Homework Solution Chapter 2

Homework Solution Chapter 2 - C H A P T E R 2...

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Unformatted text preview: C H A P T E R 2 Differentiation Section 2.1 Section 2.2 Section 2.3 Section 2.4 Section 2.5 Section 2.6 The Derivative and the Tangent Line Problem . . . . . 95 Basic Differentiation Rules and Rates of Change . . 109 Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . . . 120 The Chain Rule . . . . . . . . . . . . . . . . . . . . 134 Implicit Differentiation . . . . . . . . . . . . . . . . 147 Related Rates . . . . . . . . . . . . . . . . . . . . . 161 . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 C H A P T E R Differentiation Section 2.1 2 The Derivative and the Tangent Line Problem 0. 5 2. 5 2. 1. (a) At x1, y1 , slope At x2, y2 , slope (b) At x1, y1 , slope At x2, y2 , slope 2. (a) At x1, y1 , slope At x2, y2 , slope (b) At x1, y1 , slope At x2, y2 , slope f 4 4 f 4 4 f 1 1 f 3 3 f 4 4 5 3 5 2 3. 2 5. 4 3. 5 4. 2. 3. (a), (b) y y 6 5 4 3 2 1 f )4) 4 f )1) )x 1 4. (a) 1) 2 4.75 1 1 0.25 f 3 . 3 f )1) x 1 f )4) 5 )4, 5) f )4) f )1) )1, 2) x 1 2 3 4 5 6 f )1) 3 Thus, f 1 f 4 > 1 4 2 (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, f 1) f 4 4 f 1 < f 1. 1 (c) y f 4 4 3 x 3 1x x 1 f 1 x 1 1 1 2 2 1 5. f x 3 2x is a line. Slope g1 1 1 2 x x x 2 2 g1 4 x x x x t t t t f 0 2 6. g x 3 2x 1 is a line. Slope g2 5 5 4 h 2 4 4 2 2 x x 3 2 7. Slope at 1, 3 lim lim lim x0 8. Slope at 2, 1 3 lim lim g2 x 2 x0 1 x 2 x0 x0 x 4 4 x t t t 4 t t t 4 2 x 2 2 1 x0 lim x x 4 h 3 t 2 1 x0 lim 2 x0 lim x0 9. Slope at 0, 0 lim lim f 0 3 t t t0 10. Slope at 0 2, 7 lim lim 2 7 4 t0 t0 t0 t lim 3 t0 3 lim lim t0 t0 95 96 Chapter 2 3 lim lim f x 3 x Differentiation 12. g x x x 3 0 f x g x 5 lim gx 5 x 0 x 0 x x 5 gx 13. f x f x 5x lim f x 5x 5 5 x0 11. f x f x x0 x0 x x x x f x 5x x0 lim x0 lim lim x0 lim 0 x0 lim x0 x0 14. f x f x 3x lim 2 f x 3x x x x f x 15. h s h s 2 x 3x 2 3 lim 2 s 3 hs 3 2 3 2 3 s0 x0 s s s 2 3 hs s s 3 2 3s lim x0 lim s0 3 x lim x0 x lim 3 x0 lim 3 s0 s s 16. f x f x 9 lim 1 x 2 f x 9 1 2 x f x 2x 2x 2 4x x 1 x x x 2 x0 x x 1 2 x 1 2 f x x x 9 1 2x lim x0 lim x0 17. f x f x 2x2 lim f x x 2 2 x0 lim x x x x 2 1 x x 2x 2 x 1 x 1 2x 2 x 1 x0 lim lim 4x x 2 x x x0 x0 lim 4x x0 2 x 1 4x 1 18. f x f x 1 lim x2 f x x x x x2 2x x x x f x 2 x0 lim 1 1 1 x x 2 x2 1 x2 x0 x 2x x x 2 lim lim x0 x0 lim x0 2x x 2x Section 2.1 19. f x f x x3 lim x0 The Derivative and the Tangent Line Problem 97 12x f x x x3 x x x 3 f x 12 x x 3x x 2 lim lim lim x x 2 x3 x 3 12x 12x 12 x x3 12x x0 3x 2 x 3x 3x x x0 x 3x 2 x x x x 2 3 12 x 3x 2 12 x0 lim 3x 2 x0 12 20. f x f x x3 lim x0 x2 f x x x3 x x x 3 f x x x 3x x 2 lim lim x x 2 x3 x 3 x2 x2 x 2x x x 3x 2 2 x0 3x 2 x 3x 3x x 2 x 2 x3 x2 x0 lim 3x 2 x x x x 2 3 2x x x x0 lim 3x 2 x0 2x 2x 21. f x f x 1 x lim 1 f x 1 x 1 xx xx x 1 x 1 2 22. f x x x 1 x x x x x x 1 x 1 1 1 1 x f x 1 x x 1 x 1 1 1 f x 1 x2 lim f x 1 lim lim lim lim x x0 x0 x0 x x x2 x f x 1 x2 lim lim lim lim x x x0 x0 x2 x xx 2x x xx 2x x x2 x 2x2 x2 x 2x2 x0 x0 x0 x0 x x 2x2 x0 2x x4 2 x3 23. f x f x x lim lim lim lim x0 1 f x x x x x x x x x x x x 1 1 f x 1 x 1 1 x x x x 1 1 1 1 1 x 1 x 1 1 2 x 1 x x x x 1 1 x x 1 1 x0 x0 x0 98 Chapter 2 4 x lim f x x0 Differentiation 24. f x f x 25. (a) f x x x 4 x x f x 4 x x x 4x x x x x 4 x 2 x x (b) x x x x x x x x x x x f x x2 lim lim lim 1 f x x 2x x x x 2x 22 4. x x x 2 f x 1 x x2 1 x0 lim lim lim lim x x0 x0 x 2 x0 4 x 4 x x0 x x x x0 lim 2x x0 4x x x x x x 4 x At 2, 5 , the slope of the tangent line is m The equation of the tangent line is y y 5 5 y 4x 4x 4x 8 2 8 3. x0 x x (2, 5) -5 -2 5 26. (a) f x f x x2 lim x0 2x f x x 2x x 1 x x x 2 (b) f x 2x x x x 2 5 (-3, 4) lim x x 2 x 2x 2 1 x2 2x 1 -6 -1 3 x0 lim x0 lim 2x x0 2 3, 4 , the slope of the tangent line is m At The equation of the tangent line is y 4 y 4x 4x 3 8. 2 3 2 4. 27. (a) f x f x x3 lim f x x 3x 2 x 3x 2 x0 (b) x x x3 x 3x x 3x x x 2 10 (2, 8) f x -5 5 -4 lim lim x3 x 2 x0 x 3 x0 lim x0 3x 2 32 2 At 2, 8 , the slope of the tangent is m The equation of the tangent line is y 8 y 12 x 12x 2 16. 12. Section 2.1 28. (a) f x f x x3 lim x0 The Derivative and the Tangent Line Problem (b) 4 99 1 f x x x3 x x x3 3x 2 3x x x f x (1, 2) -6 6 lim lim x0 1 x 3x x3 x x 2 2 1 -4 x x 3x 2 3 1 x3 1 x0 x0 lim 3x 2 At 1, 2 , the slope of the tangent line is m The equation of the tangent line is y 2 y 3x 3x 1 1. 31 2 3. 29. (a) f x f x x lim lim lim f x x x x x x 1 x x0 (b) x x x x x x x f x -1 3 (1, 1) 5 -1 x x x x0 x x x x x x x0 lim 1 2 x x0 At 1, 1 , the slope of the tangent line is m 1 2 1 1 x 2 1 x 2 x lim lim lim x0 1 . 2 The equation of the tangent line is y 1 y 1 1 . 2 (b) x x x x x x x x x x 1 1 1 x 1 1 x f x -2 10 4 30. (a) f x f x 1 f x (5, 2) x x x 1 1 x 1 1 x0 x x x x 1 1 x x 1 1 -4 x0 lim x0 1 2 x 1 At 5, 2 , the slope of the tangent line is 1 1 2 5 1 4 The equation of the tangent line is m y 2 y 1 x 4 1 x 4 5 3 4 100 Chapter 2 4 x lim Differentiation 31. (a) f x f x x (b) f x x x x f x 4 x x xx x3 x2 x2 xx 4 1 x x 2x 2 x x x x x x x x x x 4 x2 4 2 10 (4, 5) -12 12 x0 x lim lim lim lim lim x2 x2 x0 x x x x 4 x x x 4x 4 x x -6 x0 4x x 2 x x x x x3 x x2 x0 x x2 x x x 4 x x x0 x0 At 4, 5 , the slope of the tangent line is m 1 4 16 3 . 4 The equation of the tangent line is y 5 y 3 x 4 3 x 4 1 x lim 1 f x 1 x 1 xx x 1 x 1 2 4 2. 32. (a) f x f x 33. From Exercise 27 we know that f x slope of the given line is 3, we have x x 1 x x x 1 x 1 x 1 f x 1 x x 1 x 1 1 1 3x 2 x 3 1. 3x2. Since the x0 lim lim lim x x x0 1, 1 the tangent Therefore, at the points 1, 1 and lines are parallel to 3x y 1 0. These lines have equations y 1 y 3x 3x 1 2 and y 1 y 3x 3x 1 2. x0 x0 At 0, 1 , the slope of the tangent line is m 0 1 1 2 1. x 1. The equation of the tangent line is y (b) (0, 1) -6 3 3 -3 Section 2.1 34. Using the limit definition of derivative, f x 3x 2 x 2 The Derivative and the Tangent Line Problem 101 3x 2. Since the slope of the given line is 3, we have 3 1x 1. Therefore, at the points 1, 3 and y 3 y 3x 3x 1 and y 1, 1 the tangent lines are parallel to 3x 1 y 3x 3x 1 4. y 4 0. These lines have equations 35. Using the limit definition of derivative, f x 1 2x x . 1 2, 36. Using the limit definition of derivative, f x we have 2x 1 1 3 2. Since the slope of the given line is 1 2x x x 1. 1 2 Since the slope of the given line is 2x 1 1 32 1 2, we have 1 2 x x 1 32 1 1 Therefore, at the point 1, 1 the tangent line is parallel to x 2y 6 0. The equation of this line is y y 1 1 y 1 x 2 1 x 2 1 x 2 1 1 2 3 . 2 1 Matches (b). 1 x 2. At the point 2, 1 , the tangent line is parallel to x 2y 7 0. The equation of the tangent line is y 1 y 1 x 2 1 x 2 2 2. 37. f x 39. f x x f x x f x 38. f x x2 f x 2x Matches (d). Matches (a). 40. f does not exist at x 0. Matches (c). decreasing slope as x 41. g 5 g 5 2 because the tangent line passes through 5, 2 . 2 5 0 9 2 4 1 2 44. The slope of the graph of f is 0 f x 0. y y 4 3 2 1 2 4 42. h 1 h 4 because the tangent line passes through 1 6 3 4 1 2 4 1 2 1, 4 . 43. The slope of the graph of f is 1 f x 1. y 45. The slope of the graph of f is negative for x < 4, positive for x > 4, and 0 at x 4. f x f x -2 -1 -1 -2 1 1 2 3 2 f 2 x -6 -4 -2 -3 -2 -1 -2 -4 -6 -8 2 4 6 -1 -2 102 Chapter 2 Differentiation 1 47. Answers will vary. Sample answer: y y 4 3 2 4 3 2 1 x 1 2 3 4 -4 -3 -2 -2 -3 -4 x 1 2 3 4 46. The slope of the graph of f is for x < 4, 1 for x > 4, and undefined at x 4. y 3 2 1 x 1 2 3 4 5 6 x 48. Answers will vary. Sample answer: y y x f 1 -4 -3 -2 -1 -1 -2 -3 -4 49. f x 53. f 0 f x 5 3x and c 1 3, 50. f x x 3 and c 2 51. f x x 2 and c 6 52. f x 2 x and c 0; f x > 0 if 9 2 and f x < x < 54. f 0 4, f 0 0; f x < 0 for x < 0, f x > 0 for x > 0 f x x2 y 12 55. f 0 0; f 0 x 0 f x x3 y 3x y 2 4 f 3 2 1 -3 -2 f 2 1 -3 -2 -1 x -1 -2 -3 2 3 10 8 6 4 x -1 -2 1 2 3 f -6 -4 -2 2 x 2 4 6 -3 56. (a) If f c 3 and f is odd, then f c f c 3. (b) If f c 3 and f is even, then f c f c 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 2x. The slope of the line through 2, 5 and x 0, y0 equals the derivative of f at x 0: 5 2 5 5 4x 0 y0 x0 y0 x0 2 58. Let x 0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x 2x. The slope of the line through 1, 3 and x 0, y0 equals the derivative of f at x 0: 3 y0 1 x0 2x 0 1 2x 0 0 0 x0 3, 1 x 0 2x 0 2x 02 4 2 8 x0 2 2x 0 x0 4 8x 0 4x 0 1 x0 2x 0 2x 0 3 3 x0 1, 3 2 3 3 x 02 x0 2x 0 3 x0 y0 x 02 3 1 0 0 x0 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: y 5 y y 7 6 5 4 3 2 1 -2 x 1 2 3 6 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: y 3 y 6x 6x y 10 2x 2x 2 1 y 5 y 2x 2x 2 9 1 9 y 3 y 2x 2x 1 1 (2, 5) (3, 3) (1, 3) (-1, 1) 8 6 4 (3, 9) -8 -6 -4 -2 -2 -4 x 2 4 6 (1, -3) Section 2.1 59. (a) g 0 (b) g 3 0 8 3, The Derivative and the Tangent Line Problem 103 3 (c) Because g 1 (d) Because g 4 g is decreasing (falling) at x g is increasing (rising) at x 1. 4. g 4 > 0. 2. g x x x x3 x 3x 2 x g x x3 3x x x 3x 2 3x x 3x x x 2 7 3, (e) Because g 4 and g 6 are both positive, g 6 is greater than g 4 , and g 6 (f) No, it is not possible. All you can say is that g is decreasing (falling) at x 60. (a) f x f x x2 lim lim f x x x2 x0 (b) g x x x x2 x 2x x x x 2x x x 1 2x 1 2x At x 2 and the tangent line is or y 2x 1. 0. 2x 1. y 2 lim x0 f x lim x2 lim x 2 x0 x x3 x0 x 2 x 3 x3 lim lim x2 x0 x0 lim x x 2 x x0 x0 lim 3x 2 x0 3x 2 lim 2x x0 1, g 1 3x 0, g 0 1, g 1 1 3x 2 1 1 3 and the tangent line is or y 3x 2. 0. At x y At x At x 1 1, f 0, f 0 1, f 1 y At x At x 0 and the tangent line is y 3 and the tangent line is 1 or y 3x 2. 0 and the tangent line is y 2 and the tangent line is y -3 3 -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 1 3 4x 3 2 4x . -2 2 By the limit definition of the derivative we have f x x f x f x 2 2 3 1.5 27 32 27 16 3 4 2 1 1 4 0.5 1 32 3 16 0 0 0 0.5 1 32 3 16 1 1 4 3 4 1.5 27 32 27 16 2 2 3 -2 62. f x 1 2 2x 3 By the limit definition of the derivative we have f x x f x f x 2 2 2 1.5 1.125 1.5 1 0.5 1 0.5 0.125 0.5 0 0 0 0.5 x. 1 0.5 1 1.5 1.125 1.5 2 2 2 -2 -1 2 0.125 0.5 104 Chapter 2 f x 2x 2 3 Differentiation 0.01 0.01 0.01 f x x 0.01 2 63. g x 64. g x 2x x 2 100 8 f x 3 x 0.01 0.01 0.01 f x 3 x 100 2x 0.01 f g f -2 -1 4 The graph of g x is approximately the graph of f x 2 2x. g -1 -1 8 The graph of g x is approximately 3 the graph of f x . 2 x 65. f 2 f 2 24 3.99 2.1 2 4 2 4, f 2.1 0.1 2.1 4 Exact: f 2 2.1 0 3.99 66. f 2 f 2 1 3 2 4 2, f 2.1 2.31525 3.1525 Exact: f 2 3 2.31525 2 2.1 2 x3 4 6 67. f x 5 1 x and f x . 2x 3 2 As x 5 1 68. f x 3x and f x 3 2 x 4 3 f f -2 , f is nearly horizontal 0. -9 9 f -5 and thus f f -6 69. f x S x 4 x x f 2 4 2 1: S 0.5: S 0.1: S 3 2 x x x x x f 2 3 2 x 3 3 2 2 2 x x 3 3 f 2 2 1 3 x 2 19 x 10 4 5 -2 3 1 x x 1 2 x 2 3 x 5 2 x S 0.1 2 3 (a) x x x x 2 x 3 x 2 19 x 10 S1 -1 f 7 x S 0.5 (b) As x 0, the line approaches the tangent line to f at 2, 3 . 1 x f 2 22 x x x2 22 x 70. f x x S x x f 2 2 x 5 x 6 4 x 5 16 x 21 2 x 52 x 2 5 2 5 2 2 5 2 2 x f 2 x 5 x 6 4 x 5 16 x 21 2 5 6 9 10 41 42 5 2 x 1 2 x 2 x 22 x 3 x x 5 2 x 2 5 2 5 2 2 (a) x 1: S 4 x x (b) As 0.5: S 0.1: S x 2 -6 S0.1 S0.5 S1 f -4 6 x x 0, the line approaches the tangent line to f at 2, 5 . 2 Section 2.1 71. f x f 2 x2 x2 The Derivative and the Tangent Line Problem 105 1, c f x x 1 gx x 2x 2 lim f x 2 f 2 2 x2 g1 1 1, c x 1 f 1 1 0 g0 0 x0 x1 x2 lim lim x2 x 1 x2 x x 1 2 3 x 2 lim x 2 x x 2 2 x2 lim x 2 4 72. g x g 1 xx x1 x, c x1 lim lim 0 1 x1 lim xx x 1 1 x1 lim x 1 73. f x f 2 x3 2 f 2 2 lim x3 2x 2 1 x 2 1 x x 2 x 2 lim x2 x 2 x 2 2 x 2 lim x 2 4 74. f x f 1 x3 x1 2x, c f x x lim lim x3 x 2x 1 3 x 1 lim x 1 x2 x x 1 3 x1 lim x 2 x 3 5 75. g x g 0 x ,c x0 lim gx x x x x x lim x x . . . Does not exist. As x 0 , As x 0 , 1 x 1 x 76. f x f 3 1 ,c x lim 3 f x x 6 2 3, x3 f 3 3 c f 6 6 6 x3 lim 1 x x 1 3 3 x3 lim 3 3x x x 1 3 x3 lim 1 3x 1 9 77. f x f 6 x x6 lim f x x x6 lim x 623 x 6 0 x6 lim x 1 6 1 3 Does not exist. 78. g x g 3 x 3 lim 1 3 ,c g 3 3 3 x 3 x 3 gx x lim x 313 x 3 0 x 3 lim x 1 3 2 3 Does not exist. 79. h x h 5 x 5 ,c x 5 5 h 5 5 x 5 lim hx x lim x x 5 5 0 x 5 lim x x 5 5 Does not exist. 80. f x f 4 x x4 4 ,c f x x 4 f 4 4 x4 lim lim x x 4 4 0 x4 lim x x 4 4 81. f x is differentiable everywhere except at x (Discontinuity) 1. Does not exist. 106 Chapter 2 Differentiation 3. 82. f x is differentiable everywhere except at x (Sharp turns in the graph) 84. f x is differentiable everywhere except at x (Discontinuities) 86. f x is differentiable everywhere except at x 83. f x is differentiable everywhere except at x (Sharp turn in the graph) 85. f x is differentiable on the interval 1, (At x 1 the tangent line is vertical.) . 3. 2. 0. (Discontinuity) 2x is differentiable x 1 for all x 1. f is not defined at x 1. (Vertical asymptote) 6 87. f x x 3 is differentiable for all x 3. There is a sharp corner at x 3. 5 88. f x 89. f x all x at x x 2 5 is differentiable for 0. There is a sharp corner 0. 5 -6 -7 -1 2 -6 -2 6 -3 6 90. f is differentiable for all x 1. f is not continuous at x 1. 3 91. f x x 1 The derivative from the left is x1 lim f x x f 1 1 x1 lim x x 1 1 0 1. -4 5 The derivative from the right is x1 -3 lim f x x f 1 1 x1 lim x x 1 1 0 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 x1 lim f x x f 1 1 x1 lim 1 x x2 1 0 x1 lim 1 x x2 1 1 1 x2 x2 x1 lim 1 1 x x2 . (Vertical tangent) 1. The limit from the right does not exist since f is undefined for x > 1. Therefore, f is not differentiable at x x x 1 3, x 1 1 2, x > 1 93. f x The derivative from the left is lim f x x f 1 1 lim x x 1 1 3 0 0. x1 x1 1 2 x1 lim x The derivative from the right is lim f x x f 1 1 lim x x 1 1 2 0 0. 1. f 1 0 x1 x1 1 x1 lim x These one-sided limits are equal. Therefore, f is differentiable at x Section 2.1 x, x 1 x 2, x > 1 The Derivative and the Tangent Line Problem 107 94. f x The derivative from the left is lim f x x f 1 1 lim x x 1 1 lim 1 1. x1 x1 x1 The derivative from the right is x1 lim f x x f 1 1 x1 lim x2 x 1 1 x1 lim x 1 2. 1. These one-sided limits are not equal. Therefore, f is not differentiable at x 95. Note that f is continuous at x 2. f x f x x f x x x2 4x f 2 2 f 2 2 1, x 2 3, x > 2 lim lim x2 x 4x x 1 2 3 2 2. f 2 5 5 lim x lim 4 4 2 4. 4. The derivative from the left is lim x2 x2 x2 The derivative from the right is lim x2 x2 x2 The one-sided limits are equal. Therefore, f is differentiable at x 1 2x 96. Note that f is continuous at x The derivative from the left is 2. f x 2x, 1, x < 2 x 2 1 1 f x f 2 2x lim x2 x2 x 2 x 2 The derivative from the right is lim 2 x2 lim 1 2 x x 2 2 1 . 2 x2 lim f x x f 2 2 x2 lim 2x 2 x 2 x 2x 2 4 2x 2x 2x 2 2 2 lim x 2x 2 2. 2 2x f 2 2 1 2 x2 lim x2 x2 lim 2 2x 2 1 . 2 The one-sided limits are equal. Therefore, f is differentiable at x 97. (a) The distance from 3, 1 to the line mx d Ax 1 m3 By1 A2 m2 (b) 5 y 4 0 is y C B2 3 2 11 1 4 3m m2 3 1 . 1 x 1 2 3 4 -4 -1 4 The function d is not differentiable at m 1. This corresponds to the line y x 4, which passes through the point 3, 1 . 108 Chapter 2 Differentiation 2x (b) g x x 3 and g x y 3 98. (a) f x x 2 and f x y 5 3x 2 f 4 3 2 1 1 x 1 2 3 4 -2 -1 -1 1 2 x (c) The derivative is a polynomial of degree 1 less than the original function. If h x x n, then n 1. h x nx g 2 g -4 -3 -2 -1 f' -3 (d) If f x f x x4, then lim lim lim lim f x x x4 x x0 x x x4 x 4x 3 x 6x 2 6x 2 f x x4 6x 2 x x 4x x 4x x 2 2 x0 4x x x 2 x 3 x 3 4 x4 x0 4x 3 x x 3 x0 lim x0 4x 3 x 4x 3. x 4, then f x 4x 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. f 2 x x f 2 99. False. The slope is lim x0 . 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, x , then the derivative from the then the derivative does not exist at that point. For example, if f x 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 2.1. x sin 1 x , 0, x x 0 0 0 f 0 and 103. f x x x sin 1 x x , x 0. Thus, lim x sin 1 x Using the Squeeze Theorem, we have x0 f is continuous at x 0. Using the alternative form of the derivative, we have f x x sin 1 x 1 f 0 0 lim lim sin . x0 x0 x 0 x 0 x Since this limit does not exist (sin 1 x oscillates between x0 lim 1 and 1), the function is not differentiable at x 0. gx x 2 sin 1 x , 0, x x 0 0 x 2 x 2 sin 1 x x 2, x 0. Thus, lim x 2 sin 1 x x0 Using the Squeeze Theorem again, we have and g is continuous at x x0 0 g0 0. Using the alternative form of the derivative again, we have 2 lim gx x g0 0 x0 lim x sin 1 x x 0 0 0, g 0 x0 lim x sin 0. 1 x 0. Therefore, g is differentiable at x Section 2.2 Basic Differentiation Rules and Rates of Change 109 104. 3 -3 -1 3 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 2.2 1. (a) y y y 1 3. y y 8 0 1 x7 7x x1 1 2x 1 2 2 Basic Differentiation Rules and Rates of Change (b) 1 2 y y y 1 x3 3x 2 3 2. (a) y y y 1 x 1 2 (b) y y y 1 x 1 1 3 2 2x 1 2 x 1 x8 8x 7 2 4. f x f x 0 2 5. y y x6 6x 5 6. y y 7. y y x 8 7 8. y 7 x8 y 1 x8 8x x 9 8 9. y 8 x9 y 5 x x1 4 5 5 10. y 5 4 x x1 3 4 4 1 x 5 1 5x 4 y 1 x 4 1 4x 3 4 11. f x f x x 1 x2 2x 1 12. g x g x 4x 3 12x 2 3x 3 1 13. f t f t t3 2t 2 4t 3 3t 6 14. y y t2 2t 2t 2 2x 3 6x 2 3 15. g x g x 16. y y 8 x3 3x 2 17. s t s t 2t 2 1 cos x 2 1 sin x 2 4 18. f x f x x2 2x 3x 3 3t 2 19. y y 2 2 sin cos cos sin 20. g t g t cos t sin t 21. y y x2 2x 22. y y 5 sin x cos x 23. y y 1 x 1 x2 3 sin x 3 cos x 24. y y 5 2x 5 8 3 2 cos x 4 5 x 8 3 2 cos x 15 8x 4 2 sin x 3x 2 sin x Function 25. y 5 2x 2 2 3x 2 3 2x Rewrite y 5 x 2 2 x 3 3 x 8 2 Differentiate y 5x 4 x 3 9 x 8 3 Simplify y 5 x3 4 3x 3 9 8x 4 26. y y 2 y 3 y 27. y 3 y 3 y 4 y 110 Chapter 2 Function Differentiation Rewrite y 9 x x 2 Differentiate y 2 x 9 1 x 2 12x 2 3 Simplify y 2 9x 3 1 2x 3 12x 2 28. y 3x x x 4 x 3 2 29. y y y 1 2 y 3 2 y 2 30. y 4x 3 y y 31. f x 3 x2 6x 6 3x 3 2 , 1, 3 6 x3 32. f t f t f 3 5 y 3 3 5t 2 5 3 2x 4x 2 3 , 5t 3 ,2 5 33. f x f x f 0 1 2 21 2 x 5 0 7 3 x , 5 0, 1 2 f x f 1 34. y y y 2 3x 3 9x 2 36 6, 2, 18 35. 1 2, 0, 1 4x 4 1 36. f x 35 3x 2 x 2, 5, 0 30x 30 75 y y 0 8x 4 2 f x f 5 6x 0 x2 2x 37. f f f 0 4 sin 4 cos 41 1 , 0, 0 1 3 38. gt g t g 3 cos t, 3 sin t , 1 39. f x f x 5 6x 3x 3 2 2x 0 6 x3 40. f x f x x2 2x 2x 3x 3 3 3x 6x 6 x3 4 x3 x3 x3 2 3 41. g t g t t2 2t 4 t3 12t t2 4 4t 2t 3 42. f x 12 t4 f x x 1 1 x 2x 2 x3 2 3 43. f x f x x3 3x 2 x2 8 x3 1 1 63 x 1 2 x 8 3 4x 2 44. h x h x 2x 2 3x x 1 x2 5x 2 45x 2 1 2x 3 x 1 1 2 2x 2 1 x2 18x 2 15x 3 45. y y x x2 3x 2 x 46. y y 3x 6x 36x 3 47. f x f x x 1 x 2 s4 4 s 5 x1 2 3 2 6x 1 1 2 x 3 48. f x 2 x2 3 f x x 5 x x1 1 x 5 3 x1 5 2x 1 x 3 2 3 4 5 1 3x 2 3 1 5x 4 5 49. h s h (s 5 s2 1 5 3 50. f t 1 3 t2 2 t 3 3 t1 1 3 3 4 2 3 2 s 3 4 5s 1 5 2 3s 1 3 f t 1 t 3 2 3t 1 3 1 3t 2 3 Section 2.2 Basic Differentiation Rules and Rates of Change 2 3 111 51. f x f x 6 x 3x 1 2 5 cos x 5 sin x 6x 1 3 2 5 cos x 5 sin x 52. f x f x x 3 cos x 4 3 2x 1 3 3 cos x 3 sin x x 2 x 3 x3 3x 2 1, 3 sin x 2 3x 4 3 53. (a) y y x4 4x 3 3x 2 6x 2 54. (a) y y x 1 2: y 3 y 4x y 1 2 2 2 At 1, 0 : y Tangent line: 41 3 61 0 2 0 2 2x 1 At 1 4x 0 4 1 y 2x y Tangent line: (b) 3 (b) -5 5 5 -2 -1 2 (1, 0) -7 55. (a) f x f x 4 2 x3 2x 7 4 3 4 56. (a) y 3 x2 x3 2x x 3x 2 6x 2x 2 31 1 3 x 2 2x 7 4 3 2 y 2 y 3x 2y 7 0 3 x 2 3 x 2 1 7 2 (b) y 3x 2 At 1, 2 : f 1 Tangent line: At 1, 6 : y Tangent line: y 2 61 11 x 11x 2 1 y 5 11 6 0 12 -3 3 -2 (b) 5 (1, 2) -2 -1 7 57. y y x4 4x 3 4x x2 8x 2 16x 4 2 x 2 58. y y x3 3x 2 x 1 > 0 for all x. Therefore, there are no horizontal tangents. 2 4x x y 0 x 0, 2 14 , 2, 14 Horizontal tangents: 0, 2 , 2, 1 x2 2x 59. y y x 3 2 60. y 2 x 3 x2 2x 1 0 x 1. 0 y cannot equal zero. At x 0, y Therefore, there are no horizontal tangents. Horizontal tangent: 0, 1 112 61. y y Chapter 2 x 1 Differentiation 62. y y sin x , At x At x 3x 3 2 cos x, 0 x < 2 2 sin x 3 x 2 3 : y 3 3 2 3 3 3 , 3 3 3 , 2 2 3 , 3 3 3 3 0 3 or 3 2 3 sin x, 0 x < 2 cos x 0 1 x : y cos x At x Horizontal tangent: 2 : y 3 Horizontal tangents: 63. x 2 2x kx k 4x 4 2x 4x 9 Equate functions. Equate derivatives. 64. k x2 2x 4x 4 7 Equate functions. Equate derivatives. 4 8 7k 3. Hence, k x2 For x k x k x2 Hence, k 3 2 4x 4 and 4x 9 x2 3, k 9x 10. 3. Hence, x 2 and k 2x 3, k 2 and for x 65. 3 x 4 3 4 3 2 4x 3 Equate functions. 66. k x k 2 x x 1 4 Equate functions. Equate derivatives. Equate derivatives. and 3 x 4 3x 3 2k 3. Hence, k 2 x x 3 3 x 4 3 x 2 2 x and x 4 2x x 4x 4k 4. x 3 x 4 67. (a) The slope appears to be steepest between A and B. (b) The average rate of change between A and B is greater than the instantaneous rate of change at B. (c) y 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. y x f B C A D E x 69. g x f x 6g x f x 70. g x 5f x g x 5f x Section 2.2 71. 3 y Basic Differentiation Rules and Rates of Change 72. 2 y 113 f 1 3 2 1 2 1 f x 2 3 -2 -1 1 1 f x 3 4 f -3 -4 If f is linear then its derivative is a constant function. f x f x ax a b If f is quadratic, then its derivative is a linear function. f x f x x 2 and y 2x 2 x2 6 6x ax 2 2ax bx b c 73. Let x1, y1 and x2, y2 be the points of tangency on y The derivatives of these functions are: y m 2x m 2x 1 x1 Since y1 m y2 x2 x 22 x 22 6x 2 2x 2 x2 3 x 22 x 22 6x 2 x2 5 5 x2 x 22 2x 22 2x 22 2 x2 1 y2 6x 2 6x 2 5 x2 x1 x2 3 6x 2 12x 2 6x 2 2 x2 9 14 4 1 x2 x2 0, x1 2 and y1 4 0 3 2 5, respectively. 2x 1 and 6 y 2x 6 m x 12 and y2 y1 x1 5: 5 y x 12 4 2x 2 2x 2 2x 2 4x 22 6 6 3 2 1 -1 )2, 3) )1, 1) x 2 3 6 2x 2 18x 2 3 18 y 5 0 1 or 2 4 3 2 1 )2, 4) x -1 -2 )1, 0) 2 3 Thus, the tangent line through 1, 0 and 2, 4 is y x2 0 2 y2 4 2 0 x 1 3, x 1 1 y 1 and y1 1 4x 4. Thus, the tangent line through 2, 3 and 1, 1 is y 1 3 2 1 x 1 1 y 2x 1. 74. m1 is the slope of the line tangent to y y x y 1 m1 1 and y x and y 1. x. m2 is the slope of the line tangent to y 1 y x 1 x are 1 m2 x2 1 . x2 1 x. Since The points of intersection of y x At x 1 x2 x 1, m 2 1 x 1. Since m 2 1 m 1, these tangent lines are perpendicular at the points of intersection. 114 Chapter 2 3x 3 sin x cos x Differentiation 2 76. f x f x x5 5x 4 4 75. f x f x 3x 3 9x 2 2 5x 5 Since cos x 1, f x 0 for all x and f does not have a horizontal tangent line. Since 5x 9x 0, f x 5. Thus, f does not have a tangent line with a slope of 3. 2 , x 2 x2 2 x2 10 10 2x 2x 2x 4x 0 5 y x x2y x2 2x 10 5 ,y 2 5 4 2, 5 77. f x f x 1 2 x 4 4 4 x x x x x, 1 x 2 1 2 4, 0 1 2 x 78. f x f x 5, 0 0 y 4 x 2 xy 2 x x 2x 4, y 2 2 x 10 The point 4, 2 is on the graph of f. Tangent line: y 4y 2 8 0 x x 0 2 x 4 4 4 4y 4 4 x The point 4 5 is on the graph of f. The slope of the 5 2 8 25 . tangent line is f Tangent line: y 25y 8x 25y 4 5 20 40 0 8 x 25 8x 20 5 2 79. f 1 3.64 1 80. f 4 1 16 0.77 3.33 1.24 -10 -1 19 81. (a) One possible secant is between 3.9, 7.7019 and 4, 8 : y y y (b) f x Tx 8 8 Sx 3 1 x 2 3x 8 4 7.7019 x 3.9 4 3.924 3 2 2 3x 4 3 4 -2 20 (4, 8) 12 -2 2.981 x 2.981x 2 f 4 4 8 S x is an approximation of the tangent line T x . --CONTINUED-- Section 2.2 81. --CONTINUED-- Basic Differentiation Rules and Rates of Change 115 (c) As you move further away from 4, 8 , the accuracy of the approximation T gets worse. 20 f -2 -2 T 12 (d) x f 4 T4 x x 3 1 1 2 2.828 2 1 5.196 5 0.5 6.548 6.5 0.1 7.702 7.7 0 8 8 0.1 8.302 8.3 0.5 9.546 9.5 1 11.180 11 2 14.697 14 3 18.520 17 82. (a) Nearby point: 1.0073138, 1.0221024 Secant line: y 1 y (Answers will vary.) -3 (b) f x 1 Tx 3x 2 3x 1 1 3x 2 1.0221024 1.0073138 3.022 x 1 1 x 1 1 2 (c) The accuracy worsens as you move away from 1, 1 . 2 (1, 1) (1, 1) 3 -2 -2 -3 3 f T (d) x f x Tx 3 8 8 2 1 5 1 0 2 0.5 0.125 0.5 0.1 0.729 0.7 0 1 1 0.1 1.331 1.3 0.5 3.375 2.5 1 8 4 2 27 7 3 64 10 x 3 2. g x 0 g x. The accuracy decreases more rapidly than in Exercise 81 because y 83. False. Let f x f x g x 85. False. If y 2 x 3 is less "linear" than y gx x 2 and g x 2x, but f x , then dy dx x2 gx. 0. 2 4. Then 84. True. If f x c, then f x is a constant. 86. True. If y x 1 xn 3, 1 x, then dy dx 1 1 1 n xn . 87. True. If g x 3 f x , then g x 3f x . 88. False. If f x x n, then f x nx n 1 1 . 89. f t f t 2t 2 7, 1, 2 90. f t f t t2 2t 2, 2.1 Instantaneous rate of change is the constant 2. Average rate of change: f 2 2 f 1 1 22 7 1 21 7 2 Instantaneous rate of change: 2, 1 f 2 2.1, 1.41 f 2.1 Average rate of change: 22 4.2 4 (These are the same because f is a line of slope 2.) f 2.1 2.1 f 2 2 1.41 1 0.1 4.1 116 Chapter 2 1 , x 1 x2 Differentiation 91. f x f x 1, 2 92. f x f x sin x, cos x 0, 6 Instantaneous rate of change: 1, 2, 1 f 1 1 2 f 2 1 1 4 Instantaneous rate of change: 0, 0 f 0 1 , 6 2 f 6 1 3 2 0.866 Average rate of change: f 2 2 93. (a) s t vt (b) s2 2 f 1 1 16t 2 32t s1 1 s t 1298 32t 1: v 1 2: v 2 1362 t2 (e) v 1362 4 0 1362 t 16 32 1362 4 295.242 ft sec 1362 4 9.226 sec 32 ft sec 64 ft sec 1346 48 ft sec 1 2 2 1362 1 1 1 2 94. Average rate of change: f 6 6 f 0 0 1 2 6 0 0 3 0.955 st vt v3 st 16t 2 32t 22t 22 220 118 ft sec 16t 2 22t 220 (c) v t 112 height after falling 108 ft 16t 2 2t 22t 2 8t 108 27 t v2 0 0 2 32 2 22 When t When t (d) 16t 2 86 ft sec 8 1362 95. st 4.9t 2 4.9t 2 vt v5 v 10 9.8t 9.8 5 9.8 10 v0 t 120t 120 120 120 s0 96. s t 4.9t 2 4.9t 2 v0 t s0 4.9 6.8 s0 0 when t 2 6.8. s0 71 m sec 22 m sec 1 2t 4.9t 2 226.6 m 97. From 0, 0 to 4, 2 , s t vt 1 2 vt 1 2 mi min. 98. 50 v Velocity (in mph) 60 30 mph for 0 < t < 4 0 for 4 < t < 6. Finally, from 6, 2 to 1 mi in 60 mph. 40 30 20 10 t 2 4 6 8 10 Similarly, v t 10, 6 , st v 60 t 4 vt Velocity (in mph) Time (in minutes) 50 40 30 20 10 t 2 4 6 8 10 (The velocity has been converted to miles per hour.) Time (in minutes) Section 2.2 99. v 2 3 Basic Differentiation Rules and Rates of Change 100. This graph corresponds with Exercise 97. s 117 40 mph 2 3 mi min 4 mi Distance (in miles) 10 8 6 s mi min 6 min 0 mph v 0 mi min 0 mi (6, 4) 4 Distance (in miles) (10, 6) 10 8 6 4 2 0 mi min 2 min v 60 mph (8, 4) 2 t 2 4 6 8 10 (10, 6) 1 mi min 2 mi (0, 0) (4, 2) 1 mi min 2 min Time (in minutes) (6, 2) t 4 6 8 10 (0, 0) 2 Time (in minutes) 101. (a) Using a graphing utility, R 0.417v 0.02. (e) dT dv For v For v 0.01114v 40, T 40 80, T 80 0.418 0.86. 1.31. 1.53. (b) Using a graphing utility, B (c) T (d) 80 0.00557v 2 B 0.0014v 0.04. 0.02 For v 100, T 100 R 0.00557v 2 T B R 0.418v (f ) For increasing speeds, the total stopping distance increases. 0 0 120 102. C gallons of fuel used cost per gallon 15,000 1.55 x 23,250 x dC dx x C dC dx 23,250 x2 10 2325 233 15 1550 103 20 1163 58 25 930 37 30 775 26 35 664 19 40 581 15 15 is larger 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. dV ds dA ds 103. V s 3, 3s 2 4 cm, dV ds 48 cm2 per cm change in s. 104. A s 2, 2s 4 m, When s When s dA ds 8 square meters per meter change in s. 118 Chapter 2 1 2 at 2 Differentiation 105. s t c and s t s t0 t0 t t at s t0 t0 t t 1 2 a t02 2at 0 t 2 t at 0 s t0 instantaneous velocity at t t0 2t0 t 1 2 a t0 t 2 Average velocity: c 2 t t 2 1 2 a t0 1 2 a t02 t 2 c t 2 2t0 t 2 t 106. C dC dQ C 351 When Q C 350 350, dC dQ 1,008,000 Q 1,008,000 Q2 5083.095 $1.93. 6.3Q 6.3 5085 $1.91 107. N f p (a) f 1.479 is the rate of change of gallons of gasoline sold when the price is $1.479 per gallon. (b) f 1.479 is usually negative. As prices go up, sales go down. dT dt 108. KT Ta 109. y ax 2 bx c 110. y y 1 , x > 0 x 1 x2 Since the parabola passes through 0, 1 and 1, 0 , we have: 0, 1 : 1 1, 0 : 0 a0 a1 2 2 b0 b1 c c 1 b 1 At a, b , the equation of the tangent line is a 1 y 1 a 1 x a2 a or y x a2 2 . a Thus, y ax 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 1 a b 2ax 2a 1 a 2 a 1 2x 2 3 3x 1. 1 a a 1 1 The x-intercept is 2a, 0 . The y-intercept is 0, 2 . a 1 bh 2 2 1 2a 2 a 2. The area of the triangle is A y Therefore, y 2 1 (a, b) = a, a 1 ( ) x 2 3 1 Section 2.2 111. y y x3 3x 2 9x 9 9: 9 x 3x 2 3x 2 3 2 3 2, 81 8 Basic Differentiation Rules and Rates of Change 119 Tangent lines through 1, y x3 9x 9 9 0 x 3x 2 3x 3 2x 3 0 or x 1 9x x 2 2x 9 3 The points of tangency are 0, 0 and Tangent lines: y 0 y 9x 112. y y x2 2x y 0 9x 9x 9x 0 and . At 0, 0 , the slope is y 0 9. At 3 2, 81 8 , the slope is y 3 2 9 4. y 81 8 9 4 x 3 2 27 4 y 4y 27 0 9 4x (a) Tangent lines through 0, a : y x2 a a a 2x x 2x 2 x2 x 0 a The points of tangency are a, At a, a , the slope is y Tangent lines: y a y 2 2 ax ax a a . At a a a, 2 and y a , the slope is y a. a y 2 2 ax ax a a a 2 a. Restriction: a must be negative. (b) Tangent lines through a, 0 : y 0 x2 0 2x x 2x 2 x2 a 2ax 2ax xx 2a 0. At 2a, 4a 2 , the slope is y 2a 4a. The points of tangency are 0, 0 and 2a, 4a 2 . At 0, 0 , the slope is y 0 Tangent lines: y 0 y 0x 0 0 and y 4a 2 y 4a x 4ax 2a 4a 2 Restriction: None, a can be any real number. 120 Chapter 2 ax 3, x 2 b, Differentiation x 2 x > 2 2 to be differentiable at x 8a b 4 b 8a 8a 4 4 b 2. b 113. f x f must be continuous at x x2 lim f x x2 lim ax 3 x2 lim f x x2 lim x 2 x < 2 x > 2 f x 3ax 2, 2x, For f to be differentiable at x 3a 2 2 2, the left derivative must equal the right derivative. 22 4 1 3 12a a b 8a 4 4 3 114. f x f 0 f x Hence, a b cos x, x < 0 ax b, x 0 cos 0 sin x, a, 0. 0, b cos x. lim lim f x x0 115. f1 x sin x is differentiable for all x an integer. 1 f2 x sin x is differentiable for all x n ,n 0. 1b x < 0 x > 0 You can verify this by graphing f1 and f2 and observing the locations of the sharp turns. Answer: a 116. Let f x f x 1 x x f x sin x sin x x 1 sin x x0 cos x cos x cos x cos x x sin x 1 cos x sin x x x0 lim 0 x0 lim sin x Section 2.3 1. g x g x x2 x2 2x 3 4x 3 3. h t h t 3 Product and Quotient Rules and Higher-Order Derivatives 1 x2 1 2x 2x 2 6x 2 4 t2 t2 4 3t 2 3 4 2x 2 2x 2x t1 3 2. f x x2 2 2 t2 1 t 3 4 2 3 6x 6x 18x 3 24x 3 5 x3 5 3x 2 15x 2 15x 2 2 x3 6x 3 12 2 6 12 2x 2x 2x 3 4x 2 f x t t2 3 4. g s g s s1 s4 2 s2 2s 2 s1 4 2 4 s2 1 2 t1 2t 3 4 1 s2 s 2 2t 4 2s3 4 s2 2s1 2 7t 2 4 3t 2 3 5s 2 2s 1 2 Section 2.3 5. f x f x x 3 cos x x3 2 Product and Quotient Rules and Higher-Order Derivatives 6. g x x sin x x cos x sin x 1 2 x x cos x 121 sin x cos x 3x 2 x sin x 3 g x 3x cos x x x2 x2 1 1 sin x 2 x 7. f x f x 8. g t 1 x2 x2 12 g t t2 2t 2t 2 7 7 2t 2t s s s s s 1 1 1 s 1 1 1 2 2 1 1 x 2x x2 1 2 3 t2 7 2 2 2 2t 2 2t 14t 7 4 2 9. h x h x x 1 x 1 3x x1 3 2 3 3 x 3 3 1 x 1 2 1 3 10. h s 3x 2 x 1 x3 h s s 1s 2 1 s 2 2 1 2 x3 1 x 9x 2 3x 2 3 x 3 1 2 1 8x 3 3x 2 3 x 3 1 sin x x2 x 2 cos x sin x 2x x2 2 x3 x3 10x 4 f 0 15 f 1 x2 x x 2x 2 x2 x f 1 1 1 4 3 3 2x x x2 3 2 2 s s 2 1 2 11. g x g x 12. f t x cos x x3 3x 3 3x 2 5 2x 2 18x 3x 15 5 3x 2 3 2 sin x f t cos t t3 t3 sin t cos t 3t 2 t3 2 x2 x2 3x 2 x x 0 x x x x 1 1 1 1 x 1 x 2 x 1 2 2 1 2 2 t sin t t4 3 cos t 13. f x f x 3x 2x 2 3x 4x 12x 3 14. f x f x 2x 2x 1 1 2 2 1 x3 1 3x 2 2x 2x 1 x3 1 2 2 1 2x x 1 2 x2 5x 2 15. f x f x 16. f x 4 1 4 f x x 1 2 1 1 6x x 2 x 32 6x 3 6 3 2 x 1 12 4 2 4 1 4 f 2 2 17. f x f x f x cos x x 2 2 sin x 2 2 cos x 1 2 4 8 cos x x sin x 4 4 122 Chapter 2 sin x x x cos x Differentiation 18. f x f x sin x 1 x2 x cos x sin x x2 f 6 6 3 3 2 3 2 2 36 18 6 2 1 2 3 3 Function 19. y x2 3 5x 2 4 7 3x3 4 5x2 4x3 x 3x2 7 2 Rewrite 2x y 1 2 x 3 5 2 x 4 7 x 3 4 x 5 3 Differentiate 2 x 3 3 4 y 2 x 3 10 x 4 7x 8 x 5 2x 6x 7 1 2 4 Simplify y 2x 3 5x 2 7 x4 8 5x3 2 x 6 x 7 2 2 3 20. y 3 y y y 21. y y y y 22. y y 2 y 3 y 23. y y 5 4 x, x > 0 3 2 x 7 5 7 y y 24. y y y y 25. f x f x 3 x2 2x x 2 x2 1 1 2 2 2 26. f x 2x x2 2x x2 1 3 1 2 x3 x2 x2 x4 3x 1 2 3 x2 3 x3 12 3x 2 2x 2x x 2 2x f x 1 3x 2 6x 2 x2 4x 12 2x 2 4x x2 1 2 x 1 2 12 12 , x 27. f x f x x 1 1 x2 x2 x x 4 x 3 x 4x x 3 28. f x f x x4 1 x4 x4 2x 3 x 2 x 2 x 1 x 1 2 1 x 1 2 x4 1 x2 x x x 1 1 x x 1 4x 3 1 3 4 4x 1 x 32 9 32 3 2 6x x 6x 3 12 1 4x3 12 2x 2 x x 1 2 2 Section 2.3 2x x x 1 2 Product and Quotient Rules and Higher-Order Derivatives 123 29. f x f x 5 2x 1 5 x 2 2 5x x 12 3 2 3 2 x 5 2 2x 5 2x x 2x 5 2x 3 2 30. f x f x 3 x 3 x 1 x 2 1 6 3 1 2 x1 3 x1 2 2 3 1 x 3 2 3 Alternate solution: f x 3 x1 5 x 6 x1 2 3 3 x x 3x 1 6 3 1 3 x f x 5 6 x 1 x2 3 2 5 x 6 5 6x1 x 1 x2 3 2 3 5 6x 1 s3 6s 5 2 x x2 6 6 31. h s h s 2 s6 4s 3 6s 2 s 3 4 2 2x x2 3 1 3x 2x 2 6x 4x 2 x 2 3x 32. h x h x x2 4x 3 1 2 x4 4x x 2 2x 2 1 1 12s 2 1 x 3 3x 2 4x 33. f x f x 2x xx 2 1 3 2x 1 2x x 3x 2 3 2 8x 2 3 2x 2 x2 2 x x 2x 3x 1 x 2x 2 2x 3 x2 x 3 2 x2 x 1 2 x2 2x x 1 1 5 16x 32x x x 2x x 1 2 34. g x g x x2 2 1 2x 1 2x x 2 1 x 12 4x x 4 x 4 x 2 1 1 2 x2 2x x2 x 2x 1 2 2 35. f x f x 3x 3 9x 2 9x 2 5 x 5 x 4x 41x 2 33x 2 1 x2 1 x 3 3x 3 3x 4 4x 1 x 3x 3 1 2 3x 3 4x 3x 8x 2 4 4x x 15x 16x 3 5 1 4x 2 20x 4x 9x 4 15x 4 36. f x f x x2 2x 2x 2x 5 6x 5 36x 3 48x 3 x x2 1 x2 4 20 20 6x 4 12x 3 1 1 x 2x 5 1 x2 x 2x 2 x 2x x 2 2 x2 x 3 1 2x 2 x2 2x x2 x x x2 x 2 1 2x x 2x 3 1 x 2 1 x x4 4x 3 x 2x 2x 5 2x 1 3x 3 3x 2 1 x4 x3 124 Chapter 2 x2 x2 x2 c2 c2 Differentiation c2 c2 c2 x2 x2 x2 4xc 2 x2 2x c2 2 37. f x f x 38. f x x2 c2 2 c 2 2x f x c 2 2x x2 2 c2 x2 2 x 2 2x 4xc2 x c2 2 c2 40. f 2t sin t 2 sin t f cos 39. f t f t t 2 sin t t2 cos t 1 cos 1 sin 1 sin cos 1 t t cos t cos t t t sin t t2 x 1 41. f t f t 42. f x cos t t sin t t2 cos t f x sin x x x cos x sin x x2 x 1 1 s 1 s2 sec x x x sec x tan x x2 sec x x tan x x2 sec x 1 cot x csc2 x cot2 x 43. f x f x tan x sec2 x tan2 x 44. y y 45. g t g t 4 t 3 4 8 sec t t1 4 8 sec t 1 4t 3 4 46. h s 8 sec t tan t h s 10 csc s 10 csc s cot s 1 t 4 8 sec t tan t 47. y y 3 1 sin x 2 cos x 3 sec x tan x 2 3 sec x tan x 2 3 sec x 2 sec 2 x tan2 x tan x 3 sec x tan x 2 1 sec x 48. y y 49. y y csc x csc x cot x cos x sin2 x sin x cos x 50. y y x sin x x cos x cos x sin x sin x x cos x cos x 1 cos x csc2 x cos x cot2 x 51. f x f x x 2 tan x x 2 sec2 x x x sec2 x 53. y y 2x sin x 2x cos x 4x cos x 52. f x 2x tan x 2 tan x 54. h x2 sin x 2x cos x h f x sin x cos x sin x cos 2x 5 sec 5 sec tan tan 5 sec sec2 tan sin x cos x cos x x 2 cos x 2 sin x 2 sin x x 2 sin x Section 2.3 x x 2x 2 x 1 2x 2 8x 2 1 2 Product and Quotient Rules and Higher-Order Derivatives x2 x2 2 x5 x 1 2x 3 x2 3 125 55. g x g x 5 (Form of answer may vary.) 56. f x f x x2 2x 2 12 x 2 1 (Form of answer may vary.) 57. g g 1 1 sin sin sin cos 1 2 58. f (Form of answer may vary.) f 1 sin cos 1 1 cos 1 1 cos 2 cos (Form of answer may vary.) 1 1 1 csc x csc x csc x 3 22 1 22 csc x cot x 1 csc x csc x cot x 1 csc x 2 4 3 2 csc x cot x 1 csc x 2 59. y y y 6 60. f x f x f 1 tan x cot x 0 1 61. ht h t sec t t t sec t tan t t2 sec tan 2 sec t 1 1 1 2 0 h sec t t tan t t2 1 62. f x f x sin x sin x sin x cos x sin x cos x sin 2x cos x sin x sin2 x sin x cos x cos x cos2 x 63. (a) f x f x x3 x3 4x 3 3x 3x 6x 2 1 x 1 1 6x 2, x 5 3 1x 1, 3 3 2 3x 2 sin x cos x cos 2x cos 2 1 (b) f 1 1; Slope at 1, 3 f 4 sin Tangent line: y 10 2 1 y x 2 - 10 (1, - 3) 10 - 10 64. (a) f x f x f 0 x x 1 x2 1 2x 2, x2 0, 2 2 1 3x 2 2x 2 65. (a) f x f x 2x 2 f 2 x x x 1 , 2, 2 1 1 2; Slope at 0, 2 2 2x y 1 1 x1 x 12 1 1 2 x 2 Tangent line: y (b) -4 4 2 1; Slope at 2, 2 2 1x 2 y x 4 Tangent line: y 4 (b) 6 (2, 2) -4 -3 6 -3 126 Chapter 2 x x x 1 , 1 Differentiation 1 3 x 1 2 66. (a) f x f x f 2 2, 67. (a) 1 1 x 2 1 2 f x f x f 4 tan x, sec2 x 4 ,1 1 1 x 2; Slope at 2 1 ; Slope at 2, 9 3 1 3 2 x 9 2 y 2 x 9 1 9 4 ,1 y y 1 1 2 2 x 2x 0 Tangent line: Tangent line: y (b) -3 4 4 2 4x 6 2y (b) - 4 ( , 1( 4 -4 -4 68. (a) f x f x f 3 sec x, 3 ,2 (b) 6 sec x tan x - 2 3; Slope at 3 ,2 -2 Tangent line: 6 3x 8 x2 x2 4 3y 6 y 2 3 2 2 3 x 0 3 69. f x f x f 2 y ; 2, 1 16x 4 70. x2 2 f x f x f 3 y 27 x2 x2 9 ; 3, 3 2 x2 54x 9 2 4 0 8 2x x2 4 2 1 2 1 x 2 1 x 2 0 2 2 9 0 27 2x x2 9 2 1 2 3 3 4 16 2 42 1 y 54 3 9 92 3 2 y 1 x 2 1 x 2 0 4 5 2y x 4 2y 8 5 256 16x 2 x 2 16 2 x 6 4x x2 x2 24 10 2 y 6 71. f x f x f 2 x2 16x ; 16 2, 72. f x f x f 2 ; 2, x2 256 8 5 y 16 16 16x 2x x 2 16 2 16 4 20 2 12 x 25 12 x 25 0 12 25 2 16 25 6 4 4x 2x x2 6 2 16 4 5 y 2 25 2 x 25 2 x 25 0 2 16 25 24 4x2 x2 6 2 y 25y 2x 16 25y 12x 16 Section 2.3 x2 x x x2 x f x 1 1 2x x 1 2x 12 x2 1 2 Product and Quotient Rules and Higher-Order Derivatives x2 x2 x2 2x x2 1 2 127 73. f x 74. f x f x 1 1 2x x 2 2x 2 2 x 1 75. f x f x 4x x2 x2 4 4x 2 2 4x x4 8x 2 x4 4x x3 4x 0 x 1. 2 2x 4x f x xx x 2 12 2. f x 0 when x 0 or x 0 when x 0. f x f 1 4 Horizontal tangents are at 0, 0 and 2, 4 . Horizontal tangent is at 0, 0 . 2 0 for 4 f has a horizontal tangent at 1, 2 . 76. f x f x f x x x2 x2 4 7 7 1 x2 x 72 4 2x 1 , f 7 2 x2 7 x2 1 14 2x2 72 8x x2 x2 8x 7 72 x 7 x x2 7 1 2 0 for x 1, 7; f 1 f has horizontal tangents at 1, 1 1 . and 7, 2 14 x x x x 1 1 1 2 77. f x f x 2y x 2 1 1 x x x x 1 1 1 x 6 y 1 2 4 2 78. f x x 1 2 1 x 1 x 2 2 1 2 f x 1 2 x x 1 1 2 3; Slope: Let x, y x, x x 1 be a point of tangency on the graph of f. 5 x x 1 1 x x 4x 5 1 x 1 5 x 10x 2 2x 1 2 1 4 1 x x x 0 0 x 2; f 1 2 1 1 1 1 1 2 -4 (-1, 5) 6 y f(x) = x x-1 x x 2 2 2 y = -x + 4 y = - 4x + 1 -2 -2 2 (2, 2) x 4 1 x 1, 3; f 1 x 2 1 x 2 y 1 0, f 3 1 x 2 1 x 2 1 2 7 2 2 (1 , - 1) 2 4x 4x 2 x f y y 0 2 2y + x = 7 6 1 y 3 y x+1 x-1 1 ,2 2 4, f 2 1 1, f 2 f(x) = (-1, 0) -6 -4 -2 2 (3, 2) x Two tangent lines: 6 -2 -4 -6 4 y y 1 2 4 x 1x 1 2 y 4x x 4 1 2y + x = -1 2 y 128 Chapter 2 x x 5x x Differentiation 6 x 4 1 x 2 2x x 4 2 2 2 79. f x g x gx 2 3 3x 1 x 22 25 x 4 2 x 5x 22 3x 80. f x 6 2 2 x cos x x cos x sin x x 3 x2 2 x2 2x sin x sin x sin x 3x x 3x 1 2x 1 5x x cos x sin x x2 x cos x sin x x2 f x 5 g x 2 gx f x f and g differ by a constant. 81. (a) p x p 1 (b) q x q 4 f xgx f 1g1 gx f x gx 3 1 32 f xg x f 1g 1 f xg x 2 f and g differ by a constant. 82. (a) p x 14 6 1 2 1 p 4 (b) q x q 7 f xgx 1 8 2 10 f xg x 4 f xg x gx 42 4 42 2 gx f x 70 1 3 1 12 16 3 4 83. Area A t At 2 3 1 t 2 2 2 2t 1 1 t 2 t 1 2 2t3 2 t1 2 84. V r 2h 1 3 t 2 2 t 2t 1 2 2 2 1 2 t 3t1 1 t 2 1 2 V t 1 3 1 t 2 2 t 1 2 3t 2 4t 1 2 cubic inches sec 6t 1 2 cm sec 2 t k V k V2 85. C dC dx 100 100 200 x2 400 x3 10: 15: 20: x x x dC dx dC dx dC dx 30 , 1 x 86. P dP dV 30 30 2 (a) When x (b) When x (c) When x $38.13 $10.37 $3.80 As the order size increases, the cost per item decreases. 4t 50 t2 t2 4 4t 2t 50 t 2 2 4t 2 t2 2 t2 t2 2 Gm1m2 d2 F d 87. Pt P t 500 1 500 500 2000 50 200 50 88. F dF dd Gm1m2 d 2 Gm1m2 d3 2 50 50 P 2 31.55 bacteria per hour Section 2.3 1 cos x 1 d dx cos x 1 sin x d 1 dx sin x cos x sin x d cos x dx sin x sec x csc x, g x sec x tan x csc x cot x csc x cot x 0, 2 sin x sin x 0 sin x 1 cos x 2 Product and Quotient Rules and Higher-Order Derivatives 129 89. (a) sec x d sec x dx cos x 0 1 cos x 2 sin x sin x cos x cos x 1 cos x sin x cos x sec x tan x (b) csc x d csc x dx cos x sin x sin x 1 sin x cos x sin x csc x cot x (c) cot x d cot x dx sin x cos x cos x sin x 2 sin2 x cos2 x sin2 x 1 sin2 x csc2 x 90. f x gx f x sec x tan x 1 sin x cos x cos x 1 1 cos x sin x sin x 1 sin3 x cos3 x 1 tan3 x 1 tan x 1 x 3 7 , 4 4 3.5806t 3 0.1361t 3 vt nt 82.577t 2 3.165t 2 603.60t 23.02t 1667.5 59.8 n(t) 91. (a) n t vt (c) A 0.05 (b) 350 12 0.1361t3 3.165t 2 23.02t 3.5806t3 82.577t 2 603.60t 59.8 1667.5 v(t) 5 0 12 5 0 12 (d) A t represents the rate of change of the average retail value per 1000 motor homes. 5 0 12 A represents the average retail value (in millions of dollars) per 1000 motor homes. r r h h h r csc r csc r r csc 1 92. (a) sin r (b) h h 30 r h csc 6 3960 2 32 x2 64 x3 cot 3 7920 3 mi radian 93. f x f x f x 4x3 6x1 3x 2 2 94. f x f x x 1 192 x4 1 2 3 x f x 130 Chapter 2 x x x 2 x 1 3 Differentiation x2 2x x 1 x2 2 x3 sec x sec x tan x sec x sec2 x sec x sec2 x tan x sec x tan x tan2 x 102. f 1 2 4 95. f x f x f x 1 1 1 x 1 x1 2 96. f x 1 1 f x f x 1 x 2 1 x x 2 1 97. f x f x f x 3 sin x 3 cos x 3 sin x 98. f x f x f x 99. f x f x x2 2x 100. f x f x 2 2x 2 2x 1 101. f x f 4 2 x 1 2x 2 1 x x x x 2x 2 0 1 2 x2 x f f 5 6 103. 4 3 y 104. 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 2 1 x 1 2 3 4 f 2 0 x 2 2. f x One such function is f x 105. f x f x f 2 2g x 2g x 2g 2 2 0 gx hx hxg x 2 hx h x h 2 4 106. f x f x f 2 4 hx h x h 2 4 107. f x f x f 2 108. f x gxh x hx 2 gxhx gxh x g2h 2 3 4 14 hxg x h2g 2 1 2 f x f 2 h2g 2 h2 1 10 2 1 g2h 2 2 3 4 2 Section 2.3 109. f 2 1 y Product and Quotient Rules and Higher-Order Derivatives 110. 3 y 131 f f f f x 2 1 1 2 -2 -1 x -1 -2 2 3 4 f It appears that f is cubic; so f would be quadratic and f would be linear. 111. f 4 3 2 1 x 1 2 3 4 5 y It appears that f is quadratic; so f would be linear and f would be constant. y f 3 2 1 1 -1 x -1 -2 -3 -4 x 112. f 113. 4 y f f -3 -2 -1 -3 -4 -5 f 2 2 -4 -3 -1 114. 2 1 y 115. v t at f 2 36 2t t 2, 0 t 6 f x v3 a3 27 m sec 6 m sec2 The speed of the object is decreasing. 116. v t at 100t 2t 15 2t 15 100 100t 2 2t 15 2 2 (a) a 5 (b) a 10 (c) a 20 1500 25 15 2 2.4 ft sec2 1.2 ft sec2 0.5 ft sec2 1500 2 10 15 1500 2 20 15 2 1500 2t 15 117. s t vt at t sec st vt at ft 8.25t 2 16.50t 16.50 2 66t 66 Average velocity on: 0, 1 is 1, 2 is 2, 3 is 3, 4 is 57.75 0 1 0 99 2 57.75 1 57.75. 41.25. 24.75. 8.25. 0 0 s t ft sec v t ft sec2 66 16.5 1 57.75 49.5 16.5 2 99 33 16.5 3 123.75 16.5 16.5 4 132 0 16.5 123.75 99 3 2 132 4 123.75 3 132 Chapter 2 y 16 12 8 4 -1 Differentiation (b) The particle speeds up (accelerates) when a > 0 and slows down when a < 0. s 118. (a) Answers will vary. t 4 5 6 7 v 1 a s position function v velocity function a acceleration function 1 x 1 n 119. f x f n xn nn nn 1 n 2 ... 2 1 2 n! 120. f f x n x Note: n! 1 ...3 1 (read "n factorial") x n n 1 n xn 1 2 ... 2 1 1 nn! x n 1 121. f x (a) gxhx f x f x gxh x gxh x gxh x f x gxh gxh f 4 hxg x g xh x 2g x h x g xh x 3g x h x g xh g x hxg x hxg x 2g x h x 3g x h x 3g x h g x 1 n 1 n 4 h xg x x x 2g x h x g xhx hxg x h xg x x g x h4 x x 3g x h x 3g x h x 3g xh x xh x xhx 4g xh x 1 g x h4 x (b) f n 4g x h nn 1 n 6g x h x g4 x h x x nn 1 n 2 . . . 2 1 g x hn 2 1 n 2 n 3 . . . 2 1 3 2 x g x hn x 2 . . . 2 1 g x hn 2 . . . 2 1 x nn 1 n 2 . . . 2 1 g x hn 3 2 1 n 3 n 4 . . . 2 1 nn 1 n 2 . . . 2 1 gn n 1 n 2 . . . 2 1 1 g x hn x n! 1! n n Note: n! 122. xf x xf x xf x nn 1! g x hn 1 1 1 x . . . xh x g x hn gn x h x 2 x n! 2! n 2! x . . . n! gn 1 !1! 2 xh x gn x h x 1 . . .3 f x f x f x f x 1 (read "n factorial") xf x xf x xf x n xf x xf n 1 2f x x 3f x 2f x n In general, x f x xf x nf x. Section 2.3 Product and Quotient Rules and Higher-Order Derivatives cos x xn x x n n 133 123. f x f x x n sin x x n cos x xn 1 124. f x nx n 1 x sin x 1 n cos x nx n 1 sin x f x sin x 2 sin x 3 sin x 4 sin x n sin x . When n When n When n When n cos x x cos x n sin x x cos x x x cos x x2 x cos x x3 x cos x xn 1 x sin x n cos x When n When n When n When n 1: f x 2: f x 3: f x 4: f x x sin x n cos x xn 1 1: f x 2: f x 3: f x 4: f x x sin x x2 x sin x x3 x sin x x4 x sin x x5 x sin x n cos x . xn 1 4 cos x 3 cos x 2 cos x cos x For general n, f x x cos x For general n, f x 125. y x3y 1 ,y x 2x2y 1 ,y x2 x3 2 x3 2 x3 2x 2 1 x2 2 2 0 126. y y y y y xy y y y 2y 2x3 6x 2 12x 12 12 sin x cos x sin x sin x 6x 6 10 x 12x 2 6x 2 6 24x2 127. y y y y y 2 sin x 2 cos x 2 sin x 2 sin x 3 128. 3 cos x 3 sin x 3 cos x 3 cos x 2 sin x 3 3 y 130. True. y is a fourth-degree polynomial. dny 0 when n > 4. dx n y 3 cos x sin x 0 129. False. If y dy dx f x g x , then gx f x. 131. True h c f cg c f c 0 0 gc f c gc 0 f xg x 132. True 133. True 134. True. If v t c then a t v t 0. 135. f x f x ax 2 2ax bx b c x-intercept at 1, 0 : 0 2, 7 on graph: 7 4a a 2b 4a b c b c Slope 10 at 2, 7 : 10 Subtracting the third equation from the second, 3 b b 2. Finally, 3 Then, 10 4 3 f x 3x 2 2x 1 b 2 c. Subtracting this equation from the first, 3 c c 1. a. 134 Chapter 2 ax3 3ax 2 0 x Differentiation bx 2 2bx cx c 2b 4b2 6a 12ac d, a 0 136. f x f x f x (a) No horizontal tangents: f x 4b2 12ac < 0 c x x2, if x 0 x2, if x < 0 2x 1, b 0: 0 (b) Exactly one horizontal tangent 4b2 12ac 0 1, b 3x 2 3x 3, c 3: (c) Exactly two horizontal tangents 4b2 12ac > 0 1, a x2 1, c 0: Example: a f x x3 Example: a f x x3 Example: b f x x3 137. f x f x f x xx 138. (a) fg fg fg fg fg f g fg True f g 2x, if x 0 2x, if x < 0 2, if x > 0 2, if x < 0 (b) fg fg fg fg fg fg fg 2f g f g fg f g False f g f 0 does not exist since the left and right derivatives are not equal. Section 2.4 y 1. y f gx 6x 1 x x2 3 tan csc3 x 3x 2 7 3 2 The Chain Rule u gx 6x 5 y y f u u4 5 4 u 2. y 1 1 x2 u x 1 y u 1 2 3. y 4. y 5. y u u u x2 x2 csc x 3x 2 1 y y y u 3 tan u u3 6. y cos u y cos u 7. y y 9. g x g x 2x 3 2x 34 8. y 2 4 3 34 15 4 9t 2 9t 3 x2 5 4 7 6 2x 7 2 y 10. f t x2 2 2x 30x 4 x2 4 9x 2 3 12 4 9x 9 108 4 9x 3 f t 2 1 3 9 3 6 9t 2 Section 2.4 11. f t f t 1 1 1 2 9x2 1 2 9x 3 24 2 t 1 2 The Chain Rule 1 2 135 12. g x 1 2 5 1 5 2 x2 1, 1, 34 2 32 3 2 4 3x 3x 5 1 2 3x 3 t 1 1 2 1 t g x 3 2 5 x 1 2 3x x 1 13. y y 4 1 3 14. g x 2 3 2x 1 4 18x 6x 9x2 4 2 3 g x x > 1 x < 1 9x 9x 9x 1 4 15. y y x2 1 4 16. f x 34 1 4 4 x2 2x x 4 f x x2 3 f x 4 3 4 9 42 27 9x 3 4 17. y y x 1x 2 1 18. s t 2 t2 t2 1 t2 1 3t 3t t2 2t 3t 1 1 3t 3 1 2 1 19. f t f t 2 t 2t 3 2 2 1 x 1 2 2 st s t 3 3 t 2 3 3 1 2t 3 20. y y y 5 t 5t 15 t 3 3 3 21. y dy dx x 1 x 2 2 1 2 22. g t 3 2 1 t2 t2 2 2 1 2 3 3 4 2 1 gt g t t 15 3 4 2x 2 3 2 1 2 t 2 t t2 2 2 3 2 2t 3 2 23. f x f x x2 x x2 4 x 2x x 2x x 2 4 24. f x 3 x 3x x 3 3x 3x 27 x 1 2 x 16 2 1 2 1 x 16 2 2 9 3 2 2 2 3 3 1 x 2 x 2 2 4 2x f x 9 9 2 2 3 3x 3 3x 9 9 3 1 2x 3x 9x 2 3 4x 25. y y x 1 x 1 1 2 x2 1 1 1 x2 x2 x2 x2 1 2 x1 1 2 x2 2x 1 1 1 2 26. y 1 x2 1 2 x2 x2 1 2 1 y 2x x2 x 16 x2 1 2 1 2 x2 1 2 x2 x2 x3 2 16 x 2 x 3x 2 2 16 32 x2 x 16 2x2 1 x2 136 Chapter 2 x x2 x 1 1 2 x 2 x2 x2 x2 1 x2 1 3 2 Differentiation x x x4 4 27. y y x x2 1 1 3 2 3 2 1 2x x2 x2 1 2 28. y x2 1 1 1 2 4 4 1 2 1 1 2 1 y 1 x4 x 4 4 2x 4 x4 4 3 2 4 x4 x4 43 2 x1 x4 2 4 4 1 2 4x 3 3 2 1 x2 29. g x g x 2 x x2 x x2 5 2 5 2 2 30. h t x2 10x 23 2 x2 x2 x 2 5 2x 2 t2 t 2 3 2 2 t2 t3 2 2 2t t 2 3t 2 t3 2 2 2t 3 4 t 3 t3 2 3 h t t3 2x 5 2 x2 2v v 2v v 3 2t 2 4t t 4 t3 2 3 3x 2 2x 3 3x 2 2x 3 31. f v f v 3 1 1 1 1 32. g x 2 2 3 2 3 2 1 2 v 1 2 v 2 1 2v 2 g x 2x 3 6x 2x 4 2 3x 2 32 2 2 9 1 2v 1 v4 3 3x 2 6 3x 2 6x 2 18x 2x 3 4 9x 4 2 2 2 3x 2 2x 3 33. y y x x2 1 1 4x3 2 12 -1 2 34. y y 5 2x x 1 2x x 1 3 2 -6 1 y 7 1 3x2 2 x x2 y -2 y y 6 -1 The zero of y corresponds to the point on the graph of y where the tangent line is horizontal. x x x 1 x 2x x 1 1 y has no zeros. 35. y y 4 36. g x y x 1 2 x 1 1 x 1 2 x 1 g x 4 -5 1 6 y y has no zeros. g has no zeros. -2 g g -2 -2 10 Section 2.4 cos x x x sin x sin 1 x x2 x cos x x2 1 -3 The Chain Rule 137 37. y dy dx 3 y 38. 5 y dy dx x2 tan 1 x 1 sec x 2 -4 6 y cos x 1 -5 y 1 2x tan x 5 y The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. 39. (a) y y y 0 sin x cos x 1 (b) y y y 0 sin 2x 2 cos 2x 2 The zeros of y correspond to the points on the graph of y where the tangent lines are horizontal. -6 40. (a) y y y 0 sin 3x 3 cos 3x 3 (b) y y y 0 sin x 2 1 x cos 2 2 1 2 1 cycle in 0, 2 2 cycles in 0, 2 The slope of sin ax at the origin is a. 3 cycles in 0, 2 Half cycle in 0, 2 The slope of sin ax at the origin is a. 41. y dy dx cos 3x 3 sin 3x 42. y dy dx sin x cos x 43. g x g x 3 tan 4x 12 sec2 4x 44. h x h x sec x2 2x sec x2 tan x2 45. y y sin cos x 2 sin 2 2x 2x 2 46. y 2x cos 1 sin 1 41 2x 2 2 cos 1 21 2x 2x 2 2x 2 2 x2 2 cos 2x 2 y 2x 2 2x sin 1 1 2 1 2 1 2 1 2 1 2 1 2 47. h x h x sin 2x cos 2x sin 2x 2 sin 2x cos 2x 2 cos 2x 48. g g sec sec 1 2 tan sec2 1 2 tan 1 2 1 2 sec 1 2 tan 1 2 1 2 2 cos2 2x 2 cos 4x 2 sin2 2x sec sec2 tan2 Alternate solution: h x h x cot x sin x sin2 x cos x sin2 x sin x 1 2 sin 4x 1 2 cos 4x 4 2 cos 4x cos v csc v 49. f x f x 50. g v cos x 2 sin x cos x sin4 x 1 cos2 x 3 sin x 52. g t 5 cos 2 10 cos 10 t t sin 5 cos sin t t t 2 cos v sin v sin v sin v g v cos v cos v cos2 v sin v 2 cos 2v sin2 x 2 cos2 x sin3 x 51. y y 4 sec2 x 8 sec x 2 53. f f 1 4 sin2 2 1 4 1 4 sin 2 2 sec x tan x g t 2 sin 2 cos 2 2 4 8 sec x tan x t cos sin 2 cos 2 1 2 sin 5 sin 2 t 138 Chapter 2 2 cot2 4 cot t t Differentiation 2 2 t csc2 2 csc2 t t 2 2 55. f t f t 3 sec2 6 sec 6 sec2 t t t 1 1 sec 1 tan t t 1 tan 1 t 1 54. h t h t 4 cot 6 sin t 1 cos3 t 1 56. y 3x 3x 5 cos 5 cos 5 sin 10 2 x 2 57. y x x 1 sin 2x 4 1 sin 4x2 4 1 2 2 2x 2 2 2 2 dy dx 3 3 x 2 2 x dy dx 1 x 2 1 2 x x sin x 1 cos 4x2 8x 4 2x cos 2x 2 58. y y sin x1 cos x1 3 sin x 1 x 3 3 2 3 1 3 59. 1 sin x 3 3 2 3 st s t t2 1 2 t 2 t t2 3 4 3 x3 2t 2t 1 2t 8 1 2, 1 2 2, 4 2t 2 3 cos x 8 1 cos x1 3 x2 3 cos x sin x 2 8 s 2 60. y y 3x3 1 3 3x 5 4x 1 5, 4 5 2, 2 9x2 4 61. f x f x f 1 4 3 x3 9 25 3 x3 4 2 4 3x2 1, 1, 9x2 x3 4 3 5 2 4x 9x2 4 5 3x3 4x 4 y 2 1 2 1 x2 2 x2 5 32 x 2x 2x 5 1 , 3 3x 2 5 62. f x f x f 4 x2 3 3x 3 2, 4, 1 16 63. f t f t f 0 3t t t 5 2 , 1 1 3 t 0, 2 3t 12 2 1 t 5 1 3x 2x 2 2x 3 x2 3x 3 2 64. f x f x f 2 2, 3 x 3 2 65. 1 2 2x 5 3 2 y y 37 sec3 2x , 0, 36 3 1 2x 3 sec2 2x 2 sec(2x tan 2x 6 sec3 2x tan 2x y 0 0 Section 2.4 1 x 1 x2 2 The Chain Rule 139 66. y y y cos x, 2 , sin x 2 cos x 2 is undefined. 3x2 1 2 3x 2 9 5 (b) 3x 3x2 2 67. (a) f x f x f 3 2, 2 3, 5 1 2 7 6x (3, 5) -5 5 -3 Tangent line: y 5 9 x 5 3 9x 5y 2 0 68. (a) f x f x 1 x x2 3 1 1 2 x x 3 2 x2 3 x2 5, 5 2, 2 1 2 (b) 2x 5 1 2 x 3 5 1 2 -9 6 9 5 1 3 3 2 1 2, 1 3 x2 13 9 13 x 9 1, 1 -6 f 2 4 33 Tangent line: y 69. (a) y y y 1 2x3 2 2x3 12 1 2 13x 9y 8 0 70. (a) f x 9 2 9 3 4 38 1 3 x2 2 3, 1 3 1, 4 2x 39 4x x2 1 3 1 6x 2 12 1 y 12x 2 2x3 1 f x f 1 x2 Tangent line: y 12 x 12x 1 11 2 3 4 y 2 x 3 2 x 3 1 14 3 Tangent line: y (b) 4 (- 1, 1) -3 -1 2 (b) 6 (1, 4) -2 -1 5 71. (a) f x f x f sin 2x, 2 cos 2x ,0 (b) 0 2 ( , 0) 2 2 -2 Tangent line: y 2x 2x y 2 0 140 Chapter 2 Differentiation 2 2 72. (a) y y y 4 cos 3x, 3 sin 3x 3 sin 3 4 4 , 73. (a) f x f x tan2 x, 4 ,1 2 tan x sec2 x 21 2 4 3 2 2 3 2 x 2 3 2 x 2 4 3 2 8 2 2 f 4 Tangent line: y 2 2 y Tangent line: y (b) - 1 4 x 4 4 4x y 1 0 (b) - 2 2 ( 4 , - 22 ( 2 ( , 1( 4 -4 -2 74. (a) y y y 2 tan3 x, 6 tan2 x 4 4 ,2 75. (a) gt g t g 1 2 3 2 y t2 3t 2 2t 1 , 1 3 , 2 2 sec2 x 12 2 y 12 x 12x 2 3t t 2 3t 2 t 2 2t 1 3 2 3 3 x 3x 5 61 2 Tangent line: y 4 3 (b) y 1 2 3 (b) 3 ( , 2( 4 - 2 -1 (c) 2 -2 (1 , 3 ( 2 2 -2 4 76. (a) f x f x f 4 (b) y 8 y (c) 12 x 2 x 9 9x 9x 4 28 x 2, 2 4, 8 77. (a) st s t s 0 4 2t 1 3 2 1 3 t t , 2 0, t 4 3 t 2 5x 2 x 3 1 0 4 3 y 0x 4 3 3 (b) y 0 (4, 8) (c) 0 0 -2 6 (0, 4 ( 3 4 -1 Section 2.4 78. (a) y y y 2 (b) y 10 y (c) -2 3 5 The Chain Rule 141 t2 9 t 2, 2, 10 79. f x f x f 3 25 x 25 3 4 4 y 3 x 4 3 x 4 8 x2, x2 3, 4 5t 2 8t 9 2 t 2 27 4 27 t 4 27 t 4 2 47 2 y 3 25 ; Tangent line 4 (3, 4) -9 9 (2, -10) -18 -4 80. f x f x f 1 y 1 y x 2 2 2 2x 2x 3 x2 2 x2 , 1, 1 for x > 0 81. f x f x 2 cos x 2 sin x 2 sin x sin 2x, 0 < x < 2 2 cos 2x 2 4 sin2 x 0 3 2 2 sin2 x 1 1; Tangent line sin x sin x 1 1 sin x sin x 0 0 1 x 1 x 2 3 5 , , 6 2 6 6 , 3 3 3 , 0 , and , 2 2 3 2 5 , 6 6 1 2 sin x (1, 1) -2 -1 2 Horizontal tangents at x Horizontal tangent at the points 5 , 6 x 2x 2x 2x 2x x 2x x 2x 1 13 2 3 3 2 82. f x f x 1 1 1 1 1 2 83. x 2x 2x 1 1 1 2 f x f x 2 x2 6 x2 12x x 4 12x5 1 1 3 2 2x 1 12x 12 1 2x2 24x3 72x2 1 x2 x 3 2 f x 2 60x 4 12 5x2 1 13 0 x 1 Horizontal tangent at 1, 1 142 Chapter 2 x x 2x sec2 2 sec 2 sec2 f x 2 sec2 2 2 2 2 2 2 Differentiation 1 84. f x f x f x 86. f x f x 2 2 2 x x 85. 2 f x f x sin x 2 2x cos x 2 2x 2x 2 cos x2 sin x2 2 cos x2 x 3 1 2 2 2 3 2 f x x 2x2 sin x2 sec x tan x sec2 x 4 x tan x x 2 x 2 tan sec2 x tan2 x x x 2 sec2 x tan x sec4 sec2 sec2 x sec2 x 3 sec2 x x 2 tan2 2 87. hx h x h x h 1 1 3x 9 1 3 3x 9 2 3x 24 1 3, 1 2 1, 3 64 9 3x 1 1 2 88. f x f x f x f 0 1 x 1 x 2 3 x 4 3 128 4 4 4 x 3 2 4 1 2, 0, 1 2 1 3 6 3x 5 2 3 4x 4 5 2 89. f x f x f x cos x 2 , sin x2 0, 1 2x 2x sin x2 90. gt g t g t tan 2t , 2 sec2 2t 4 sec 2t 8 sec2 6 , 3 2x cos x 2 2x 4x 2 cos x 2 2 sin x 2 2 sin x 2 g sec 2t tan 2t 2 2t tan 2t f 0 0 6 32 3 91. f 3 2 1 y 92. f 3 2 1 y f f x 2 2 2 3 3 -3 -2 -1 x -1 -2 -3 1 2 3 f f The zeros of f correspond to the points where the graph of f has horizontal tangents. f is decreasing on f is increasing on 1, , 1 so f must be negative there. so f must be positive there. Section 2.4 93. 3 2 y The Chain Rule 143 94. 4 y f 3 2 f x 4 x 3 1 2 -1 -2 f f -3 -4 The zeros of f correspond to the points where the graph of f has horizontal tangents. 95. g x g x 97. (a) f x f x f 5 (c) f x f x f 5 f 3x f 3x 3 g x gxhx gx h x 3 gx hx hxg x hx 3 6 3 f x 2 The zeros of f correspond to the points where the graph of f has horizontal tangents. 96. g x f x2 f x 2 2x g x ghx g hx h x g 3 2 2g 3 2x f x 2 3 f 3x g x (b) f x g x hx 6 3 24 f x f 5 2 Need g 3 to find f 5 . (d) gx h x 2 f x f x gx 3gx 3 3 3 2g 2 x 162 3 2 12 9 f x 4 3 f 5 6 98. (a) g x (b) h x (c) r x 2g x 2f x h x f 3x r x 2f x f 3x 3x . 1 3 x 3 3f 3x f x g x h x 12 2 2. r x s x 2 4 4 8 2 3 2 3 4 3 1 0 1 3 1 3 2 3 1 1 1 2 2 2 2 4 3 4 4 8 Hence, you need to know f r 0 r (d) s x 1 f x 3f 0 3f 3 2 s x 3 3 1 4 12 1 3 1 2 4 f x 1 Hence, you need to know f x s 99. (a) h x h 1 2 f 0 f gx , g1 f g1 g 1 f 4g 1 1 (b) sx s x s 5 1 2 1 2 1 3, etc. 4, g 1 1 2, f 4 1, h x f gx g x g f x , f 5 g f x f x g f 5 f 5 6, f 5 1, g 6 does not exist. g 6 1 Since g 6 does not exist, s 5 is not defined. 144 Chapter 2 hx h x h 3 f gx Differentiation (b) sx s x s 9 g f x g f x f x g f 9 f 9 g 8 2 1 2 2 100. (a) f gx g x f g3 g 3 f 5 1 1 2 101. (a) F F 132,400 331 v 1 (b) F v 2 132,400 331 v 1 1 132,400 331 132,400 331 v 2 1 F 1 132,400 331 132,400 331 v 2 v 2 1 When v 1 cos 12t 3 y 1 3 30, F 1.461. When v 30, F 1.016. 102. y v 1 sin 12t 4 12 sin 12t 3 cos 12t 0.25 feet and v 4 feet per second. 1 12 cos 12t 4 103. 0.2 cos 8t The maximum angular displacement is 1 cos 8t 1 . d dt When t second. 0.2 8 sin 8t 1.6 sin 8t 1.6 sin 24 1.4489 radians per 0.2 (since 4 sin 12t When t 8, y 3, d dt 104. y A cos t 3.5 2 1.75 105. S dS dt C R2 C 2R r2 dR dt 2r dr dt 0 and 2 (a) Amplitude: A y Period: 10 y (b) v y 1.75 cos t 2 10 t 5 t 5 0.35 sin t 5 5 Since r is constant, we have dr dt dS dt 1.76 4.224 105 2 1.2 10 2 10 10 5 1.75 cos 1.75 0.04224. 5 sin 106. (a) Using a graphing utility, or trial and error, you obtain a model similar to T t (b) 100 (c) T t 15 11 cos t 6 2.1 65 21 sin t 6 2.1 . 0 13 -15 0 0 13 (d) The temperature changes most rapidly in the spring (AprilJune) and fall (SeptemberNovember). Section 2.4 107. (a) x (b) C 1.6372t3 60x 60 dC dt 60 1350 1.6372t3 4.9116t2 19.3120t2 38.624t 2317.44t 0.5082t 0.5082 30.492 0.6162 1350 19.3120t2 0.5082t 0.6162 The Chain Rule 145 294.696t2 The function dC dt is quadratic, not linear. The cost function levels off at the end of the day, perhaps due to fatigue. 108. f x sin x cos x 2 3 4 2 109. f x p f x for all x. (a) f x f f f x x 4 sin x cos x (a) Yes, f x p f x , which shows that f is periodic as well. (b) Yes, let g x f 2x , so g x Since f is periodic, so is g . 2 2 f 2x . sin x sin x 2 (b) f x (c) f f 2k 2k 1 f x 1 1 k 2k k 1 sin x 0 x x sin x 2k 1 cos x (b) s x s 4 5 6 0. 0 2 5 . 4 g f x f x g f 4 f 4 5 ,g 2 5 2 6 6 1 5 2 4 4 2 5 . 8 1 and 2 110. (a) r x r 1 f gx g x f g1 g 1 4 and f 4 0. Thus, r 1 Note that g 1 Also, g 1 Note that f 4 f 4 5 . Thus, s 4 4 1 1 sec2 x f x 111. (a) g x g x sin2 x cos 2 x 1g x 2 cos x sin x 0 0 (b) tan2 x gx 2 sin x cos x 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. 112. (a) If f x x 1 x f x , then d dx f x f (b) If f d f dx x f x x 1 x f x , then d f x dx f x f x. d f dx f x f f x f x. Thus, f x is even. 113. u d u dx uu u2 d dx u u2 u2 u , u u 1 2 u 2 0 1 2 Thus, f is odd. 114. g x 2uu g x 2x 2 2x 2x 3 3 , 3 x 3 2 146 Chapter 2 x2 2x 4 x x2 2 Differentiation 116. h x 4 , 4 x 4 sec2 sec2 x 4 x 4 1 1 2 115. f x f x x cos x x sin x x cos x, x x 0 117. f x f x sin x cos x sin x , x sin x k x 2 h x 118. (a) f x f x f x P1 x P2 x (b) 2 tan f 1 f 1 tan x 4 4 f 1 2 x 1 1 1 2 1 2 2 4 2 4 8 2 2 f 1 2 1 4 f 1 x 1 2 x 2 4 f 1 x f 1 8 x 1 2 2 x 1 1 P1 P2 f 0 0 3 (c) P2 is a better approximation than P1. (d) The accuracy worsens as you move away from x 119. (a) f x f x f x sec 2x 2 sec 2x tan 2x 2 2 sec 2x tan 2x tan 2x 4 sec 2x tan2 2x f f f 6 6 6 sec 3 3 2 tan 23 4 3 56 2 2 c 1. (b) 6 P 2 2 sec 2x sec2 2x 2 0 0 P1 f 0.78 sec3 2x (c) P2 is a better approximation than P1. (d) The accuracy worsens as you move away from x 6. 2 sec 423 4 3 x 1 56 x 2 28 x 3 P1 x P2 x 6 6 2 4 3 x 4 3 x 6 6 2 2 6 x 1 2, 1 120. False. If y 1 y x 1 2 1 then 2 1. sin2 2x, then 121. False. If f x f x 2 sin 2x 2 cos 2x . 122. True Section 2.5 123. f x f x f 0 a1 2a2 ... nan a1 sin x a1 cos x a1 f 0 x0 Implicit Differentiation 147 a2 sin 2x 2a2 cos 2x ... nan ... ... an sin nx nan cos nx 2a2 lim f x x f x sin x f 0 0 sin x x x 0 lim x0 lim f x 1 sin x 1P n 124. d dx Pn x x 1 k xk n 1 1 n x Pn x n x 1 2n k 1 xk 2 1P n 1 n kx k 1 xk 1 Pn x xk n 1 1 kx k n 2 x Pn x Pn Pn x 1 1, xk xk n 1 1 n 1 dn 1 dx n x k 1 1 n 2 d dn 1 dx dx n x k 1 xk 1 Pn x n 1 kx k 1P n x 1 1 kPn 1 kx k xk 1 1 2 For n d 1 dx x k 1 P1 x xk 1 2 P1 1 k. Also, P0 1 1. We now use mathematical induction to verify that Pn 1 Pn 1 k nn! for n 0. Assume true for n. Then 1 n n k n 1 k Pn 1 1k 1 k nn! 1 !. n Section 2.5 1. 2x x2 y2 2yy y 36 0 x y Implicit Differentiation 2. 2x x2 y2 2yy y 16 0 x y 3. 1 x 2 1 2 x1 2 y1 1 2y 2 9 0 x y 1 2 1 2 1 y 2 y y x 4. 3x2 x3 y3 3y2y y 8 0 x2 y2 5. 3x 2 x3 xy y xy y2 2yy xy y 4 0 y 3x2 2y y 3x2 2y x 148 6. x2y Chapter 2 x2y 2xy y2 x 2 Differentiation y2x 2yxy 2xy y y 0 y 2 2 7. 3x3y2y 2xy x3y3 3x2y3 3x y 3 2 y y x 1 1y y 0 0 1 3x2y3 y y 2x x x 2y 2y 2y 2y 0 0 0 0 2 xy 4 xy 1 0 cos x cos y sin x sin y cot x cot y y x 11. sin x cos x 2 cos 2y 4 sin 2y y y 1 0 9. 3x 2 3x 2y x3 6xy 1 3x2y3 3x3y2 1 2xy 2 2y2 3x 2 y y 12 0 6xy 6xy 4xy 3x 2 3x 2 3x 2 2y 2 2y 2 8. 1 xy 2 x 2 xy xy 1 2 xy xy y 1 2 x 1 1 3x 2y 4xyy 4xy y y y 2 xy 2 xy 4 xy y y 10. 2 sin x sin y y 2 sin x cos y cos y cos x y cos x 4 sin 2y 12. 2 sin x cos y cos sin x x cos sin y y y 2 2 0 0 cos sin x y 13. sin x cos x y x1 tan y 1 1 tan y 1 x sec2 y y cos x tan y x sec2 y cos x sin y y y 14. cot y csc2 y y y x 1 1 y y 1 csc2 y 1 cot 2 y tan2 y 15. y y y x cos xy y y sin xy xy y cos xy 16. x 1 y sec 1 y y cos xy y cos xy 1 x cos xy y 1 1 sec tan y2 y y y2 sec 1 y tan 1 y y2 cos 1 1 cot y y 17. (a) x2 y2 y2 y 16 16 (b) x2 16 x2 -6 -2 6 y y1 = 16 - x 2 2 x 2 6 (c) Explicitly: dy dx 1 16 2 x 16 x2 1 2 2x x 16 x2 x y -6 y2 = - 16 - x 2 (d) Implicitly: 2x 2yy y 0 x y x2 Section 2.5 18. (a) x2 4x 4 x 2 y2 2 Implicit Differentiation 149 6y y y 9 3 3 2 2 9 4, 4 3 4 Circle x 4 2 9 (b) -1 -2 y x 1 2 3 4 5 2 y = -3 + 4 - (x - 2) 2 y (c) Explicitly: dy 1 4 dx 2 x 4 x x 2 2 -3 -4 -5 x 2 x 2 2 x x 4 2 3 2 2 1 2 2 x 2 y = -3 - 4 - (x - 2)2 (d) Implicitly: 2 2x 4 3 x y 2 x 2 2yy 4 2y 6y 6y y 0 2x x y 2 2 3 2 2 2 3 19. (a) 16y2 y2 y 144 1 144 16 9x2 9x2 x2 9 16 16 x2 (b) 6 4 2 -6 -2 -4 -6 y y1 = 3 4 16 - x2 3 4 x 2 6 16 (c) Explicitly: dy dx 3 y2 = - 4 16 - x2 3 16 8 3x 4 16 x2 1 2 2x 3x 44 3y 9x 16y (d) Implicitly: 18x 32yy y 0 9x 16y x2 20. (a) 9y 2 y2 y x2 x2 9 9 1 x2 3 x2 9 9 9 (b) -6 4 6 -4 (c) Explicitly: dy dx 2 x2 x 3 x2 1 (d) Implicitly: 9 3 9 1 2 9y 2 18yy x2 2x 18yy 9 0 2x 2x 18y x 9y 2x x 9y x 3 3y y 21. xy xy y1 xy y At 4, 4 0 y y x 1: y 1 4 22. 2x x2 y3 3y2y y 0 0 2x 3y2 2 3 At 1, 1 : y 150 Chapter 2 x2 x2 x2 4 4 Differentiation 23. y2 2yy 2yy y 24. x2 42 4 2x x3 3x2y x 3xy2 3x2y x2y y 3 x3 x3 0 0 0 y2 y3 y3 4 2x x2 2 y3 3xy2 xy 2 y2 16x x2 4 8x y x2 x 2y 2 2xy 2xyy x2 4 2xy y y 2xy 2x 2y At 2, 0 : y is undefined. At 25. 2 x 3 1 3 y y xx 1, 1 : y x3 y3 3y 2y 4x y y 4xy 4xy 4y 1 1 4y 3x 2 3x 2 4x 12 8 1 0 cos y x sin y 1 cot y x cot y x 8 5 x2 3 y2 1 3 3 5 0 x y 1 3 3 1 3 26. 3x 2 3y 2 y x 2 y 3 y y At 8, 1 : y 1 2 4y 3y 2 4 3 x cos y At 2, 1 : y 27. 1 y tan x sec2 x y y y x 1 1 sec2 x y sec2 x y tan2 x y x y 1 y 28. x y sin y cos y y tan2 sin2 x x2 x2 At 0, 0 : y 29. x2 4y x2 0 4y y 2x y 8 0 2xy x 4 2 1 At 2, 3 : y 1 2 3 30. 4 x 2yy 4 y2 x y2 1 y x3 3x2 3x2 2y 4 y2 x 2x 8 x2 4 x2 4 x2 At 2, 1 : y 32 64 16x 4 1 2 8 x2 4 2 At 2, 2 : y 2 Or, you could just solve for y: y Section 2.5 31. 2 x2 4x3 4x2yy 4x2yy x2 y2 2x 4xy2 4y3y y3 y2 2yy 4y3y 4x2y x2 y At 1, 1 : y 33. 2 y y 2 2y y At 4, 0 : y Tangent line: y 2 2 Implicit Differentiation 6xy 6y 6x y 0 0 6y 6y 3y2 16 9 8 3 3x2 3x2 6x 32 40 2y y2 4 5 x2 2x 151 4x2y 4x2y 4x2y 8xy 4 2xy 2xy x2y y 8x 8xy 4x3 x3 x3 y3 4xy2 xy2 xy2 x2 32. 3x2 3y2y x3 y3 6xy y 3y2 4y x2y At 4 8 , : y 3 3 16 3 64 9 0 4x 4 2 y 2 1 0 y 1x x 4 4 At 3, 4 : y Tangent line: y 2 4 y 7x 2 14x y x 1 1 y 1x x 2 1 Tangent line: y At 3, 1 : y 2x 2x 13y 2 3 10 y y 3, 4, 0 34. 2x x 1 1 2 y 2 y 2 y 2 2y 2y 2 20, 0 2x x 3, 4 1 1 2 35. xy xy y y 1, 0 1, 1 36. 6 3 xy 16 26yy y 0, 0 3, 1 6 3 xy 6 3y 6 3 y 14x 26y 6 3 x 8 3 8 3 4 3 At 1, 1 : y Tangent line: y 6 3 14 3 26 6 3 3 1 y 3x 3x 5, 0 x y 1 3 1 3 37. x 2 2yy x2y 2 2xy2 9 x2 18x 4y 2 8yy y 0, 0 18x 2x2y 4, 2 3 38. 2 x 3 1 3 x2 3 y2 1 3 3 8, 1 2xy2 8y 2 y 3 y y At 8, 1 : y Tangent line: y 1 2 1 y At 4, 2 3 : y 18 4 2 2 16 2 3 24 48 3 1 2 3 3 x 6 3 x 6 4 12 16 3 3 6 y x 1 3 1 x 2 1 x 2 8 5 Tangent line: y 2 3 y 4 8 3 3 152 39. Chapter 2 3 x2 6 x2 y 2 2x Differentiation y2 2yy 2 100 x 2 100 2x y2 , 2yy 4, 2 40. y2 x2 y 2x 2 2yy x 2 2xy 2 y2 y4 4y3y 2x 2, 2x 2 4x 1, 1 At 4, 2 : 6 16 4 8 960 4y 480y 880y y Tangent line: 11y 2x y 100 8 800 160 2 11 4y 400y At 1, 1 : 2y 2 4y 6y y 4 2 1 3 2 30 y 0 2 11 x 4 Tangent line: y 1 y 1 3 x 1 2 3 1 3x 2 11 x 30 11 41. (a) x2 2 x y2 8 yy 4 y 1, 0 4x y 1, 2 (b) x2 a2 y y0 y b2 y2 b2 y0 y02 b2 x02 a2 1 2x a2 2yy b2 x0 , x02 a2 0 y b2x a2y b2x0 x a2y0 x0 x a2 y02 b2 Tangent line at x0, y0 At 1, 2 : y Tangent line: y 2 2 y 2x 2x 2 y 8 1 4 1 1 y 4 1 x 2 1 y Since 1, you have y0 y b2 x0 x a2 1. Note: From part (a), 1x 2 2x 4, Tangent line. 42. (a) x2 6 x 3 y2 8 y y 4 y y 4 y 1, 0 x 3 4x 3y 2: y 3, 2 (b) x2 a2 y yy0 b2 Since y2 b2 y0 y02 b2 x02 a2 1 2x a2 2yy b2 x0 , 0 y xb2 ya2 x0 b2 x y0a2 x0 x a2 y02 b2 x02 a2 Tangent line at x0, y0 1, you have x0 x a2 yy0 b2 1. At 3, 43 3 2 2 y 2x 2x 2 3 4 1 1 x 2 y 4 1 y 2x 4, Tangent line. Tangent line: y Note: From part (a), 3x 6 2 y 8 Section 2.5 43. tan y y sec2 y y sec2 y y x 1 1 sec2 y 1 1 1 x2 cos2 y, 1 2 x2 < y < Implicit Differentiation 153 44. cos y sin y y y sin2 y cos2 y sin2 y sin y y x 1 1 , sin y 1 1 1 1 1 x2 cos2 y cos2 y , 1 x2 0 < y < 2 tan2 y 1 < x < 1 45. 2x x2 y2 2yy y y 36 0 x y y 1 xy y2 y x x y y2 x2y2 2x2yy x2yy 2xy2 xy2 y2 x2 y3 2x 2 1 y 2xyy x2 y 4xyy 4 4xy2 x 1 2xy2 1 2 36 y3 46. 3 0 0 1 0 0 0 0 2x2y 4 2x 2y 4 xy2 x2y x2yy x2 y x2y2 x2y 4 2 2xyy x2yy x2yy y2 y2 y2 x 2y 4 xy2 2 4xy2 4x2y 4 x 4y 3y x 4y 3y y 2xy2 2xy2 x 4y3 1 1 47. 2x x2 y2 2yy y 16 0 x y 0 0 0 x2 y2 y3 x2 16 y3 48. 1 y xy xy y y y x x x 1 0 0 y 1 1 x 1 49. y2 2yy y x3 3x2 3x2 2y 3y 2x 3x2 2y x3 y2 4x2 2x 3 3y 4x2 3y 2x 4x2 3x 4y 6y xy xy 3y 2x 3y 2 x 1 1 yy yy y2 yy y x y 2 2 y 2x 3y y3y y 154 50. Chapter 2 y2 2yy y y 4x 4 2 y 2y 2y Differentiation 51. 1 x 2 1 2 x 1 y 2 y 1 2y 4 0 9 (9, 1) y 2 y2 2 y 4 y3 At 9, 1 : y Tangent line: 1 3 y y x -1 -1 14 1 y 1 x 3 1 x 3 0 9 4 x 3y 12 52. y2 2yy x x2 x 2 1 1 1 1 x2 1 2x2 x2 1 2 x 12 2x 1 2x -1 1 (2, ) 5 5 5 x2 -1 y At 2, 1 2x x2 2y x2 1 2 5 : y 5 1 4 4 1 5 5 10 8 2 2 5 5 4 y 10 5y x 10 5y 1 10 5 1 x 10 5 x 0 2 2 Tangent line: 53. 2x x2 y2 2yy y 25 0 x y 6 At 4, 3 : Tangent line: y Normal line: y At 3, 4 : Tangent line: y Normal line: y 4 4 3 x 4 4 x 3 (-3, 4) 3 3 4 x 3 3 x 4 4 4x 4 3x 4y (4, 3) 3y 0 25 0 -9 9 -6 6 3 3x 3 4x 4y 3y 25 0 0 -9 9 -6 Section 2.5 54. x2 y2 y At 0, 3 : Tangent line: y Normal line: x At 2, 5 : 5 5 2 x 5 5 x 2 2 2x 2 5x 5y 2y 9 0 0 -6 Implicit Differentiation 155 9 x y -6 4 (0, 3) 6 3 0 -4 4 (2, 5 ) 6 Tangent line: y Normal line: y -4 55. 2x x2 y2 2yy y y x r2 0 x y slope of tangent line 56. y2 2yy y 4x 4 2 y 1 at 1, 2 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 y y0 y y0 x x0 y0 x x0 x0 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, x 1 2 3 x 2x 2 1 2 2 16 16 1 2 2. 2 2 and 2 2 2 2 2 2 2 2 x Centers of the circles: 1 1 2 2, 2 2 2 Equations: x x 1 1 2 2 2 2 2 2, 2 2 2 which passes through the origin. y y 16 16 57. 25x2 16y2 50x 200x 32yy 160y 200 400 160y y 200 160 0 0 50x 32y y (- 4, 10) 10 Horizontal tangents occur when x 25 16 16y2 200 4 160y 4: 400 10 0 0y 0, 10 (- 8, 5) (- 4, 0) -10 - 8 - 6 - 4 x -2 2 6 (0, 5) 4 y y Horizontal tangents: 4, 0 , 4, 10 5: 400 8 Vertical tangents occur when y 25x2 400 200x 800 25x x Vertical tangents: 0, 5 , 8, 5 0 0x 0, 8 156 58. 4x2 Chapter 2 y2 8x 8x 2yy Differentiation 4y 8 4 4y y 0 0 8 8x 2y 4 1: 0 y y 4 2: 4 8x 2 4x in the equation 2x2 0 2x 2 + y 2 = 6 4 y 2 = 4x -1 -1 y (1, 0) 1 2 3 4 x 4 y 4x 2 (0, - 2) -3 -4 -5 (2, - 2) Horizontal tangents occur when x 41 2 (1, - 4) y2 81 4y y2 4 4y 4 0y 0, 4 Horizontal tangents: 1, 0 , 1, Vertical tangents occur when y 4x2 2 2 8x 4 2 4x 2 0 4x x 2 0x 0, 2 Vertical tangents: 0, 2 , 2, 59. Find the points of intersection by letting y2 2x2 4x 6 and x 3 x 1 y2 6. The curves intersect at 1, 2 . Ellipse: 4x 2yy y 0 2x y Parabola: 2yy y 4 2 y -6 (1, 2) 6 (1, - 2) -4 At 1, 2 , the slopes are: y At 1, y 1 2 , the slopes are: 1 y 1 y 1 Tangents are perpendicular. 60. Find the points of intersection by letting y2 2x2 3x3 5 1. -2 x3 in the equation 2x2 5 0 3y2 5. 2 and 3x3 2x2 2x2 + 3y2 = 5 (1, 1) 4 Intersect when x Points of intersection: 1, 1 y2 2yy y x3: 3x2 3x2 2y 2x2 4x 3y2 6yy y 5: (1, - 1) -2 0 2x 3y y 2= x 3 At 1, 1 , the slopes are: y At 1, y 3 2 1 , the slopes are: 3 2 y 2 3 y 2 3 Tangents are perpendicular. Section 2.5 61. y x and x sin y -6 4 Implicit Differentiation 157 x = sin y Point of intersection: 0, 0 y y x: 1 x 1 y At 0, 0 , the slopes are: y 1 y 1 sin y: y cos y sec y (0, 0) 6 -4 x+y=0 Tangents are perpendicular. 62. Rewriting each equation and differentiating: x3 y y 3 y x3 3 x2 1 1 x 3y 29 y y 3 1 3 3 x 1 x2 29 -15 -3 12 x (3y - 29) = 3 15 x 3 = 3y - 3 For each value of x, the derivatives are negative reciprocals of each other. Thus, the tangent lines are orthogonal at both points of intersection. 63. xy xy y y C 0 y x 2x x2 y2 2yy y K 0 -3 2 2 C=4 C=1 K = -1 -2 3 -3 3 x y K=2 -2 At any point of intersection x, y the product of the slopes is y x x y 1. The curves are orthogonal. 64. 2x x2 y2 2yy y C2 0 x y y y Kx K -3 2 2 K=1 3 -3 K = -1 C=1 3 C=2 -2 -2 At the point of intersection x, y , the product of the slopes is x y K x Kx K 1. The curves are orthogonal. 65. 2y2 3x 4 0 12x3 4yy y (b) 4y dy dt 12x3 y 0 12x3 12x3 4y dx dt dy dt 0 3x3 dx dt 3x3 y (b) 2x dx dt 3y 2 dx dt 6xy 66. x2 3xy2 3y2 y3 10 6xyy 6xy 3y2y 3y y y dy dt 2x 2 (a) 4yy (a) 2x 0 3y2 3y 3y2 3y2 3y2 2 2x 2x 6xy 0 6xy 3y2 dy dt dy dt dx dt 158 67. cos (a) Chapter 2 y 3 sin sin y y x Differentiation 1 3 cos x y 0 3 cos x sin y dx dt dy dt 0 3 cos x dx dt (b) 4 sin x sin y dy dt 4 cos x 68. 4 sin x cos y (a) 4 sin x 1 sin y y 4 cos x cos y y 0 cos x cos y sin x sin y 0 sin x sin y dy dt (b) sin y dy dt 3 cos x sin y dx cos y dt dx dt cos x cos y 69. 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. 70. 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 . 71. 72. Highest wind speed near L 1671 18 00 B A 1994 18 00 Use starting point B. 73. (a) x 4 4y2 y2 y (b) y 4 4x2 16x2 4x2 y2 x4 - 10 10 10 1 4 x 4 4x2 1 4 x 4 4x 2 16x2 0 16 28 7, 256 2 144 8 1 7 x 8 x2 . 23 28 1 4 x 4 x4 - 10 - 10 39 36 10 10 x4 16x2 36 x2 y1 y3 - 10 y2 y4 Note that x2 1 8 7, 1 82 7 1 8x 7, 1 x3 y 7 2. Hence, there are four values of x: To find the slope, 2yy --CONTINUED-- Section 2.5 73. --CONTINUED-- For x y1 For x y2 For x y3 For x y4 1 1 3 1 3 1 1 3 1 1 3 7 7, y 7 7 x 7 1 7 7, y 7 x 7, y 7 x 7, y 7 x 1 3 1 1 3 1 1 3 1 7 1 3 1 7 7 7 7 , and the line is 3 1 3 7 7x 8 7 23 . Implicit Differentiation 159 7 , and the line is 7 7 7 3 1 3 7 7x 23 8 7. 7 , and the line is 3 1 3 7 7x 23 8 7 . 7 , and the line is 7 3 1 3 7 7x 8 7 23 . (c) Equating y3 and y4: 1 3 7 7 x 7 7x 7 7 1 7 x 7x 1 7 7 3 7 7 7 16 7 x If x 8 7 , then y 7 x 1 2 x 1 y 0 qy q y x y0 y0 x0 x x0 Thus, if y c 1 3 7 7x 14x 8 7 7 8 7 ,5 . 7 75. y yq 1 7 7 x 1 7 1 7 7x 7 3 7 x 7 7 7 7 5 and the lines intersect at 74. x p q; p, q integers and q > 0 xp px p p q p q 1 dy 2 y dx dy dx y y xp yq 1 1 p q q x p 1y yq p p x q q 1 Tangent line at x0, y0 : y x-intercept: x0 y-intercept: 0, y0 Sum of intercepts: x0 x0 y0 y0 xp 1 p x xp x n, n x0 y0, 0 x0 y0 p q, then y nx n 1. x0 y0 x0 2 x0 y0 x0 c 2 y0 y0 c 2 160 Chapter 2 Differentiation 3 4 76. 2x x2 y2 2yy y 25, slope 0 x y 25 25 3 77. y4 4y3y y2 2yy 2y x2 2x 4y3 y 2x 4y3 0 x 0 1 3 y 4 4 x 3 2x y x2 16 2 x 9 25 2 x 9 x 2y Horizontal tangents at 0, 1 and 0, Note: y4 y2 x2 y2 3, 4 0 1 1 2 4x2 Points: 3, 4 and If you graph these four equations, you will see that these are horizontal tangents at 0, 1 , but not at 0, 0 . 78. 2x 4 x2 4 y2 9 2yy 9 y 9x 4y 1, 0 9x 4y y x 4y 2 4, 0 0 4 9x x But, 9x 2 4 4y 2 36 4y 2 1, 3 2 3 36 9x 2. Hence, 9x 2 36x 4y 2 36 9x 2 x 1. Points on ellipse: At 1, At 1, 3 2 3 2 3 : y 3 : y 9x 4y 3 2 3 x 2 3 x 2 9 4 3 2 3 3 2 Tangent lines: y y 4 4 3 x 2 3 x 2 2 3 2 3 79. x 1 y y2 2yy 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 2y y2 x0 y2 x0 y y2 1 2 1 2 0 x0 --CONTINUED-- Section 2.6 79. --CONTINUED-- (a) If x0 (b) If x0 (c) If x0 1 4, 1 2, Related Rates 161 then y 2 then y2 1 4 1 2 1 4, which is impossible. Thus, the only normal line is the x-axis y 0. 0 y 1 2 0. Same as part (a). 1, then y 2 x and there are three normal lines: 1 1 1 , and the line joining x0, 0 and , , 2 2 2 1 and 1. Thus, 1 x 3 4. The x-axis, the line joining x0, 0 and 1 2 3 . 4 If two normals are perpendicular, then their slopes are 2y 1 y y2 0 y x0 1 2 and x 1 2 1 4 x0 3 4 1 4 x0 1 x0 2 The perpendicular normal lines are y x2 32 2x 32 y2 8 2yy 8 and y 80. (a) 1 0 y 4 42 x 4y 1 2 (c) x2 x2 32 4 4x 2 17x 2 17x Second point: 2x 8 24x 96x 28 x 28 , 17 6 2 1 32 0 0 x 4, 28 17 36 112 4 46 17 At 4, 2 : y Slope of normal line is 2. y 2 y (b) -6 2x 2x 4 4 6 6 -4 Section 2.6 1. y dy dt dx dt x dx 2 x dt 2 x dy dt 1 Related Rates 2. y dy dt dx dt 4 and dx dt 3 . 4 2, 3, 2 x2 4x 1 4x 3x 6 dx dt dy 6 dt 3 and dy dt 43 1 and dx dt 1 41 6 5 5 . 2 6 2 12. (a) When x dy dt (a) When x dx dt 2, 1 3 2 4 (b) When x dx dt 25 and dy dt 20. (b) When x dy dt 5, 2 25 2 162 3. x Chapter 2 xy dy dt y dx dt dy dt dx dt (a) When x dy dt (b) When x dx dt 1 4 8, y 4 0 Differentiation 4. 2x y dx x dt x dy y dt 1 2, and dx dt 5 . 8 4, and dy dt 3 . 2 6, 10, dx dt x2 2y y2 dy dt dy dt dx dt (a) When x dy dt (b) When x dx dt 1 1 2 2x x2 dx dt 2, 2 2 2 25 0, 8 cm sec. 25 x2 3 4 3, y 3 8 4 4, y 2 25 0 x dx y dt y dy x dt 4, and dx dt 6. 3, and dy dt 3 . 2 2, 8, 1 2 10 8 1, y 6 5. y dx dt dy dt x2 2 2x 1 6. y dx dt dx dt 1, 1 2 0, 0 cm sec. 4 cm sec. dy dt 1 2 (a) When x dy dt 2 (a) When x dy dt (b) When x dy dt (b) When x dy dt 20 2 1, 0 cm sec. 2, 22 2 25 8 cm sec. 25 (c) When x dy dt (c) When x 4 cm sec. dy dt 8. y dx dt sin x 2 cos x dx dt 21 2 7. y dx dt dy dt tan x 2 sec2 x dx dt 3, 2 dy dt (a) When x dy dt 2 2 (a) When x dy dt cos 6 6, 2 4, 4 2 3, 3 2 1 cm sec. 2 cm sec. 3 cm sec. 8 cm sec. 4, (b) When x dy dt (c) When x dy dt 1 2 (b) When x 4 cm sec. dy dt cos 2 2 2 0, 2 2 cm sec. (c) When x dy dt cos Section 2.6 dx dy negative positive dt dt dy dx positive negative dt dt dx dy negative negative dt dt dx dy positive positive dt dt Related Rates 163 9. (a) (b) 10. (a) (b) 11. Yes, y changes at a constant rate. dy dt a dx dt 12. Answers will vary. See page 149. No, the rate dy dt is a multiple of dx dt. 13. D dx dt dD dt 2 1 4 x 2 2x3 x4 4x3 x4 3x2 1 1 2 x2 y2 x2 x2 1 2 x4 3x2 1 14. D dx dt 2 x2 y2 x2 sin2 x 4x3 6x dx dt dD dt 1 2 x 2 x 2x sin2 x 1 2 2x 2 sin x cos x dx dt 3x dx 3x2 1 dt 6x 3x2 1 sin x cos x dx x2 sin2 x dt 2 sin x cos x x2 sin2 x 15. A dr dt dA dt 3 r2 16. A dA dt r2 2 r dr dt 2 r dr dt dA 6, dt 24, dA dt 2 2 6 3 24 3 36 cm2 min. 144 cm2 min. If dr dt is constant, dA dt is not constant. dA dt depends on r and dr dt. (a) When r (b) When r 17. (a) sin cos 2 2 A 1 2b b s h h s 1 bh 2 s cos 2s sin 2 s s h 2 s cos 2 b 1 2s sin 2 2 s2 2 sin cos 2 2 2 (b) dA dt When When s2 cos 2 , d d where dt dt s2 2 s2 1 2 2 3 2 1 2 s2 sin 2 1 rad min. 2 1 2 s2 . 8 3s 2 . 8 (c) If dA d is constant, is proportional to cos . dt dt dA 6 dt dA , 3 dt 164 Chapter 2 4 3 r 3 2 4 r2 dr dt 6, 6 2 Differentiation 4 3 dV r , 3 dt 4 r2 dr dt 1 800 4 r2 18. V dr dt dV dt 19. V dV dt dr dt 800 1 dV 4 r 2 dt 30, 1 30 2 (a) When r dV dt 4 (a) When r 2 288 in3 min. dr dt 4 800 2 cm min. 9 When r dV dt 4 24, 24 2 (b) When r 2 4608 in3 min. dr dt 4 60, 1 60 2 800 1 cm min. 18 (b) If dr dt is constant, dV dt is proportional to r 2. 20. V dx dt dV dt x3 3 3x 2 dx dt 1, 2 21. s dx dt ds dt 6x2 3 12x dx dt 1, 36 cm2 sec. (a) When x dV dt 31 (a) When x 9 cm3 sec. ds dt 3 12 1 3 10, (b) When x dV dt 3 10 10, 2 (b) When x 900 cm3 sec. ds dt 3 12 10 3 360 cm2 sec. 22. V dr dt dV dt 1 2 r h 3 2 3 r2 dr dt 1 2 r 3r 3 r3 23. V 1 2 r h 3 3 3 h 4 1 3 9 2 h h 4 since 2r 3h dV dt 6, dV dt 2 216 in3 min. 10 9 2 dh dh h 4 dt dt 15, 2 (a) When r dV dt 3 4 dV dt 9 h2 6 2 When h dh dt 9 (b) When r dV dt 3 24, 24 2 4 10 15 8 ft min. 405 2 3456 in3 min. h r Section 2.6 1 2 r h 3 10 dh 25 2 dh h 144 dt dt 8, dh dt 144 dV 25 h 2 dt 12 Related Rates 165 24. V dV dt dV dt 1 25 3 h 3 144 25 h3 3 144 By similar triangles, r 5 h r 12 5 h. 12 5 r When h 144 10 25 64 9 ft min. 10 h 25. 6 12 1 3 1 (a) Total volume of pool Volume of 1 m of water % pool filled 18 144 1 2 2 12 6 1 2 1 6 12 18 m3 144 m3 (see similar triangle diagram) 12 2 h=1 1 6 6 12.5% 100% (b) Since for 0 h 2, b V dV dt 1 bh 6 2 36h dh dt 3bh 6h, you have 3 6h h 1 144h 18h2 1 144 1 1 m min. 144 b=6 1 dh 4 dt 26. V 1 bh 12 2 6bh 6h2 since b 1 dV 12h dt h 12 ft 3 ft 3 ft h ft dV (a) dt dh dh 12h dt dt dV 1 and dt 3 in./min 8 y2 dy dt dy dt 252 0 When h (b) If dh dt x2 2x dx dt 2y dh 2, dt 1 2 12 1 1 ft min. 6 2 feet, then dV dt (12) 2 1 32 3 3 ft min. 4 1 ft/min and h 32 27. y 25 x y 7, y 15, y 24, y dx dt 576 400 7, dy dt 2x dx since y dt 24, dy dt dy dt 27 24 2. 7 ft sec. 12 3 ft sec. 2 x (a) When x When x When x 20, 2 15 20 48 ft sec. 7 2 24 7 --CONTINUED-- 166 Chapter 2 Differentiation 27. --CONTINUED-- (b) A dA dt 1 xy 2 1 dy x 2 dt y dx dt 7, y 24, 527 24 dx dt 2, and dy dt 7 . Thus, 12 From part (a) we have x dA dt (c) sec2 1 7 2 tan d dt d dt Using x d dt x2 2x dx dt 2y 24 25 y2 dy dt dx dt When x 2.5, y x y 1 y dx dt 1 y x y2 dx dt 7 12 24 2 21.96 ft2 sec. dy dt x y2 dy 2, dt 2 y 25 cos2 7, y 2 dy dt x dx 24, dt 7 24 7 and cos 12 1 rad sec. 12 24 , we have 25 1 2 24 7 12 28. 25 0 y 5 y x dy dt 18.75, dx dt 0.15y x dy since dt 18.75 0.15 2.5 0.15 x 0.26 m sec. 29. When y 6, x 122 62 6 3, and s x2 108 12 36 y 2 12. 12 - y s x ( x, y ) x2 dx 2x dt x Also, x2 2x dx dt 2 12 dx dt y2 2y dx dt x x dx y dt dy dt y 12x y y y 12 y 2 s2 ds 2s dt s ds dt y 12 dy 1 dt 12 dy dt 122. 0 12 s 6 3 6 dy dt x dx y dt ds dx dt dt 3 15 y x dx dt s ds . dt ds dt 12 6 12 6 3 0.2 1 5 3 3 15 m sec (horizontal) Thus, x dx x dt dy dt sy 12x 1 m sec (vertical) 5 Section 2.6 30. Let L be the length of the rope. 4 ft/sec Related Rates 167 (a) 2L L2 dL dt dx dt When L x dx dt 144 2x L x dx dt dL dt 13: L2 4 13 5 x2 13 ft 12 ft (b) If dx dt 4, and L x dx L dt dL dt 5 13 x dx L dt L12 13: 4 L2 L 4 L L2 144 20 ft sec 13 144 4 0 dL dt 4L x since dL dt 4 ft sec 144 52 5 169 144 5 L12 lim dL dt lim 10.4 ft sec Speed of the boat increases as it approaches the dock. 31. (a) s2 dx dt dy dt 2s ds dt ds dt When x (b) t 250 750 y2 0 dx dt When s dx dt y x2 y2 450 200 y 600 2x dx dt 2y dy dt y dy dt s 150 and y 1 hr 3 s2 2s ds dt since dy dt 0 200, s 250 and ds dt 150 450 200 250 150 100 50 y s x -50 50 100 150 200 x x dx dt 600 750 mph. 20 min 32. 2x x2 dx dt 33. s2 x dx dt 902 30 28 2x x2 s ds x dt 10, x 100 480 3 25 160 3 75 5 3, 2s ds dt dx ds dt dt 30, s 28 x s 902 dx dt 302 28 10 30 10, 8.85 ft sec. 10 240 5 3 277.13 mph. When x ds dt 30 30 10 2nd x 5 mi s 30 ft 3rd x x s 90 ft Home 1st 168 Chapter 2 Differentiation 15 6 y y y 5 x 3 5 5 3 dx dt 5 5 3 25 ft sec 3 x 34. s2 x dx dt ds dt 902 60 28 x s dx dt x2 35. (a) 15y 15x 6y dx dt 902 56 13 602 30 13, dy dt When x ds dt 60, s 60 28 30 13 2nd 30 ft x 15.53 ft sec. 15 6 1st x y 3rd s 90 ft Home (b) d y x dt dy dt dx dt 25 3 5 10 ft sec 3 36. (a) 20y 20 6 20x 14y y dx dt dy dt y y 6y 20 x 20x 10 x 7 5 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 6 x y (b) d y x dt 37. x t 1 t sin , x 2 2 6 2 6 1 ,y 2 y2 1 (c) When x 1 2 2 (a) Period: (b) When x 12 seconds 12 0, 3 2 3 m. 2 1 ,y 4 dx dt 1 1 4 2 15 and t 4 12 cos t 6 1: 1 t cos 2 6 6 1 0 1 4 15 4 1 15 12 dy dt 12 3 2 y cos Lowest point: x2 2x dx dt dy dt 2y y2 dy dt x dx dt 6 1 5 5 . 120 Thus, 24 Speed 5 120 5 m sec 120 Section 2.6 3 sin 5 2 Related Rates 169 38. x t t, x2 y2 1 (a) Period: (b) When x 2 seconds 3 ,y 5 0, 4 5 1 3 cos 5 1 0 3 10 15 4 9 25 5 3 5 dy dt cos x dx y dt 6 t 1 4 2 1 3 5 2 4 m. 5 Lowest point: (c) When x 3 ,y 10 dx dt x2 2x dx dt dy dt 2y y2 dy dt 3 15 and 4 10 3 sin 5 t sin t 1 t 2 1 : 6 Thus, 9 5 . 125 0.5058 m sec Speed 9 5 125 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 dV dt dr 4 r2 . dt dr 4 r2 k dt dr . dt 40. 1 R dR1 dt dR2 dt 1 R2 dR dt 1 R1 1 1.5 1 R12 1 R2 Therefore, k 4 r 2 dR1 dt 1 R22 75: dR2 dt When R1 R dR dt 41. 1.3pV 0.3 V 0.3 dV dt pV 1.3 V 1.3 dp dt k 0 0 dp V dt 42. 30 30 2 50 and R2 1 50 2 1 1 75 2 1.5 0.6 ohms sec rg tan 32r tan 32r sec2 d dt dv dt Likewise, d dt v2 v2, 2v r is a constant. dv dt d dt dv . dt dV 1.3p dt dp V dt dV 1.3p dt 16r sec2 v v cos2 16r 170 Chapter 2 y 30 Differentiation 43. tan dy dt sec2 d dt d dt When y 30, y 3 m sec 1 dy 30 dt 1 cos2 30 4 dy dt 2 d . Thus, 2 dt 1 1 3 30 2 y x 30 , and cos 1 rad sec. 20 44. sin dx dt cos d dt d dt 10 x 1 ft sec x 10 x2 dx dt 10 10 dx sec x 2 dt 10 252 y ,y x 1 25 252 102 10 1 25 5 21 2 25 21 2 21 525 0.017 rad sec 45. tan dx dt sec2 d dt d dt 5 L y=5 600 mi hr 5 x2 cos2 52 L2 dx dt 5 dx x2 dt 1 dx 5 dt x2 L2 sin2 5 dx x2 dt 1 5 600 120 sin2 60 , 120 3 4 x (a) When d dt 120 4 30 , (b) When d dt 1 rad min. 2 (c) When d dt 3 rad min. 2 75 , 120 sin2 75 111.96 rad hr 1.87 rad min. 30 rad hr 90 rad hr 46. tan d dt sec2 d dt dx dt (a) When x 50 30 2 1 dx 50 dt 50 sec2 30 , dx dt d dt 200 ft sec. 3 (b) When 60 , dx dt 60 rad min rad sec 200 ft sec. Police 50 ft x (c) When 70 , dx dt 427.43 ft sec. Section 2.6 d dt (a) sin Related Rates 171 47. 10 rev sec 2 rad rev cos d dt dx dt x 30 1 dx 30 dt 30 sin 30 sin 600 sin d dt 20 20 rad sec (b) P 30 0 x - 2000 4 2000 x (d) For dx dt 30 , 600 sin 30 600 1 2 300 cm sec. (c) dx dt sin 600 sin 1 2 is greatest when n n or 90 or n n 180 . 180 . For dx dt 60 , dx dt is least when 600 sin 60 600 3 2 300 3 cm sec. 48. sin 22 0 dx dt x y x y2 x y y 49. tan dy dt 1 y dx dt 89.9056 mi hr dx dt 2 d dt 22 x x 50 50 sec2 50 sec2 1 cos2 , 25 50 tan d dt d dt 4 dy dt sin 22 240 x 4 50. (a) dy dt 3 dx dt means that y changes three times as fast as x changes. 0 or x L. y changes more rapidly when x is near the (b) y changes slowly when x middle of the interval. 51. x2 y2 25; acceleration of the top of the ladder dx dt dx dt d 2x dt 2 2y y dy dt dy dt dx dt 0 0 dx dt y d 2y dt 2 dy dt d 2y dt 2 First derivative: 2x x Second derivative: x dy dt d 2y dt 2 0 1 y x d 2x dt 2 dx dt 2 dy dt 2 When x d 2y dt 2 1 24 7, y 70 24, dy dt 2 2 7 dx , and 12 dt 7 12 2 2 (see Exercise 27). Since 1 24 4 49 144 1 24 625 144 dx d 2x is constant, 2 dt dt 0.1808 ft sec2 0. 172 Chapter 2 Differentiation d 2x dt 2 52. L2 144 x2; acceleration of the boat dL dt dL dt d 2L dt 2 2x x dx dt First derivative: 2L L dx dt dL dt dL dt d 2x dt 2 x d 2x dt 2 1 x dL dt L dx dt d 2L dt 2 dx dt dL dt 2 Second derivative: L dx dt 2 When L d 2x dt 2 13, x 5, dx dt 2 10.4, and 10.4 1 5 2 4 (see Exercise 30). Since d 2L dL is constant, 2 dt dt 0. 1 13 0 5 1 16 5 4 108.16 92.16 18.432 ft sec2 53. (a) m s (b) dm dt If t 0.3754s3 1.1262s2 10 and ds dt 20 18.780s2 37.560s 313.23s 313.23 ds dt 1707.8 0.75, then s 17.8 and dm dt 1.1154 million year. 54. yt dy dt y1 y 1 4.9t 2 9.8t 4.9 9.8 y 20 15.1 20 y 12 (0, 0) x By similar triangles: 20x When y 15.1: 20x 20 20 x 240 240 15.1 x x y x xy x 15.1 240 240 4.9 12 x 20x 240 20 dx dt dx dt xy x dy dt x 20 y dx dt dy y dt 9.8 97.96 m sec. At t 1, dx dt 240 4.9 20 15.1 Review Exercises for Chapter 2 173 Review Exercises for Chapter 2 1. f x f x x2 lim lim x0 2x f x x x2 2x x 3 x x x 2 f x 2x x x x 2 2 x x 2x 3 x2 2 x x 2x 3 3 x2 2x 3 x0 lim lim 2x x x0 2 x x0 lim 2x x0 x 2 2x 2 2. f x f x x lim lim lim lim x0 1 f x x x x x x 1 1 f x x x2 x x x x x f x 1 x xx x 1 x x x x x x 1 1 x x x 1 x x x 2 1 1 1 1 x 1 x x x x x x x x x f x 1 x x x x 1 2 x x x x 1 x x x x x0 x0 x0 lim x0 3. f x x x f x lim lim lim x0 lim x0 x0 x x 1 x 1 x 1 x xx 1 1 x lim x0 x0 x2 x x 1 x 1 x x 2 1 x lim x0 xx 2 x x 1 x 1 x 2 4. f x 2 x 2 f x 2x x x 2x xx 2 x xx 2 x xx xx 2 x2 f x 2 x xx x x x 2 x f x lim x0 lim x0 lim x0 lim lim x0 x0 174 Chapter 2 Differentiation 1. 6. f is differentiable for all x x2 1 3. 5. f is differentiable for all x 7. f x 4 x 2 2 2 because of the sharp turn 8. f x 4x 4x 2, x2, if x < if x 2 2 2 (a) Continuous at x (a) Nonremovable discontinuity at x (b) Not differentiable at x discontinuous there. y (b) Not differentiable at x in the graph. y 7 6 5 4 3 2 x 1 2 3 4 5 6 2 because the function is 5 4 1 -5 -4 -2 -1 -1 -2 x 1 2 -1 -2 -3 9. Using the limit definition, you obtain g x x 1, g 1 4 3 1 6 3 . 2 4 3x 1 6. At 10. Using the limit definition, you obtain h x x 2, h 2 3 8 4 2 67 . 8 3 8 4x. At 11. (a) Using the limit definition, f x f 1 3. The tangent line is y 2 y (b) -4 3x 2. At x 1, 12. (a) Using the limit definition, f x At x x 2 . 12 2. The tangent line is 0 2. 3x 3x 0 1 1. 2 0, f 0 2 y 2x 2x 4 y (-1, -2) (b) -4 -6 (0, 2) 6 -4 13. g 2 x2 lim gx x x2 x x x3 x x x2 g2 2 1 2 4 2 2 x2 x x 2 x 2 8 2 4 14. f 2 x2 lim f x x 1 x x 3 x x 1 f 2 2 1 3 2 x 2 x 1 13 1 13 1 9 x2 lim x2 lim lim x2 lim x2 x2 lim x2 lim x 2 x2 lim 15. y y 25 0 16. y y 0 12 17. f x f x x8 8x7 18. g x g x x12 12x11 Review Exercises for Chapter 2 19. h t h t 3t 4 12t 3 20. f t f t 8t 5 40t 4 21. f x f x x3 3x2 3x x 23. h x h x 6 x 3x 1 2 175 3x2 6x 2 x 1 2 1 2 22. g s g s 4s4 16s3 5s2 10s 33 x x 2 3 6x1 2 3x1 3 3 24. f x f x x1 1 x 2 2 3 x 1 x2 2 x 9 1 x 2 3 2 x 1 2x3 2 25. gt g t 2 t 3 2 26. h x 3 2 27. f 3 2 2 3 sin 3 cos 4 t 3 4 3t 3 h x 4 x 9 4 9x3 sin 4 cos 4 f 28. g g 4 cos 4 sin 6 29. f f 3 cos 3 sin 30. g g 5 sin 3 5 cos 3 2 2 31. 2 y 32. f f 1 y 33. f F F t 200 T 100 T 1 - x -1 1 2 2 x -1 f (a) When T 4, F 4 50 vibrations/sec/lb. (b) When T 9, 331 vibrations/sec/lb. F 9 3 34. s 16t 2 s0 Second ball: First ball: 16t 2 100 t 0 100 16 10 4 2.5 seconds to hit ground 16t 2 75 t2 0 75 16 5 3 4 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. st s 9.2 s0 16t 2 16 9.2 1354.24 s0 2 36. s t s0 0 16t 2 t 16t 2 14,400 30 sec 14,400 0 The building is approximately 1354 feet high (or 415 m). Since 600 mph 6 mi/sec, in 30 seconds the bomb 1 5 miles. will move horizontally 6 30 1 176 37. (a) Chapter 2 y Differentiation (c) Ball reaches maximum height when x (d) y y y 0 x 25. 15 x 1 1 0.6 0 0.02x2 0.04x 10 5 y 10 y 25 20 40 60 Total horizontal distance: 50 (b) 0 0 x x 1 0.02x2 x implies x 50 50. y 30 y 50 (e) y 25 0 0.2 1 38. y v02 64 ( v02 128 v02 v02 , 64 128 ) ( v32 , 0 ) x 2 0 (a) y x 32 2 x v02 x 1 32 x v02 v02 32 v02 32. v0 64 2 (b) y 1 64 x v02 v02 ,y 64 1 64 v02 v02 64 0. 0 if x 0 or x When x (d) v0 Projectile strikes the ground when x 70 ft sec v02 32 70 2 32 v02 128 153.125 ft 70 2 128 38.28 ft Projectile reaches its maximum height at x (one-half the distance). (c) y x 32 2 x v02 x 1 32 x v02 0 Range: x Maximum height: y 50 when x 0 and x s02 32. Therefore, the range is x v02 32. When the initial velocity is doubled the range is x 2v0 32 2 4v02 32 0 0 160 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 2 v02 64 v02 128 v02 . 128 If the initial velocity is doubled, the maximum height is y 2v0 64 2 2v0 2 128 4 v02 128 or four times the original maximum height. Review Exercises for Chapter 2 39. x t (a) v t (c) v t x 3 2 177 t2 3t x t 0 for t 2 2 2t 3 2 t 3 2 t 1 (b) v t < 0 for t < (d) x t 0 for t 21 22 3 2 1, 2 3 3 1 1 3 2 1 1 2 1 2 1 4 v1 v2 The speed is 1 when the position is 0. 40. (a) y (c) 12 0.14x2 4.43x 58.4 (b) 320 0 0 0 0 60 60 (d) If x 65, y 362 feet. (e) As the speed increases, the stopping distance increases at an increasing rate. 41. f x f x 3x2 3x2 2 6x3 7 x2 7 2x 9x2 2x 2 16x 3 x2 7 2x 3 6x 42. g x g x x3 x3 x3 4x3 x1 2 sin x x cos x 44. f t f t t 3 cos t t3 sin t t 3 sin t x x2 x2 x2 x2 47. f x f x 4 4 1 3x2 3x2 1 1 2x 1 2 3x x 3x 1 3x 6x 2 3x3 2 x 2 3x 2 6x 2 6 3x 3 6 6x 43. h x h x x sin x 1 2 x sin x cos t 3t 2 3t 2 cos t 45. f x f x x2 1 1 x2 x2 12 x 1 2x 46. f x f x 6x x2 x2 23 x 5 1 1 6 x2 5x 2 6x 12 3x2 2 5 2x 1 1 48. f x 2 9 3x2 9 3x2 2x 2x 1 6x 4 6x 3x2 2 f x 2 6x 2 18 1 3x2 3x 2x 2 49. y y x2 cos x cos x 2x cos2 3x 2 sec x 3x 2 sec x tan x 6x sec x x2 x sin x 2x cos x x 2 sin x cos2 x 50. y y sin x x2 x2 cos x x4 2x 2 x 2 tan x x 2 sec2 x 2x tan x sin x 2x x cos x x3 2 sin x 51. y y 52. y y 178 53. y y Chapter 2 x cos x x sin x Differentiation sin x cos x cos x x sin x 54. g x g x 3x sin x 3x cos x 5x cos x x 2 cos x 3 sin x 3 1 , 2 x 1 8 3 y 1 1 1 1 1 1 y sin x , sin x 8 x 8x 1 1 2 2 x 2 sin x sin x 2x cos x x2 55. f x f x f 1 2x3 1 x2 2 4 1 y 2x 3 2x x 2, 1, 1 56. f x f x x x x 2 1 4 1 , 1 1 x 3 1 x 2 1 2 Tangent line: y 4x 4x 1 3 f 1 2 Tangent line: y 57. f x f x f 0 0 x tan x, x sec2 x 0, 0 tan x 58. f x f x ,1 1 sin x sin x 2 cos x sin x cos x 1 1 2 Tangent line: y 0 y 0x 0 0 f Tangent line: y 2x 2x 2 1 59. v t at v4 a4 36 v t 36 t 2, 0 t 6 2t 16 sec2 20 m sec 8m 60. v t at 4t 90t 10 10 90 90t 4 4t 10 2 900 10 90 14 225 49 2 4t 4t (a) v 1 a1 2t 225 5 2 6.43 ft sec 4.59 ft sec2 (b) v 5 a5 90 5 30 225 152 12x1 3x 4 15 ft sec 1 ft sec2 (c) v 10 a 10 90 10 50 225 252 18 ft sec 0.36 ft sec2 61. gt g t g t t3 3t 2 6t 3t 3 2 62. f x f x f x 63. f 9 4x7 4 f f 3 tan 3 sec2 6 sec 6 sec 2 3 4 9 x 4 7 4 sec tan tan Review Exercises for Chapter 2 64. h t h t h t 4 sin t 4 cos t 4 sin t 5 cos t 5 sin t 5 cos t y 65. y y y y 0 10 10 0 sin x h x 2 cos x x x x2 x x2 3 1 3 1 3 x2 5 2 179 2 sin x 2 cos x 2 sin x 2 sin x 3 cos x 3 sin x 3 cos x 3 cos x 2 sin x 3 cos x 66. xy xy y xy y cos x sin x y 67. h x x2 x2 1 3 1 1 x2 6x 1 x 12 3 2x 2x 68. f x f x x2 5 x2 1 x 1 x 69. f s 4 s2 s2 s s2 s s2 1 1 5 2 s3 3s2 5 s3 1 3s 5 5 2 s 2 5 s3 25 5 1 3 2 2x 1 x2 f s 5 2 2s 1 1 3 2 3 2 3s s2 8s3 70. h h 1 1 1 3 3 71. y 31 1 6 2 3 cos 3x 3 sin 3x 9 sin 3x 1 1 3 1 1 2 1 1 4 y y 2 1 6 3 1 72. y y 1 cos 2x 2 cos2 x 4 cos x sin x 4 sin x cos x 73. y y x 2 1 2 sin 2x 4 1 cos 2x 2 4 1 1 2 cos 2x sin2 x 2 sin 2x 2 2 sin x cos x 0 sec7 x 7 sec5 x 5 74. y y 75. y sec4 x sec x tan x 1 y 2 32 sin x 3 2 72 sin x 7 sin5 2 x cos x sin2 x sec6 x sec x tan x sec5 x tan x sec2 x sec5 x tan3 x 3x x2 1 3 x2 1 1 2 sin1 2 x cos x cos x cos3 x sin x x 2 x 2 sin x 1 sin x 76. f x f x 77. y 3x 1 x 2 2 x2 1 3x 2 3 2 1 3 1 2 2x y cos x x 22 sin x 3 x2 1 x2 1 x2 1 3 2 180 Chapter 2 cos x 1 x 1 x 1 x 1 2 Differentiation 78. y y 79. 1 x x 1 2 f x f x f 2 1 1 1 2 12 23 x3 x3 1 2 1 sin x cos x 1 1 3x 2 3x2 2 1 x3 1 sin x 1 cos x 1 2 80. f x f x f 3 3 x2 1 1 1 2 cot 3x 3 csc2 3x 2 3 81. 2x 2x 3 x2 1 2 3 y y y 4 1 csc 2x 2 csc 2x cot 2x 0 1 2 x 3 23 34 y y csc 3x 82. 83. g x g x 2x x x x 1 1 2 3 2 1 2 4 g 3 csc 3x cot 3x 0 3 3 y -2 7 6 g does not equal zero for any value of x in the domain. The graph of g has no horizontal tangent lines. 2 g -2 84. f x f x x 4 x3 4x 2 x 3x2 2 x 4 6x x2 8 4 2x 8 2 85. f t f t f t t t 1 5 6t 1 3 t t 1 1 1 3 1 2 t 1 5 6 1 x The zeros of f correspond to the points on the graph of f where the tangent line is horizontal. 100 1 1 6 f does not equal zero for any x in the domain. The graph of f has no horizontal tangent lines. 5 f -7 f 5 f f - 60 -2 -1 7 86. y y 3x x 3x 2 2 3 87. f t 2 t2 t 1 5, tt 24 1 2, 4 4 2 7x 2 3x (a) f t f 2 (b) y 4 y 7t 2 y does not equal zero for any x in the domain. The graph has no horizontal tangent lines. 75 24 t 24t y 2 44 y y (c) 5 4 3 -3 - 25 (2, 4) 3 2 1 -2 -1 x 1 3 4 5 f g Review Exercises for Chapter 2 88. g x x x2 1, x2 19 10 10 3 10 y (c) 12 10 8 6 4 2 -2 -2 y 181 3, 3 10 1 x2 x 2 89. y 1 (a) tan y y 2 1 x, 2 1 2, tan 1 x cos2 3 3 3 6 cos2 3 6 y 3 1 x (a) g x g 3 (b) y 6 cos2 3 y 11.1983 x 2 tan 3 3 3 cos2 3 19 10 x 10 19 10 x 10 3 27 10 10 (b) y tan 3 3 x cos2 (c) y1 y 8 (3, 3 10 ) y 3( 2 x g x 1 2 3 4 5 6 (- 2, tan -4 -8 90. y (a) 2 csc3 y y 1 x 2 sin3 3 cos x sin4 x x x , 1, 2 csc3 1 (c) 6 5 4 3 y y (1, 2 csc3 1) 3 cos 1 sin4 1 2 csc3 1 y 3 cos 1 x sin4 1 3 cos 1 x sin4 1 1 2 sin3 1 3 cos 1 sin4 1 x 1 (b) y 2 1 x -1 1 2 3 4 5 6 91. y y y 2x2 4x 4 sin 2x 2 cos 2x 4 sin 2x 92. y y y tan x 2 3 93. x f x f x f x cot x csc2 x 2 csc x csc x cot x x 2x 2 x3 sec2 2 sec x sec x tan x 2 sec2 x tan x 2 csc2 x cot x 94. y y y sin2 x 2 sin x cos x 2 cos 2x sin 2x 95. f t t 1 t 1 2t 1 t 2 96. g x g x g x 6x x2 2 5 1 3x2 x2 5x 12 3 5 f t f t 1 t3 2 t4 98. hx h x x x2 2x2 x2 1 1 1 2 6x3 15x2 18x x2 1 3 97. g g g tan 3 3 sec2 3 sin cos 1 1 sin 1 18 sec2 3 tan 3 h x x 2x2 3 x2 1 3 2 182 Chapter 2 Differentiation 99. T T T t2 700 4t 10 4t 10 1 700 t 2 t2 1400 t 2 4t 10 2 1, 1400 1 2 4 10 2 5, 1400 5 20 2 10 2 (a) When t T 1 (b) When t 18.667 deg hr. T (d) When t 3.240 deg hr. T 9 3, 1400 3 2 12 10 2 10, 7.284 deg hr. (c) When t T 25 2gh 4 h 1400 10 100 40 2 10 2 0.747 deg hr. 100. v dv dh 2 32 h 8 h (a) When h (b) When h 9, 4, dv dh dv dh 4 ft sec. 3 2 ft sec. 101. 2x 3xy x2 3xy 3y 3x y3 3y2y y2 y y 10 0 2x 2x 3x y x 3y 3y y2 x y y1 2 102. 2x x2 9y2 18yy 4x 4 3y 3y 1y y 0 0 4 2x 3 6y 4 2x 3 6y 1 103. y 1 x 2 1 2 16 0 y y 2 x x1 2y x 1 y 2 1 2 y x x y 2 y 2 xy x y 2 y y 2 xy y 2 x 2 xy y 2 x 2 y 2 xy x 105. x cos y y 2y x 2x y y y x x x sin y sin y cos x y y cos x y sin x y sin x y cos x sin y 104. y2 0 0 x2 x 4y y y x3 x3 3x2 3x2 3x2 x2 x2y x2y x2y 2xy xy xy 2xy y y2 2y2 xy y 4yy y x cos y 2xy y x 4y y sin x sin y cos x x cos y Review Exercises for Chapter 2 106. 1 y cos x sin x y y y y x 1 1 1 sin x y 107. 2x x2 y2 2yy y At 2, 4 : y Tangent line: x Normal line: y 2x 108. 2x x2 y2 2yy y At 5, 3 : y Tangent line: 5x Normal line: 3x 5y 3y y 16 0 x y 5 3 y 3 16 3 30 0 5 x 3 0 3 x 5 5 5 (b) When x (c) When x -10 10 183 20 0 -9 6 (2, 4) 9 y sin x sin x y sin x y 1 1 x y -6 csc x 1 2 y 2y 4 y 4 10 2x 0 0 2 1 x 2 2 10 109. y dy dt dy dt x 2 units sec dx 1 dx dt 2 x dt 1 dx , 2 dt 1, 4, dx dt dx dt 2 x dy dt 4 x -10 (a) When x 2 2 units/sec. 4 units/sec. 8 units/sec. 110. Surface area dx dt dA dt s h s dV dt 5 12x dx dt A 6x 2, x length of edge 12 4.5 5 270 cm2 sec 111. 1 2 2 1 h 4 1 Width of water at depth h: w V dV dt dh dt 2 5 2 2 5 4 2 2s 4 2 h dh dt 2 h 2 h 1 h 4 5 8 4 4 2 hh h 1 2 2 1 2 2 dV dt 54 h 1, dh dt 2 m min. 25 s 2 h 2 When h 184 112. Chapter 2 tan d dt sec2 d dt dx dt When x dx dt 6 1 , 2 1 4 1 x 32 dx dt Differentiation 113. rad min st s t s 4.9t2 1 6 6 x2 1 t tan 30 15 km min 2 xt 450 km hr. dx dt 3 ds dt 60 9.8t 35 25 5 4.9 1 3 3s t 3 9.8 5 4.9 38.34 m sec st xt 60 4.9t2 4.9t 2 tan2 1 s (t) x 30 x(t ) Problem Solving for Chapter 2 1. (a) x 2 y r 2 r2, Circle y, Parabola -3 3 x2 Substituting: y y2 y2 y y 2r y 2r y 2r r 2 3 -1 r2 r2 0 0 y y r2 y 1 Since you want only one solution, let 1 (b) Let x, y be a point of tangency: x2 y y b 2 2r 0r 1 2. Graph y x 2 and x 2 y 1 2 2 1 4. 1 2x 2x, 2 y by 0y x b y , Circle x2 y Parabola Equating: 2x 2b b Also, y y y x2 y b x b y 1 2 x2 y imply: 1 2 2 3 1 1 b 2 y 2 y 2 -3 -1 3 b 1 and y y 1y 1y 1 4 1y 3 and b 4 5 4 Center: 0, Graph y 5 4 x 2 and x 2 y 5 2 4 1. Problem Solving for Chapter 2 2. Let a, a2 and b, b2 2b 5 be the points of tangency. For y x 2, y 2x 2b 2 a b 1, and for y x 2 2x 5, y 2x 2. Thus, 2a or a 1 b. Furthermore, the slope of the common tangent line is a2 b2 2b a b 5 1 1 b 2 y 10 8 6 4 -8 -6 -4 -2 -4 -6 185 1 2b 4b 2b b 2 b 1 b2 b b2 1 6 4 2 1 0 2b b b2 2b 4b2 0 5 2b 6b 5 2b 2 2b 2 x 2 4 6 8 10 2b2 2b b 2 2 2 b b 2, 0 For b 2, a 1 b 1 and the points of tangency are The tangent line has slope 2: y 1 2 x 1 y 1, 1 and 2, 2x 1 1, 5. 8. For b 1, a 1 b 2 and the points of tangency are 2, 4 and The tangent line has slope 4: y 4 4 x 2 y 4x 4 3. (a) f x f 0 f 0 P1 x cos x 1 0 1 P1 x P1 0 P 1 a0 a1x 1 0 (b) f x f 0 f 0 f 0 P2 x cos x 1 0 1 1 1 2 2x P2 x P2 0 P P 2 2 a0 a1x a2x 2 1 0 1 2 a0 a0 a1 a1 a0 a0 a1 a1 2a2 a2 0 0 0 (c) x cos x P2 x 1.0 0.5403 0.5 0.1 0.9950 0.9950 0.001 1 1 0 1 1 0.001 1 1 0.1 0.9950 0.9950 1.0 0.5403 0.5 P2 x is a good approximation of f x (d) f x f 0 f 0 f 0 f 0 x sin x 0 1 0 1 1 3 6x cos x when x is near 0. a0 a1x a2 x 2 0 1 0 1 6 P3 x P3 0 P3 0 P P 3 3 a3x3 a0 a0 a1 a1 2a2 a2 6a3 a3 0 0 P3 x 4. (a) y x 2, y 2x, Slope 4 y 4 at 2, 4 4x 4x 2 4 (b) Slope of normal line: Normal line: y 4 y y 1 4 1 4 Tangent line: y x 2 9 2 9 2 1 4x 1 4x (c) Tangent line: y Normal line: x 0 0 x2 18 2 0 0 4x 2 4x x 9 x --CONTINUED-- x 2, 9 4 9 81 4 , 16 Second intersection point: 186 Chapter 2 Differentiation 4. --CONTINUED-- (d) Let a, a2 , a 0, be a point on the parabola y x 2. Tangent line at a, a2 is y 2a x 1 2a x a a2. To find points of intersection, solve: Normal line at a, a2 is y x2 x2 x2 1 x 2a x x x x 1 x 2a 1 16a2 1 4a 2 a a2. 1 x 2a a2 a2 a a a a2 1 2 1 2 1 4a 1 16a2 2 1 4a 1 4a 1 4a 1 4a 1 x 4a a Point of tangency a 1 2a 2a2 1 2a a a 1 x 4a The normal line intersects a second time at x 2a2 1 . 2a 6. f x f x a b cos cx bc sin cx a b 1 c 4 c 4 1 3 2 1 Equation 1 Equation 2 Equation 3 5. Let p x p x At 1, 1 : Ax3 3Ax 2 A 3A Bx 2 2Bx B 2B B 2B Cx C. C C C C D D 1 14 Equation 1 Equation 2 3 2 Equation 3 Equation 4 2 At 0, 1 : At At 1, 3: A 3A D 3 , : a 4 2 b cos bc sin Adding Equations 1 and 3: 2B 2D 2C From Equation 1, a 4 1 b b cos c 4 b. Equation 2 becomes 3 2 b b cos c 4 1 . 2 Subtracting Equations 1 and 3: 2A Adding Equations 2 and 4: 6A 2C 12 16 Subtracting Equations 2 and 4: 4B 1 2 From Equation 3, b 1 c sin c 2 2B 5. Subtracting Hence, B 4 and D 2A 2C 4 and 6A 2C 12, you obtain 4A 8 A 2. Finally, C 1 4 2A 0. Thus, 2 2x3 4x 2 5. px 1 . Thus: c sin c 4 1 2 1 c c sin 2 4 4 1 c cos c sin c 4 4 1 cos c 4 Graphing the equation gc c 1 c sin 2 4 cos c 4 1, you see that many values of c will work. One answer: c 2, b 1 ,a 2 3 f x 2 3 2 1 cos 2x 2 Problem Solving for Chapter 2 7. (a) x4 a2 y 2 y a2x 2 a2x 2 187 a2y 2 x4 a2x 2 a a2x 2 a x4 x4 a2x 2 a x4 8. (a) b2y 2 y2 x3 a x ; a, b > 0 x3 a x b2 x3 a b x and y2 x3 a b x . Graph y1 . Graph: y1 (b) -3 2 and y2 (b) a determines the x-intercept on the right: a, 0 . b affects the height. (c) Differentiating implicitly: a= 1 2 3 a=2 a=1 -2 2b 2 y y y 3x 2 a x x3 0 3ax 2 4x 3 a, 0 are the x-intercepts, along with 0, 0 . 3ax 2 4x3 2b2y 4x3 4x 3a 4 3 3ax 2 3a x b2y 2 y2 3a 4 (c) Differentiating implicitly: 4x 3 y 2a2 x 2a2 y y 2a2x 4x3 2a2y x a2 a2y 2x 2 0 2x 2 a2 x a a 3a 4 27a3 1 a 64 4 3 3a2 16b 3 3a2 16b 2 a2 2 2 a2 a2 2 a4 2 a4 4 a2 4 a2y2 a2y 2 27a 4 y 256b 2 a4 4 a2y 2 y2 y Two points: 3a 3 3a2 3a , , , 4 16b 4 a 2 a a , , 2 2 a , 2 a , 2 a a , , 2 2 a , 2 a 2 (b) y Four points: 9. (a) 30 y (0, 30) (90, 6) (100, 3) x 90 100 30 (0, 30) (60, 6) (70, 3) x 60 70 Not drawn to scale Not drawn to scale Line determined by 0, 30 and 90, 6 : y 30 30 6 x 0 90 24 x 90 When x y 100: 30 10 > 3 Shadow determined by man 3 0 4 xy 15 4 x 15 30 Line determined by 0, 30 and 60, 6 : y 30 30 6 x 0 60 70: 30 0 2 xy 5 2 x 5 30 When x y 2 70 5 4 100 15 2 < 3 Shadow determined by child --CONTINUED-- 188 Chapter 2 Differentiation 9. --CONTINUED-- (c) Need 0, 30 , d, 6 , d 30 0 6 d 6 d d 3 10 10, 3 collinear. 24 d 3 d 10 80 feet 5. (d) Let y be the distance from the base of the street light to the tip of the shadow. We know that dx dt For x > 80, the shadow is determined by the man. y 30 y 6 x y dy 5 x and 4 dt 5 dx 4 dt 25 4 For x < 80, the shadow is determined by the child. y 30 y x 3 10 y 10 x 9 dy 100 and 9 dt 10 dx 9 dt 50 9 Therefore: dy dt 25 4, 50 9, x > 80 0 < x < 80 80. dD dt 1 2 x 2 dx dt dy dt dy dt is not continuous at x dy dt dx dt 1 x 3 dx dt 10. (a) y 1 dx dt x1 3 1 8 3 2 3 (b) D x2 y2 y 2 2x 2y 2 3 x dx dt y dy dt x2 y2 8 12 21 64 4 d dt 2 12 cm sec y sec2 x 68 8 (c) tan x dy dt x2 y dx dt 98 68 49 cm sec 17 From the triangle, sec d dt 81 2 12 64 68 64 Lx Lx L x x lim L x x L0 x x L x x 16 68 Lx 68 8. Hence 4 rad sec. 17 Ex x x Ex Ex 1 x E x x E x x E0 lim E x x 1 x0 11. L x lim lim x0 12. E x lim lim x0 Lx x0 x0 ExE x x E x lim Also, L 0 But, L 0 L0 L0 x0 lim E x x0 x0 . E x lim But, E 0 1 x0 0 because 0 L0 L 0 L 0 0. lim x0 1. Thus, L x L 0 for all x. The graph of L is a line through the origin of slope L 0 . Thus, E x ExE 0 e x. E x exists for all x. For example: E x Problem Solving for Chapter 2 sin z z z0 189 13. (a) z (degrees) sin z z d sin z dz 0.1 0.0174524 0.01 0.0174533 0.0001 0.0174533 (b) lim z0 0.0174533 sin z z 180 90 180 c . sin 1 1 cos cz Cz In fact, lim (d) S 90 C 180 d Sz dz 180 (c) lim sin z sin z z 0 z z cos z cos z z cos z sin z sin z z 1 cos z sin z sin z z sin 2 cos lim 180 z 0 lim z 0 sin z lim z 0 cos z d sin cz dz 180 sin z 0 180 180 cos z (e) The formulas for the derivatives are more complicated in degrees. 14. (a) v t at (b) v t S5 27 5t 27 5 27 5t 27 10 27 ft sec ft sec2 27 5 2 15. j t a t (a) j t is the rate of change of acceleration. 0 27 5 27 5t 27 t 73.5 feet 5 seconds (b) s t vt at a t 8.25t 2 16.5t 16.5 jt 0 66t 66 6 (c) The acceleration due to gravity on Earth is greater in magnitude than that on the moon. The acceleration is constant, so j t (c) a is position. b is acceleration. c is jerk. d is velocity. 0. ...
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This note was uploaded on 09/01/2008 for the course BIOL 1362 taught by Professor Richardknapp during the Spring '08 term at SUNY Adirondack.

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