 2 30. g g 4 cos 4 sin 6 R eview Exercises for Chapter 3 ft 31. ft 5 3 32. g s 4e t 3 sin t F 33. 4e t 3 cos t 2e s sin s 5 3 gs 227 2e s cos s 200 T 100 T Ft (a) When T 4, F 4 50 vibrations sec lb. (b) When T 9, F 9 331 vibrations sec lb. 3 16t 2 34. s s0 First ball: 16t 2 100 0 100 16 t 10 4 2.5 seconds to hit ground Second ball: 16t 2 75 0 53 4 75 16 t2 2.165 seconds to hit ground Since the second ball was released one second after the first ball, the first ball will hit the ground first. The second ball will hit the ground 3.165 2.5 0.665 second later. 35. 16t 2 st s 9.2 s0 16 9.2 2 s0 16t 2 0 1354.24 s0 16t 2 36. s t 30 sec Since 600 mph 1 6 move horizontally 37. (a) y mi/sec, in 30 seconds the bomb will 1 6 y x 0.02x2 10 y 1 0.04x 5 y0 1 (d) y 10 40 0.6 y 25 x 20 0 60 Total horizontal distance: 50 x x1 y 30 0.02x2 x implies x 50 50. 0.2 y 50 0 30 5 miles. (c) Ball reaches maximum height when x 15 (b) 0 0 14,400 t The building is approximately 1354 feet high (or 415 m). 14,400 1 (e) y 25 0 25. 228 Chapter 3 38. Differentiation y v02 64 ( v02 v02 , 64 128 ) 2 0 ( v32 , 0 ) x v02 128 (a) y 32 2 x v02 x 0 if x 32 x v02 x1 v02 . 32 0 or x v02 32. Projectile reaches its maximum height at x (one-half the distance). 32 2 x v02 x 32 x v02 x1 (d) v0 v0 64 Range: x 0 2v0 32 2 v02 ,y 64 64 v02 v02 64 1 v02 32 70 2 32 v02 128 Maximum height: y 153.125 ft 70 2 128 50 4v02 32 0 160 0 or four times the initial range. From part (a), the maximum height occurs when x v02 64. The maximum height is y v02 64 v02 64 32 v02 v02 64 v02 64 2 v02 128 v02 . 128 If the initial velocity is doubled, the maximum height is y 2v0 64 2v0 2 128 2 4 v02 128 or four times the original maximum height. t2 39. x t 3t 2 t (a) v t xt 2t at vt 2t 1 (b) v t < 0 for t < 2 (c) v t x 2 (d) x t 3 2 0 for t 3 2 3 3 2 0 for t 3 2 1, 2 v1 1 1 2 1 2 21 3 1 v2 1 4 22 3 1 The speed is 1 when the position is 0. 40. (a) y 0.14x2 4.43x 58.4 (b) 320 0 60 0 —CONTINUED— 0. 70 ft sec 2 when x 0 and x x02 32. Therefore, the range is x v02 32. When the initial velocity is doubled the range is x 64 x v02 When x Projectile strikes the ground when x (c) y 1 (b) y 38.28 ft R eview Exercises for Chapter 3 40. —CONTINUED— (c) (d) If x 12 0 65, y 362 feet. 60 0 (e) As the speed increases, the stopping distance increases at an increasing rate. 41. f x 3x2 7 x2 2x fx 3x2 7 2x 2 3 2 2 6x 9x 42. g x x2 16x 2x x3 3x x gx 3 x3 3x 1 3 6x x3 7 43. h x hx 45. f x fx x1 2 sin x x sin x 1 sin x 2x t3 ft 2x x 2 2 2x 3 1 x3 x2 x fx x2 fx 1 1 2x x2 x2 12 1 x2 x 1 2x 2 1 1 x2 fx 16 x2 23 2 fx 51. y y 53. y y 3x2 4 1 3x2 4 2 6x 4 6x 3x2 x2 cos x cos x 2x 52. y x2 cos2 x sin x 3x 2 sec x 3x 2 sec x tan x 2x cos x x 2 sin x cos2 x 54. y 6x sec x y y 2x 2 sec2 x 6x 12 5 2x 2 2x 2x 1 2 6x 2 18 1 3x2 3x 2x 2 sin x x2 x2 cos x x 2 tan x x2 1 9 3x2 fx 2 11 2 3x2 5x 9 3x2 50. f x x 1 2 x2 49. f x 6 5 1 1 x2 6 3t2 cos t 11 x 6x x2 48. f x 1 3x 1 1 x x x2 3 cos t 3t2 sin t x x 46. f x 21 6x 6x t3 sin t 2 x3 1 x3 47. f x 2 2 t 3 cos t 44. f t x cos x 3x 6x 2 3x 2 x 3 3x 4x3 2 sin x 2x 2 sin x x3 55. y 2x tan x x cos x x4 y x tan x x sec2 x tan x 229 230 Chapter 3 1 1 56. y sin x sin x 1 y 1 58. v t Differentiation 57. y sin x cos x 1 2 cos x sin x at vt v4 36 4xex 59. g t 16 t3 gt 20 m sec 3t2 gt 2t 3 3t 2 1 12x1 60. f x 3 fx 3x 62. h t 63. y 2 cos x 6 4 cos t 5 sin t tan ht y 3 cos x xy 2 sin x cos x y sin x 0 3 cos x xy sec x tan x g x on [0, 2 66. sin x r 2h V t 1 2 1 dV dt sin x cos x cos x sin x 1 sin3 x cos3 x 2 tan3x 1 2 31 t 2 2 2t1 2 1 2 t 12 t 1 3t 2 cubic inch sec 4t 37 , 44 x x3 1 1 2 y 2 1 tan x t3 1 1 y csc x cot x sec x tan x csc x cot x 1 cos x 1 sin x csc x, f x sin x y xy sec x, g x cos x x 10 0 65. f x 10 xy 3 cos x 2 sin x 5 cos t y 3 sin x 2 sin x 4 sin t 64. 3 cos x y y sec tan sec2 12 x3 68. f x 12 3x2 2 1 x3 3x2 74 5 cos t ht 2 sin x y 4 sin t 4 34 9 x 4 fx sec2 6 sec f fx 4ex x 6t 3 tan f 67. f x 4ex 8 m sec f 61. y cos x 2 t2, 0 ≤ t ≤ 6 36 a4 1 sin x sin x 2 4xex fx x2 1 12 x 3 13 1 23 2x 3 x2 1 23 2x y sin x 9 4x7 4 R eview Exercises for Chapter 3 x x2 69. h x hx 2 2 3 1 x x2 x2 3 1 2x x2 3 x 71. f s fs s2 1 s2 1 11 x2 2 52 52 6x s3 3 1 3 1 x2 2x 1 s3 32 3s s2 1 32 8s3 52 s 2 5 5 s3 1 3s 32 1 2s h 2 1 25 y 1 1 1 2 sin 2x 1 2 75. y 4 cos x sin x 2 2 sin x cos x 2 1 3 6 2 cos2 x cos 2x y 4 sin x cos x csc 3x 77. y cot 3x 3 csc 3x cot 3x 3 csc2 3x 3 csc 3x cot 3x y csc 3x y 78. y sec7 x 7 sec5 x 5 79. y sec6 x sec x tan x sec4 x sec x tan x sec5 x tan x sec2 x y 1 y csc 2x cot 2x 2 sin 2x 4 1 2 1 cos 2x 2 4 fx 3 x2 1 81. y 12 3 x2 1 x2 1 82. y y 3x 1 2 x 2 x2 1 3x 2 1 12 2x y x2 1 1 x x 1 2 sin1 2 x cos x sin5 2 x cos x x sin2 x sin x 1 sin x sin x x2 x 2 cos x x 22 32 83. g t 1 sin x 1 2 72 sin x 7 3 32 cos x 1 x1 x 2 32 sin x 3 cos3 x 3x x2 1 80. f x sin2 x cos 2x cos x sec5 x tan3 x cos x 1 gt 11 2 1 sin x 1 cos x 1 t 2e t 4 12t te 4 4 csc 2x cot 2x x 2 1 1 2 1 csc 2x 1 2 0 76. y 1 6 1 74. y 2 31 1 5 1 9 sin 3x 1 4 1 x 72. h 3s2 s s2 y 5 x2 fx 5 1 3 cos 3x 3 2x 3 1 s s2 73. y x 12 5 1 x x2 70. f x 4 1 t4 te [t 4 2te t 8] 4 sin x 231 232 Chapter 3 84. h z Differentiation z2 2 e hz e2x 2x 1 2x e 2 z2 2 ze e e 2x 85. y y e2x 12 e 2e2x 3t 3e y 2x 2e 3e 86. y 2x 1 2 3t 9e 3 t2 2 3t t e2x e 2x e2x e 2x 87. g x gx x2 ex ex 2x x2 hx 1 2 ln x 1 x 2 x y 1 a b2 1 b b2 y 1 ax 1 ax b ln a a2 91. f x 2 bx ab bx 2 93. 1 x2 1 a 1 ax2 b b a2 a bx b a a2 x a bx x 2x 1 ax2 4 x3 3x2 6x 2x 1x The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. x2 ab bx 1 x 2 a 1 ln a a 1 x bx bx 2 ln x 1 xa bx 4 7t 2 f′ b ax a − 0.1 bx 1.3 f − 0.1 bx 2 a bx 0.1 1 4 a 5 1 tt 2 ln x 2 ln x The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 1 x x2 a 4x fx t2 t 1 a bx 1 b a a bx 97. f t ln x ln x 1 a bx ln a x y dy dx bx 1 x ln x b 1 b2 a bx ft a bx bx ax2 a bx 98. f x 95. x bx 12 1 ln a b2 y dy dx bx a x ln x 2 1 2 ln x 6 6x a ln a x ln x fx a bx b ln a2 x dy dx 2 2x 3 x2 2 a 1 2x gx 2 2x 2 ln x2 3 7x2 3x3 1 x ln x 4x 3x2 ln x 23 dy dx 96. 1 x2 x3 2 fx 94. ln x 1 1 ln x x2 cos 2 esin 2 f 1 ln x 2 ln x e xx x 1 x x 89. g x x e ln 92. f x x2ex 2x 90. h x 1 sin 2 e 2 88. f 2x 8 2 99. g x 2x x 8 gx 4 100 f′ −7 f 5 1 x x 12 2 1 32 g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines. 4 g′ −2 7 g − 60 −2 R eview Exercises for Chapter 3 100. g x x x2 1 12 gx 2x2 x2 1 1 101. f t ft t 1 12 t 13 1 t 1 233 56 5 6t 1 16 f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines. g does not equal zero for any value of x. The graph of g has no horizontal tangent lines. 5 5 f g′ −6 f′ 6 −2 7 g −1 −3 102. y y 3x x 2 3 103. y 2 2 7x 2 3x 3x 2 tan 1 3 y x −3 3 106. y sin 2x y 4x 2 cos 2x y 4 y 1 x y 4 sin 2x 2 x 2x 109. f t 107. fx fx 2 sec x sec x tan x csc2 x 2 csc x t 111. g 2 110. g x ft 2 cos 2x tan 3 t t 1 1 t3 gx ft y sin 2t 1 2 t4 gx 112. g 3 sec2 3 cos 1 g 18 sec2 3 tan 3 sin hx x x2 hx 1 csc x cot x 2 csc2 x cot x 2 sec2 x tan x 1 sin 2x cot x fx sec2 x 3 8 −4 tan x 2 x3 sin2 x y′ −1 y′ −4 2x2 2 sin x cos x x y 2 − 25 y x cot y y − 20 108. y csc3 10 5 y′ x The zero of y corresponds to the point on the graph of y where the tangent line is horizontal. y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. 75 105. y 2 csc3 104. y sec2 1 x 21 x y y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. x 2x2 x2 hx x 2x2 3 x2 1 3 2 1 1 1 1 6x x2 5 1 2 3x2 x2 2 6x3 5x 12 3 15x2 18x x2 1 3 5 234 Chapter 3 Differentiation x3 ln x 113. g x 114. f x 3x2 ln x gx gx 2x 3 ln x 700 t2 T 2x2 e 2x2 3 8x 3 x x2 x3 T 1 10 1, 1 T (b) When t 1400 1 2 4 10 2 (c) When t 18.667 deg hr. T 5, 25 3, 9 (d) When t 1400 5 20 2gh 2 10 3.240 deg hr. 2 2 32 h 1400 3 2 12 10 2 10, 1400 10 100 40 T 7.284 deg hr. 2 10 0.747 deg hr. 2 8h 4 dv dh h (a) When h 9, dv dh 4 ft sec. 3 (b) When h 117. (a) You get an error message because ln h does not exist for h 0. 4, (f) dv dh 2 ft sec. h 0.8627 25 0 5, p 20, p 6.4474 ln p 1 dp (implicit differentiation) p dh p 6.4474 For h 6.4474 ln p. 6.4474 For h (c) 0.8627 1 dp dh (b) Reversing the data, you obtain h 0.55 and 1 dp dh 0 (d) If p 0.75, h (e) If h 13 km, p 118. y x3 12 e 1400 t 2 4t 10 2 t2 v 12x 1 x3 5x 4t (a) When t 116. fx fx x2 3 ln x 1 6x ln x 115. T 1 x x3 6x2e 10 ln (a) 10 2.72 km. 100 x 20 0 10 0 ––CONTINUED–– 100 dp dh 0.00931 atmos km. As the altitude increases, the rate of change of pressure decreases. 0.15 atmosphere. x2 0.06 and 0.0853 atmos km. x2 10 ln 10 100 x2 ln x 100 x2 R eview Exercises for Chapter 3 118. ––CONTINUED–– (b) dy dx x 10 x2 10 100 10 10 100 x2 x2 10 10 100 x2 x 100 x2 x 100 x 100 x 10 5, dy dx x2 119. 100 x 3y2y y2 y 2x 2x 3x x2 0 2x xe y y ey ey 2x sin x2 xey ey x1 2y x 1 y 2 1 x 2e y 0 1y 4 y1 2x 4 2x 3 6y 1 xe y xy xe y y ey xy xe y xy y y y ey ye x y ex e y ye x xe y x 16 0 2 x y 2y y y 2x 2 xy x y 2y 2 xy y 2x y 2 xy y 2x y2 x3 x2y xy y2 0 x3 x2y xy 2y2 0 x 3y sin x2 xy 1 2y x x2 ye x ex yx 0 ye x y ex 123. 124. 4 3y y 2x sin x2 y 18yy 122. xe y y 12 4x 3 6y xe y sin x2 2x 1 x 2 9y2 3y y2 cos x2 y 19 9. 3y y 121. 9, dy dx 120. 3y 3xy x2 x2 10 3x 2x 0 10 x 3. When x y3 3xy x → 10 10 x 1 10 x x2 x2 When x x 100 dy dx (c) lim x2 10 x x2 100 x 100 10 x 100 x2 100 x2 x2 10 x 100 10 1 x x2 100 3x2 x2y 2xy 4y y 3x2 2xy y y 3x2 x2 2xy y x 4y 2y 2 xy x 2y x 2x y x sin y 125. x cos y y xy y 4yy yy xx y x cos y sin y cos x y y cos x y sin x y cos x y sin x sin y y sin x sin y cos x x cos y 235 236 Chapter 3 Differentiation cos x y x y sin x y 1 y sin x y 1 126. x2 127. y2 20 6 (2, 4) 1 2x 2yy 0 −9 sin x 1 y y y sin x y sin x y csc x 1 −6 1 2 At 2, 4 : y 1 y Tangent line: x 1 x 2 4 y2 2x 16 2yy 10 4 2x 2 y 0 129. 10 10 0, 2yy y ln x e, 1 0 4 y 2y −6 x y 5x 3y 5 x 3 3 16 x ln x −4 2y 5 At e, (e, −1) y y Tangent line: 1 1: y e 0 y 3x 3 5y 30 1 x e y 3 x 5 1 1 x e 1 ex e y Normal line: ex e2 1 ln x2 2 1 ln x Tangent line: y 5 0 Normal line: y 130. ln x 1 x y 1 y x, 0, 1 y 1 131. 1 y ln y y x x y e 1 x x2 1 x4 ln x 4 y y At 0, 1 : y y y 1 1 x y x x2 1 1 x4 x 0 Tangent line: y Normal line: x 0⇒y 1 0 Graph y 6 −10 5 3 At 5, 3 : y y x y ln x x y y y2 y ln x 0 −10 2 0 2x x2 2y Normal line: y 128. 9 x y ex x3 x x. 8 (0, 1) −7 5 −1 1 x x2 1 1 8x2 4 2 x 4 x x2 4 x2 1 1 1 x 4 R eview Exercises for Chapter 3 2x y 132. ln y 3 1 x 3 ln 2x y y 1 x 1 ln x x2 2 fx 1 12x3 45x2 x 32 xx f4 1 x 2 4x 6 2x x2 1 1 1 x 8x f4 1 1 2 1 f4 4 1 xx 2 3 x f x 1 0 139. y y 3 137. y 141. y x arcsin x 2x arcsin x 1 x2 y 142. y x2 x2 y 1 x2 1 x2 12 4 2x 4 6 4 3 f x2 3 4 6 arctan x2 138. y 12 1 y 32 2x x2 2x 2x2 1 arctan e2x 2 1 1 2e2x 2 1 e4x y 1 e2x e4x x2 arcsin x 21 arcsin x 1 x4 x2 1 1 x2 arcsec x 2 2 21 1 x2 x2 2x arcsin x 1 x2 arcsin x 2 x 2 arcsec , 2 < x < 4 2 x y sec2 x x2 1 140. y 2 x2 1 x2 x 2 1 x2 4 x 4 x2 x2 4 x 4 x2 4 x2 x 4 (a) When x 1 dx dx ⇒ dt 2 x dt 1, dx dt 4 units/sec. (c) When x 4, dx dt 8 units/sec. 2 units sec dy dt 1 dx , 2 dt (b) When x x dy dt 2x dy dt 23 3 3 x tan arcsin x 1 1 1 3 0.160 6 3 3 1 f y x x2 3 tan x f x arcsec x x 23 2 fx 1 3 e0 1 fx ex f 23 2 1 35 3 17 ex 1 1 3 13 12 ln x 1 1 x 3 3 f fx f 1 f 135. 3 2 3 3 x x x 1 4 fx x 1 f 3 f 1 2 fx 1 x2 3 3 1 2x 143. 2 ln x2 x3 1 133. x2 1 2x 136. 2 4x 2x 134. 1 1 6 y f x2 3 4x 2 2 units/sec. 1 1 2 2 237 238 Chapter 3 Differentiation surface area dx 12x dt 1 h 4 5 dA dt 12 2 6x 2. A dx dt 145. s h s 144. Let x be the length of an edge. Then dV 1 dt Width of water at depth h: 270 cm2 sec 12 4.5 5 w 2 V 5 2 2 dV dt 5 4 2 dh dt tan 146. 32 dx dt 147. st 1 1 dx , 2 dt 6 x2 16 s 1 4 6 1 15 km min 2 4.9t2 dx dt h 1, h 5 8 4 hh dh dt 1 2 1 2 2 s 2 h dh dt 2 m min. 25 2 θ 148. (a) A s (t) 60 30˚ 4.9t2 dA (b) dt 25 xt h 2 450 km hr. 4.9t2 35 tan 30 h 2 1 9.8t t 4 x tan2 60 st rad min dx dt d dt When x 4 2 x d dt sec2 2 1 h 4 2 dV dt 54 h When h 2s 2x x2 2 xe 2x2 1 3 2e For x 3s t 3 38.34 m sec 9.8 5 4.9 2 and dA dt st xt ds dt 2xe 2e x2 2 x2 2 dx dt x(t ) 5 4.9 3 base height 6e 2 dx dt 4 x2 2 dx dt 4, 24 e2 3.25 cm2 min. R eview Exercises for Chapter 3 x3 149. f x 3x 1 y From the graph you can see that f x has three real zeros. 3x2 fx 1 x 3 −2 −1 1 2 1 2 −1 n f xn xn f xn f xn f xn xn f xn f xn 1 1.5000 0.1250 3.7500 0.0333 1.5333 0.0049 4.0530 0.0012 1.5321 f xn f xn f xn f xn −3 1.5333 2 −2 xn n xn f xn f xn 1 0.5000 0.3750 2.2500 0.1667 0.3333 2 0.3333 0.0371 2.6667 0.0139 0.3472 3 0.3472 0.0003 2.6384 0.0001 0.3473 f xn f xn f xn f xn n xn xn f xn f xn 1 1.9000 0.1590 7.8300 0.0203 1.8797 2 1.8797 0.0024 7.5998 0.0003 1.8794 The three real zeros of f x are x x3 150. f x 2x 1.532, x 0.347, and x 1.879. 1 y From the graph you can see that f x has one real zero. 3x2 fx f changes sign in 3 2 2 1 1, 0 . x −2 f xn xn n f xn f xn f xn xn f xn f xn 1 0.5000 0.1250 2.7500 0.0455 0.4545 0.0029 2.6197 0.0011 −1 0.4545 2 −1 0.4534 On the interval 1, 0 : x xe x gx 4 x 151. g x 0.453. 1 ex 152. f x 3 fx From the graph, there is one zero near 1. Newton’s Method gives: x ln x 1 ln x From the graph, there is one zero near 3. Newton’s Method gives: x1 1 x1 3 x2 1.23576 x2 2.8590 x3 1.20297 x3 2.8574 x4 1.20217 x4 2.8574 To three decimal places, x 1.202. To three decimal places, x 2.857. 239 240 Chapter 3 Differentiation x4 153. Find the zeros of f x 4x 3 fx x 3. y 4 1 3 2 From the graph you can see that f x has two real zeros. f changes sign in xn n 2, 1 x 1. −4 −3 −2 f xn f xn f xn f xn f xn f xn xn 1 1.2000 0.2736 7.9120 0.0346 1.1654 0.0100 7.3312 0.0014 2 3 4 −3 −4 1.1654 2 1 1.1640 On the interval 2, 1: x 1.164. f changes sign in 1, 2 . f xn f xn f xn 0.5625 12.5000 0.0450 1.4550 1.4550 0.0268 11.3211 0.0024 1.4526 1.4526 0.0003 11.2602 0.0000 1.4526 n xn 1 1.5000 2 3 f xn On the interval 1, 2 : x 154. Find the zeros of f x fx cos x f xn f xn xn 1.453. sin x x 1. y 6 1 4 From the graph you can see that f x has three real zeros. 2 f xn f xn f xn f xn f xn f xn 0.2000 0.2122 3.5416 0.0599 0.2599 2 0.2599 0.0113 3.1513 0.0036 0.2635 3 0.2635 0.0000 3.1253 0.0000 0.2635 n xn f xn 1 1.0000 0.0000 n xn 1 n xn 1 xn f xn f xn f xn 2.1416 0.0000 f xn f xn f xn f xn 1.8000 0.2122 3.5416 0.0599 1.7401 2 1.7401 0.0113 3.1513 0.0036 1.7365 3 1.7365 0.0000 3.1253 0.0000 1.7365 The three real zeros of f x are x 0.264, x xn f xn f xn 1.0000 xn 1, and x f xn f xn 1.737. −π π −4 −6 2π x Problem Solving for Chapter 3 241 Problem Solving for Chapter 3 1. (a) x 2 y 2 r r 2 Circle x2 3 y Parabola y Substituting: 2 r r2 y r2 y 0 −3 y2 2r y r2 y2 2r y y yy 2r 1 3 −1 0 Since you want only one solution, let 1 1 . Graph y 2 0⇒r 2r x 2 and x 2 y 1 2 2 1 . 4 (b) Let x, y be a point of tangency: x2 y b 1 ⇒ 2x 2 x2 ⇒ y y 2y x 0⇒y by b circle y 2x (parabola) Equating: 3 x 2x b y −3 2b y 3 1 −1 b 1 ⇒b 2 y Also, x 2 y y 2 b 2 b 1⇒y Center: 0, x 2 and x 2 y 1 2 y 2 1 4 1⇒y 1⇒y 3 and b 4 5 . 4 5 4 Graph y x 2 imply 1 and y y y 1 2 5 4 y 2 1. 2. Let a, a2 and b, b2 2b 5 be the points of tangency. For y 2b 2 ⇒ a y x 2 2x 5, y 2x 2. Thus, 2a Furthermore, the slope of the common tangent line is a2 b2 2b ab 5 1 b 2 1 ⇒ 1 2b b2 b 2b b 5 b2 1 b2 2b 2b ⇒ 2b2 4b 6 4b2 ⇒ 2b 2b 4 ⇒ b2 b ⇒b 2b b y 2, a 1 1 2x For b y 1, a 4 4x 2, −8 −6 −4 −2 −4 −6 5 2b 2 6b 2 2 0 1 0 1 1 and the points of tangency are b 1 2 y 10 8 6 4 0 2 For b 2b x 2, y 2x and for b 1, or a 1 b. 1 ⇒y b 2x 1, 1 , 2, 4x 4 2. 1 2 and the points of tangency are 2, 4 and 2 ⇒y 5 . The tangent line has slope 1, 8 . The tangent line has slope 4. x 2 4 6 8 10 242 Chapter 3 Differentiation cos x P1 x a0 f0 1 P1 0 cos x P2 x a0 1 f0 1 P2 0 a0 ⇒ a0 1 0 a0 ⇒ a0 f0 0 a1 ⇒ a1 P1 x f0 0 0 1 P 1 0 a1x f0 1.0 0.1 cos x 0.5403 P2 x 0.5 x 0.001 0 0.001 0.9950 1 1 0.9950 1 1 P2 x is a good approximation of f x (d) 0.1 0.9950 0.9950 0 a1 ⇒ a1 P 2 0 2a2 ⇒ a2 1 2 12 2x 0.5403 1 2 1.0 1 1 a1x P 1 P2 x (c) a2x 2 fx (b) fx 3. (a) 0.5 cos x when x is near 0. fx sin x P3 x a0 f0 0 P3 0 a0 ⇒ a0 0 f0 1 P 3 0 a1 ⇒ a1 1 f0 0 P 3 0 2a2 ⇒ a2 0 3 0 6a3 ⇒ a3 f 0 P3 x 4. (a) y 1 x x 2, y P a1x a2 x 2 a3x3 1 6 13 6x 2x. Slope Tangent line: y 4 4x y 4x (b) Slope of normal line: 4 1 x 4 2 x2 x2 ⇒ 4x 2 x 18 ⇒ 4x (c) Tangent line: y 1 x 2a 9x 1 x 2a a2 a a2. a2 1 2 1 4a x 0 2 x 9 4 9 81 , 4 16 0; Normal line: x 1 4a ±a 1 4a a x 0 1 16a2 1 4a a x 2 a2 a 1 2 1 16a2 x2 9 2 9 2 Second intersection point: a2. a 1 2a x 1 x 2a x2 1 x 4 2, x 2. To find points of intersection, solve: 1 x 4 x 2a x Normal line at a, a2 is y 4 1 4 y 0, be a point on the parabola y Tangent line at a, a2 is y 2 y Normal line: y (d) Let a, a2 , a 4 at 2, 4 . 1 4a 1 4a 1 ⇒x 4a a a point of tangency 1 4a 0 ⇒x 2 a 1 2a The normal line intersects a second time at x 2a2 1 . 2a 2a2 1 2a P roblem Solving for Chapter 3 Ax3 5. Let p x Bx 2 3Ax 2 px Cx 2Bx D C. D At 1, 1 : A B C 3A 2B C A B C 3A 2B C At 1, 3: 1 Equation 1 14 Equation 2 D Adding Equations 2 and 4: 6A a fx 2 2C 2C Subtracting Equations 2 and 4: 4B 6. f x Equation 4 2D Subtracting Equations 1 and 3: 2A 1 Hence, B 4 and D 2 you obtain 4A 8 ⇒ A Equation 3 2 Adding Equations 1 and 3: 2B 3 4 12 16 2 2B 5. Subtracting 2A 2C 4 and 6A 2C 2. Finally, C 1 4 2A 0. Thus, p x 2x3 4x 2 2 b cos cx bc sin cx At 0, 1 : a b 1 Equation 1 3 , :a 42 b cos c 4 3 2 Equation 2 bc sin At c 4 1 Equation 3 From Equation 1, a 1 Equation 2 becomes 1 1 c sin c b. b b cos 1 c cos c sin c 4 4 4 1 cos 3 ⇒ 2 b 1 ,a 2 2, b x4 a2 x 2 a2 y 2 a2x 2 cos 3 ⇒fx 2 3 2 c 4 Graph y1 1 . 2 1, you see that many values of c will work. 1 cos 2x 2 x4 ± c 4 1 c c sin 2 4 c 4 a 2 y2 y b cos 1 2 1 c c sin 2 4 Graphing the equation g c One answer: c c 4 1 . Thus: c sin c 4 From Equation 3, b 7. (a) 12, 5. a2x 2 a a2x 2 a ––CONTINUED–– x4 x4 and y2 a2x 2 a x4 . 243 244 Chapter 3 Differentiation 7. ––CONTINUED–– (b) 2 a= 1 2 ± a, 0 are the x-intercepts, along with 0, 0 . −3 3 a=2 a=1 −2 (c) Differentiating implicitly: 4x 3 2a2 x 2a2 y y 2a2x 4x3 2a2y y 2 a2 a2 2 a4 4 2x 2 2 ay 0 ⇒ 2x 2 a2 ⇒ x ±a 2 a2y2 2 a4 2 a2y 2 a2 x a2 a4 4 a2 4 a ± 2 y2 y Four points: 8. (a) b2y 2 x3 a x3 y2 a2y 2 aa ,, 22 a , 2 a , 2 aa ,, 22 x ; a, b > 0 x3 a b x x3 a b and y2 (c) Differentiating implicitly: y 3x 2 a x3 x 3ax 2 4x3 2b2y ⇒ 3ax 2 3a 3ax 2 4x 3 0 4x3 4x x 3a 4 b2y 2 y2 3a 4 3 a 27a 4 ⇒y 256b 2 Two points: 27a3 1 a 64 4 3a 4 ± a 2 (b) a determines the x-intercept on the right: a, 0 . b affects the height. ax b2 Graph y1 2b 2 y y a , 2 3 3a2 16b 3a 3 3a2 3a , , , 4 16b 4 3 3a2 16b x . P roblem Solving for Chapter 3 9. (a) Line determined by 0, 30 and 90, 6 : y (0, 30) 30 y 30 6 x 0 90 30 (90, 6) (100, 3) When x x 90 100 24 x 90 0 4 100 15 100, y 4 x⇒y 15 4 x 15 30 10 > 3 ⇒ Shadow determined by man. 3 30 Not drawn to scale (b) Line determined by 0, 30 and 60, 6 : y 30 (0, 30) y (60, 6) (70, 3) x 60 When x 70 30 6 x 0 60 30 2 70 5 70, y Not drawn to scale (c) Need 0, 30 , d, 6 , d 30 0 6 6 d d 2 x⇒y 5 0 2 x 5 30 2 < 3 ⇒ Shadow determined by child. 30 10, 3 collinear. 3 d ⇒ 10 24 d 3 ⇒d 10 80 feet (d) Let y be the distance from the base of the street light to the tip of the shadow. We know that For x > 80, the shadow is determined by the man. y 30 y x dy 5 x and 4 dt ⇒y 6 5 dx 4 dt 25 . 4 For x < 80, the shadow is determined by the child. y 30 y x 3 10 10 x 9 ⇒y 100 dy and 9 dt 10 dx 9 dt 50 . 9 Therefore, 25 , 4 50 , 9 dy dt x > 80 0 < x < 80 dy is not continuous at x dt y x1 3 ⇒ 1 10. (a) 1 8 3 23 dx dt (b) D dy dt 1 x 3 80. 23 dx dt dx dt 12 cm sec x2 y2 ⇒ dD dt 12 x 2 x y 2 2x 2y dy dt dx dy y dt dt x2 y2 8 12 21 64 4 —CONTINUED— dx dt 98 68 49 cm sec. 17 dx dt 5. 245 246 Chapter 3 Differentiation 10. —CONTINUED— y ⇒ sec2 x (c) tan x dy dt d dt y dx dt 68 x2 2 θ y 11. ln x 16 68 12. b 1 x a If x 0, c b ea x a y a ex e ax ae a b, 0, eax aea b. If y b 1, b Tangent line 1. Thus, b c bx b b 1 1. ab b, x a a 0.0001 0.0174524 0.0174533 0.0174533 d sin z dz lim sin z z z sin z cos z z →0 lim z →0 cos z sin z z lim z →0 sin z 0 14. (a) v t at (b) v t S5 27 5t 27 5 cos z 1 lim z →0 180 27 5 2 sin 180 180 . 90 cos d Sz dz sin z cos z z d sin cz dz 180 sin 180 1 2 1 c cos cz 180 Cz (e) The formulas for the derivatives are more complicated in degrees. cos z 15. j t ft sec sin z z C 180 sin z 2 0.0174533 z→0 (d) S 90 sin z z 180 sin z z In fact, lim sin z 27 ft sec 27 5t 27 10 cos z 1 always. z→0 0.01 sin z z (c) c (b) lim 0.1 ea b 1 Thus, a z (degrees) Tangent line 1 c 13. (a) 8 ex y y 1 x a y 4 rad sec. 17 y 1 x y y 81 2 12 64 68 64 68 d . Hence 8 dt From the triangle, sec at (a) j t is the rate of change of acceleration. 0⇒ 27 5 27 5t 6 27 ⇒ t 5 seconds 73.5 feet (c) The acceleration due to gravity on Earth is greater in magnitude than that on the moon. 8.25t 2 66t vt 16.5t 66 at 16.5 (b) s t at jt 0 The acceleration is constant, so j t 0. (c) a is position. b is acceleration. c is jerk. d is velocity. ...
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This note was uploaded on 11/13/2010 for the course MATH MAT 231 taught by Professor Thurber during the Spring '08 term at Thomas Edison State.

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