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ODD02 - CHAPTER 2 Differentiation Section 2.1 Section 2.2...

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Unformatted text preview: CHAPTER 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 . . . 53 Basic Differentiation Rules and Rates of Change . 60 The Product and Quotient Rules and Higher-Order Derivatives . . . . . . . . . . . . . . 67 The Chain Rule . . . . . . . . . . . . . . . . . . . 73 Implicit Differentiation . . . . . . . . . . . . . . . 79 Related Rates . . . . . . . . . . . . . . . . . . . . 85 . . . . . . . . . . . . . . . . . . . . . . . . . 92 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . 98 CHAPTER Differentiation Section 2.1 2 The Derivative and the Tangent Line Problem Solutions to Odd-Numbered Exercises 1. (a) m (b) m 0 3 3. (a), (b) (c) y f4 4 3 x 3 1x f1 x 1 1 1 2 2 1 f 1) y y 6 5 4 3 2 1 f )4) 4 f )1) )x 1 1) f )1) x 1 f )4) 5 )4, 5) x f )4) f )1) 3 1 f )1) )1, 2) 2 x 1 2 3 4 5 6 5. f x 3 2x is a line. Slope 2 7. Slope at 1, 3 lim lim g1 1 1 x x→0 x x 2 g1 4 x x x 2 2 3 1 x→0 lim 2x 2x x→0 lim 2 x→0 9. Slope at 0, 0 lim lim f0 3t t t t t t 3 f0 2 t→0 11. f x 0 fx 3 lim fx 3 x 0 3 x→0 t→0 x x fx lim 3 t→0 lim x→0 lim 0 x→0 13. f x fx 5x lim fx 5x 5 5 x→0 15. h s x x x x fx hs 5x 3 lim 2 s 3 hs 2 s 3 s→0 s s hs 2 s 3 lim lim x→0 3 lim s→0 s s 3 x→0 2 s 3 lim s→0 s 2 3 53 54 Chapter 2 Differentiation 17. f x fx 2x2 lim lim x→0 x fx 2x 2x2 4x x 1 x x x 2 fx x 2x 2 x x 2 1 x x x 2x2 1 x 1 2x2 x 1 x→0 lim lim 19. f x fx x3 4x x 2x x x→0 x x→0 lim 4x x→0 2x 1 4x 1 12x fx x x3 x x x 3 lim lim lim fx 12 x x 3x x 2 2 x→0 x x x 3 x3 3 12x 12x 12 x x3 12x x→0 3x2 x x→0 x 3x2 x 3x x 3x x 12 x 3x2 12 x x 2 lim x→0 lim 3x2 x→0 12 21. f x fx 1 x lim x→0 1 fx 1 x x x 1 x x x x x 1 2 fx 1 x 1 lim x x→0 lim lim lim x xx xx x 1 1 x→0 x 1x x 1x 1 1 1 x→0 x→0 1 1x 1 x 23. f x fx x lim x→0 1 fx x x x x 1 1 x x x x x x x x 1 1 1 fx 1 x 1 1 x 1 2x x x x 1 1 1 1 1 x x x x 1 1 x x 1 1 lim lim lim x→0 x→0 x→0 x Section 2.1 The Derivative and the Tangent Line Problem 55 25. (a) f x fx x2 lim x→0 1 fx x x x 2 17. (b) fx −5 8 (2, 5) 5 −2 lim x 2x x x→0 1 x x 2 x2 1 lim x→0 x x 2x lim 2x x→0 At 2, 5 , the slope of the tangent line is m 22 4. The equation of the tangent line is y y 5 5 y 4x 4x 4x x3 lim fx x 3x2 x x→0 2 8 3. 27. (a) f x fx 18. (b) x x x3 x fx −5 10 (2, 8) 5 −4 lim lim x3 2 x→0 x→0 3x x x 3x x x x 2 3 lim 3x2 x→0 3x2 32 2 At 2, 8 , the slope of the tangent is m The equation of the tangent line is y 8 y 29. (a) f x fx x lim lim lim fx x x x x x 1 x x→0 12. 12 x 12x 2 16. 18. (b) x x x x x x x fx −1 3 (1, 1) 5 −1 x x x x→0 x x x x x x x→0 lim x→0 1 2x At 1, 1 , the slope of the tangent line is m 1 21 1 . 2 The equation of the tangent line is y 1 y 1 x 2 1 x 2 1 1 . 2 56 Chapter 2 Differentiation 31. (a) f x fx 4 x lim fx x→0 (b) x x x fx 4 x x xx x3 xx 2x 2 x 2 10 (4, 5) − 12 12 x lim lim lim lim lim x2 x2 x→0 x x x 4 x x 4x 4x x −6 x→0 4x x 2 x x xx x x3 x x→0 x x2 x xx 4x x x2 x xx x2 x x→0 x xx x2 xx 4 1 xx x 4 x2 4 x→0 At 4, 5 , the slope of the tangent line is m 1 4 16 3 4 The equation of the tangent line is y y 5 3 x 4 3 x 4 2 3x2. Since the 35. Using the limit definition of derivative, fx 1 2x x . 1 2, 4 33. From Exercise 27 we know that f x slope of the given line is 3, we have 3x2 x 3 ± 1. Since the slope of the given line is 1 2x x x 1. 1 2 we have Therefore, at the points 1, 1 and 1, 1 the tangent 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. 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 37. g 5 g5 2 because the tangent line passes through 5, 2 2 5 0 9 2 4 1 2 39. f x x⇒f x 1 b S ection 2.1 The Derivative and the Tangent Line Problem 57 41. f x x ⇒ f x matches (a) 43. 4 3 2 1 −4 −3 −2 −1 −1 −2 −3 −4 y decreasing slope as x → x 1 2 3 4 Answers will vary. Sample answer: y 45. (a) If f c (b) If f c 3 and f is odd, then f 3 and f is even, then f c c fc fc 3 3 4 2x. The slope of the x 47. Let x0, y0 be a point of tangency on the graph of f. By the limit definition for the derivative, f x y line through 2, 5 and x0, y0 equals the derivative of f at x0: 5 2 5 5 4x0 y0 x0 y0 x02 0 0 7 4 2 8 x0 2 2x0 x0 4 8x0 4x0 1 x0 2x0 2x02 3 3 ⇒ x0 1, 3 −2 6 5 4 3 2 1 x 1 2 3 6 (2, 5) (3, 3) (1, 3) x0 Therefore, the points of tangency are 1, 3 and 3, 3 , and the corresponding slopes are 2 and lines are y 5 y 49. (a) g 0 (b) g 3 0 8 3, 2. The equations of the tangent 2x 2x 3 2 1 y 5 y 2x 2x 2 9 (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. 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 51. f x 13 4x 32 4x . 2 By the limit definition of the derivative we have f x x fx fx 2 2 3 1.5 27 32 27 16 −2 2 1 1 4 3 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 58 Chapter 2 Differentiation 53. g x fx 2x 3 0.01 0.01 0.01 fx x 0.01 2 55. f 2 2x x2 100 f2 24 3.99 2.1 2 4 2 4, f 2.1 2.1 4 2.1 0 3.99 0.1 Exact: f 2 g f −2 −1 4 The graph of g x is approximately the graph of f x . 57. f x 1 x 5 and f x 1 . 2x3 2 As x → 5 f −2 , f is nearly horizontal and thus f 0. f′ −5 59. f x S x 4 x x f2 4 2 3 2 x x x x x f2 3 2 x 3 2 x x 3 2 3 f2 2 1 3 x 2 19 x 10 4 5 −2 3 1 x x 1 2 x 2 3 x 5 2x 2 3 (a) x x x 1: S 0.5: S 0.1: S x 2 2 3 S 0.1 x 3 x 2 19 x 10 S1 S 0.5 − 1 f 7 x (b) As x → 0, the line approaches the tangent line to f at 2, 3 . 61. f x f2 x2 x→2 1, c fx x 2x2 x→ 2 2 f2 2 1, c fx x 0 g0 0 x→0 x→2 lim lim x2 x 1 2 3 x →2 lim x 2x x2 2 x→2 lim x 2 4 63. f x f 65. g x g0 2 x3 2 f2 2 x→ 2 lim lim x2 x x 2 2 x→ 2 lim x3 2x2 1 x2 67. f x 1 x→ 2 lim x2 23 4 6 x ,c x→0 x x→6 6 ,c f6 6 lim gx x x x x x lim x x . Does not exist. f6 lim lim fx x x As x → 0 , As x → 0 , 1 → x 1 x → x→6 623 x6 1 6 13 0 x→6 lim x Does not exist. S ection 2.1 The Derivative and the Tangent Line Problem 59 69. h x h 5 x 5 ,c x→ 5 5 h 5 5 x 5 5 5 0 5 lim hx x x x x 71. f x is differentiable everywhere except at x (Sharp turn in the graph.) 3. x→ 5 lim x→ 5 lim Does not exist. 73. f x is differentiable everywhere except at x (Discontinuity) 77. f x is differentiable on the interval 1, (At x 1 the tangent line is vertical) 81. f x x 1 . 1. 75. f x is differentiable everywhere except at x (Sharp turn in the graph) 79. f x is differentiable everywhere except at x (Discontinuity) 83. f x x x 1 3, x ≤ 1 1 2, x > 1 3. 0. The derivative from the left is x→1 The derivative from the left is x x 1 1 0 1. x→1 lim fx x f1 1 x→1 lim lim fx x f1 1 x→1 lim x x 1 1 3 0 0. 1 2 The derivative from the right is x→1 x→1 lim x lim fx x f1 1 x→1 lim x x 1 1 0 1. The derivative from the right is x→1 The one-sided limits are not equal. Therefore, f is not differentiable at x 1. lim fx x f1 1 x→1 lim x x 1 1 2 0 0. 1 x→1 lim x These one-sided limits are equal. Therefore, f is differentiable at x 1. f 1 0 85. Note that f is continuous at x 2. f x fx x fx x x2 4x f2 2 f2 2 1, x ≤ 2 3, x > 2 x→2 The derivative from the left is lim x→2 lim lim x2 x 4x x 1 2 3 2 5 5 x→2 lim x lim 4 4 y 2 4. 4. The derivative from the right is lim x→2 x→2 x→2 The one-sided limits are equal. Therefore, f is differentiable at x 87. (a) The distance from 3, 1 to the line mx d Ax1 A m3 By1 2 2. f 2 y 4 0 is C B 2 3 2 11 m2 1 4 3m m2 3 1 . 1 x (b) 5 1 2 3 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 . −4 −1 4 60 Chapter 2 Differentiation 89. False. the slope is lim f2 x→0 x x f2 . 91. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does not exist at that point. For example, if f x x , then the derivative from the left at x 0 is 1 and the derivative from the right at x 0 is 1. At x 0, the derivative does not exist. x sin 1 x , 0, x x 0 0 0. Thus, lim x sin 1 x x→0 93. f x x ≤ x sin 1 x ≤ x , x Using the Squeeze Theorem, we have x 0. Using the alternative form of the derivative we have x→0 0 f 0 and f is continuous at lim fx x f0 0 x→0 lim x sin 1 x x0 0 x→0 lim sin 1 . x 0. Since this limit does not exist (it oscillates between gx x2 sin 1 x , x 0, x 0 0 1 and 1), the function is not differentiable at x Using the Squeeze Theorem again we have x2 ≤ x2 sin 1 x ≤ x2, x 0. Thus, lim x2 sin 1 x x→0 ous at x 0. Using the alternative form of the derivative again we have x→0 0 f 0 and f is continu- lim fx x f0 0 x→0 lim x2 sin 1 x x0 0 x→0 lim x sin 1 x 0. Therefore, g is differentiable at x 0, g 0 0. Section 2.2 1. (a) y y y1 x1 2 Basic Differentiation Rules and Rates of Change (b) y y y1 x6 6x5 y 7x 8 x3 2 (c) y y y1 1 x7 x2 2x 2 (d) y y y1 5 x3 3x2 3 x1 45 5 1 12 2x 1 2 3 12 2x 3 2 3. y y 8 0 5. y y 7. y x 7 9. y 7 x8 y x 1 x 5 1 5x4 5 11. f x fx x 1 1 13. f t fx 2t2 4t 3t 3 6 15. g x gx x2 2x 4x3 12x 2 17. s t st 1 x 1 x2 t3 3t2 2t 2 4 19. y y 2 2 sin cos cos sin 21. y y x2 2x 1 cos x 2 1 sin x 2 23. y y 3 sin x 3 cos x Section 2.2 Basic Differentiation Rules and Rates of Change 61 Function 25. y 5 2x 2 3 2x x x 3 x2 6x 6 Rewrite y 5 x 2 3 x 8 x 2 Derivative y 5x 9 x 8 1 x 2 3 Simplify y 5 x3 9 8x4 1 2x3 73 x, 5 0, 2 27. y 3 y y 3 y 4 y 29. y 12 y 32 y 1 2 21 2 x 5 0 31. f x 3x 2, 1, 3 3 33. f x fx f0 1 2 35. y 2x 4x2 1 2, 0, 1 4x 4 1 fx f1 6 x3 y y0 8x 4 37. f f f0 4 sin 4 cos 41 x3 1 3x2 x2 8 x3 x 1 x 2 12 , 0, 0 1 3 4 x3 x3 3 39. f x fx x2 2x 5 6x 3x 3 2 41. g t 6 x3 gt t2 4 t3 12t t2 4t 3 2x 2t 4 2t 12 t4 43. f x fx 47. f x fx x 8 3 4x 2 45. y y x x2 3x2 1 1 x3 x 1 6 x x1 23 2 6x1 1 2x 3 49. h s 2 x2 3 h (s s4 4 s 5 5 s2 45 3 2x 2 s 3 13 4 5s1 5 2 3s1 3 51. f x fx 6x 3x 12 5 cos x 5 sin x 6x1 3 2 5 cos x 5 sin x x 53. (a) y y x4 4x3 3x2 6x 2 55. (a) f x fx 1 4 2 x3 3 x 2 2x 74 34 At 1, 0 : y Tangent line: 41 3 61 0 2 0 2. 2x 3 2x7 4 3 2 y y 2 3 x 2 7 0 3 x 2 7 2 1 y 2x y At 1, 2 , f 1 Tangent line: (b) 3 −2 −1 2 (1, 0) 3x (b) 5 2y (1, 2) −2 −1 7 62 Chapter 2 Differentiation 57. y y x4 4x3 4x x2 4x x 8x2 16x 4 2x 2 59. y y 2 1 x2 2x x 3 2 2 cannot equal zero. x3 Therefore, there are no horizontal tangents. y 0⇒x 0, ± 2 14 , 2, 14 63. x 2 2x kx k 4x 4 2x 4x 9 Equate functions Equate derivatives 4 and 4x 9⇒ x2 3, k 9⇒x 10. ± 3. Horizontal tangents: 0, 2 , 2, 61. y y cos x At x x 1 sin x, 0 ≤ x < 2 cos x 0 1⇒x ,y . , Hence, k x 2 2x 3, k Horizontal tangent: For x 2 and for x 65. k x k x2 3 x 4 3 4 3 Equate functions Equate derivatives 32 x 32 4 x and 4 x Hence, k 3 x 4 3 3⇒ x 4 3 x 4 3 3⇒ x 2 (c) 3⇒x 2⇒k 3. 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. y f BC A D E x 69. g x fx 6⇒g x fx 71. 3 y f 1 3 2 1 2 1 f x 2 3 If f is linear then its derivative is a constant function. fx fx ax a b S ection 2.2 Basic Differentiation Rules and Rates of Change 63 73. Let x1, y1 and x2, y2 be the points of tangency on y these functions are y m 2x ⇒ m 2x1 x1 Since y1 m y2 x2 x22 x22 6x2 2 x2 and y 2x2 x2 6. 6x 5, respectively. The derivatives of 2x1 2x2 x2 3 and 6 y 2x 6⇒m x1 and y2 y1 x1 6x2 x2 5 x22 x22 6x2 x2 6x2 5 x1 x2 3 6x2 12x2 3 5, 5 y x12 2 4 2x2 6 6 2x2 18x2 6. 3 2 1 )2, 3) )1, 1) x 2 3 5 x2 x22 2x22 2x2 2 x2 2 2x2 2x2 4x22 0 0 1 or 2 −1 9 14 4 1 x2 3 18 y 6x2 2 x2 5 4 3 2 1 )2, 4) x2 1 ⇒ y2 0, x1 2 and y1 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, x1 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. 75. f x fx 1 2x 4 4 4 x x x x 1 x 2 x, 12 4, 0 1 2x 77. f 1 3.64 1 0y 4x 2 xy 2xx 2x 4, y 2 0.77 3.33 1.24 The point 4, 2 is on the graph of f. Tangent line: y 4y 2 8 0 x x 02 x 44 4 4y 4 4 64 Chapter 2 Differentiation 79. (a) One possible secant is between 3.9, 7.7019 and 4, 8 : y y y (b) f x Tx 8 8 Sx 31 x 2 3x 8 4 7.7019 x 3.9 4 4 −2 20 (4, 8) 12 −2 2.981 x 2.981x 3.924 2 ⇒f 4 4 8 3 2 2 3x 4 3 S x is an approximation of the tangent line T x . (c) As you move further away from 4, 8 , the accuracy of the approximation T gets worse. 20 f −2 −2 T 12 (d) x f4 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 81. False. Let f x fx gx 85. True. If g x x2 and g x 2x, but f x 3f x , then g x x2 4. Then gx. 3f x . 83. False. If y 2, then dy dx 0. 2 is a constant. 87. f t ft 2t 2 7, 1, 2 89. f x fx 1 x2 1 , 1, 2 x Instantaneous rate of change is the constant 2. Average rate of change: f2 2 f1 1 22 7 1 21 7 2 Instantaneous rate of change: 1, 2, 1 ⇒f 1 1 2 ⇒f 2 1 1 4 (These are the same because f is a line of slope 2.) Average rate of change: f2 2 f1 1 12 2 1 1 1 2 S ection 2.2 Basic Differentiation Rules and Rates of Change 65 91. (a) s t vt (b) s2 2 16t2 32t s1 1 st 1362 93. st 4.9t2 4.9t2 v0t 120t 120 120 s0 1298 32t 1: v 1 2: v 2 1346 48 ft sec vt v5 v 10 9.8t 9.8 5 9.8 10 71 m sec 22 m sec (c) v t 120 When t When t (d) 16t2 t2 (e) v 32 ft sec. 64 ft sec. 0 1362 4 1362 4 295.242 ft sec 97. v 2 3 1362 1362 ⇒t 16 32 9.226 sec 1362 4 8 1362 95. 60 v 40 mph 2 3 mi min 4 mi Distance (in miles) 10 8 s Velocity (in mph) 50 40 30 20 10 t 2 4 6 8 10 mi min 6 min 0 mph v 0 mi min 0 mi (10, 6) 6 (6, 4) 4 0 mi min 2 min v 60 mph Time (in minutes) (8, 4) 2 t 2 4 6 8 10 1mi min (0, 0) 1 mi min 2 min 2 mi Time (in minutes) (The velocity has been converted to miles per hour) 99. (a) Using a graphing utility, you obtain R (c) T dT (e) dv For v For v For v R 0.167v B 0.02. 0.1431v 0.44 (d) 60 (b) Using a graphing utility, you obtain B 0.00586v2 T 0.0239v 0.46. 0.00586v2 0.1431 0.01172v 40, T 40 80, T 80 B R 0.612. 1.081. 1.315. 0 0 100 100, T 100 (f) For increasing speeds, the total stopping distance increases. 1,008,000 Q 1,008,000 Q2 5083.095 dC dQ $1.93. 101. A s2, dA ds 2s 4 m, 103. C dC dQ C 351 When Q C 350 350, 6.3Q 6.3 5085 $1.91 When s dA ds 8 square meters per meter change in s. 105. (a) f 1.47 is the rate of change of the amount of gasoline sold when the price is $1.47 per gallon. (b) f 1.47 is usually negative. As prices go up, sales go down. 66 Chapter 2 Differentiation 107. y ax2 bx c Since the parabola passes through 0, 1 and 1, 0 , we have 0, 1 : 1 1, 0 : 0 Thus, y y 1 1 a b ax2 2ax 2a 1 a 2 a 1 2x 2 a0 a1 2 2 b0 b1 a 1x 1 a 1 c⇒c 1⇒b 1 a 1. x 1, we know that the derivative is 1 at the point 1, 0 . 1. From the tangent line y a 1 3 3x 1. Therefore, y 109. y y x3 3x2 9x 9 Tangent lines through 1, y x3 9x 9 9 0 x 3x2 3x3 2x3 9: 9x 3x2 3x2 3 2 3 2, 81 8 1 9x x2 2x 9 3 0 or x The points of tangency are 0, 0 and Tangent lines: y 0 y 9x y 0 x≤2 x>2 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 111. f x ax3, x2 b, f must be continuous at x x→2 x→2 2 to be differentiable at x 8a b 4 b 8a 8a 4 4 b 2. b lim f x x→2 x→2 lim ax 3 lim f x lim x2 x<2 x>2 fx 3ax2, 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 Section 2.3 The Product and Quotient Rules and Higher-Order Derivatives 67 113. Let f x fx cos x. lim lim fx x→0 x x fx sin x sin x x 1 sin x x→0 cos x cos x cos x cos x x sin x 1 cos x sin x x x→0 lim 0 x→0 lim sin x Section 2.3 1. g x gx x2 x 2 The Product and Quotient Rules and Higher-Order Derivatives 1 x2 1 2x 2x 2 6x 2 2x 2 2x 2x 2 2 x 2 3. h t 2x 2x 2x3 4x 2 ht 3 t t2 3 4 t2 t2 4 3t2 3 t1 32 t 4 23 t1 2t 3 2x3 4x3 1 4t 3 2t 4 7t2 4 3t2 3 5. f x fx x 3 cos x x3 2 7. f x cos x 3x 2 x3 sin x fx x x2 x2 1 11 x 2x x2 1 2 1 x2 x2 12 sin x 3x cos x 3 9. h x x3 x3 x 1 x1 3 x3 1 1 1 x 2 3 x1 3 x3 1 2 3 11. g x 3x 2 gx sin x x2 x 2 cos x x2 sin x 2x 2 x cos x x3 2 sin x hx x3 1 x 9x 2 2 3 x3 3x 12 1 3x2 3 x3 8x 3 1 2 13. f x fx x3 x3 10x4 3x 2x2 3x 4x 12x3 3x 3 3x2 5 2x2 18x 3x 15 5 3x2 3 15. f x fx x2 x x 4 3 3 2x x x2 3 2 41 2x 2 x2 x 6x x 2 x 32 6x 3 4 2 4 f0 15 1 1 6 3 4 2 f1 1 4 17. fx fx f x cos x x 2 2 sin x 2 2 cos x 1 2 4 8 cos x x sin x 4 4 68 Chapter 2 Differentiation Function 19. y 21. y x2 3 7 3x3 4x3 x 2 Rewrite 2x y 12 x 3 7 x 3 3 Derivative 2 x 3 y 2 x 3 7x 2x 2 3 4 Simplify y 2x 3 7 x4 2 x 2 y y y y 23. y 25. f x fx y 2x x x2 1 1 2 4 x, x > 0 12 y 3 x2 2 2 2 2x x2 2x x2 1 3 1 2 2x x2 2x 2x2 4x x2 1 2 x 27. f x fx x1 1 x 1 2, 12 12 x 4 x 3 x 4x x 3 x2 x2 x 6x x 6x 3 2 29. f x 9 32 3 12 fx 2x x x 12 5 2x1 5 x 2 2 5x x 12 3 4 4x 1 x 32 32 32 x 5 2 2x 5 2x3 2 2x 5 2x x 31. h s hs s3 6s5 1 x 3 2 2 s6 4s3 4 2 12s2 6s2 s3 2 33. f x fx x x2 2x xx 3x 2 x2 2x 3x 1 3 2x x2 1 3x 3 2x 2 6x 4x 2 x 2 3x 8x 2 2x 1 2x 3x 2 3 3 2x 2 x2 35. f x fx 3x3 9x2 9x2 9x 4 15x 4 x2 x2 x2 c2 c2 2 2x 2 2x 3 x2 x 3 2 5x 1 1 5 16x 32x 3x3 3x 4 20 20 4x 1 x 3x 3 6x 4 4x 2 12x 3 1 4x 3x3 3x 4 8x 2 4x x 15x3 16x 51 4x2 20x 4x x 4x 4 x2 36x3 48x 3 5x 4x 41x 2 33x 2 37. f x fx 39. f x x2 c2 2 c2 2x ft t2 sin t t2 cos t t t cos t 2t sin t 2 sin t c2 2x x2 4xc c2 2 2 x2 S ection 2.3 The Product and Quotient Rules and Higher-Order Derivatives 69 41. f t ft 45. g t gt cos t t t sin t t2 4 43. f x cos t t sin t t2 t1 4 x 1 tan x sec2 x tan2 x cos t fx t 34 8 sec t 8 sec t 1 4t3 4 47. y 8 sec t tan t y 3 1 sin x 2 cos x 3 sec x tan x 2 3 sec x tan x 2 3 sec x 2 sec2 x tan2 x tan x 3 sec x tan x 2 1 sec x 1 t 4 8 sec t tan t 49. y y csc x csc x cot x cos x sin2 x sin x cos x 51. f x fx x2 tan x x2 sec2 x xx sec2 x 2x tan x 2 tan x cos x 1 cos x csc2 x cos x cot2 x 53. y y 2x sin x 2x cos x 4x cos x 57. g g x 2 cos x 2 sin x 2 sin x x2 x2 sin x sin x 2x cos x 55. g x gx x x 2x2 x 1 2x 2 8x 2 1 2 5 (form of answer may vary) 1 1 sin sin sin 1 1 1 csc x csc x csc x 22 3 1 22 csc x cot x 1 csc x csc x cot x 1 csc x 2 43 2 csc x cot x 1 csc x 2 cos 1 2 (form of answer may vary) 59. y y y 6 61. ht ht sec t t t sec t tan t t2 sec t t tan t t2 1 1 1 2 sec t 1 h sec tan 2 70 Chapter 2 Differentiation 63. (a) f x fx x3 x3 4x3 3x 3x 6x2 1x 11 6x 2, x 5 3. 1x 1, 3 3 53. (b) − 10 10 2 3x2 (1, − 3) 10 f1 1 slope at 1, 3 − 10 Tangent line: y 65. (a) f x fx f2 x x x 1 1 ⇒y x 2 51. (b) , 2, 2 x1 2 6 11 x1 1 1 2 x 1 1 1 (2, 2) 2 −3 6 2 slope at 2, 2 . 1x 2 ⇒y x 4 −3 Tangent line: y 67. (a) fx fx f 4 tan x, sec2 x 2 2 ,1 4 55. (b) 4 (( π ,1 4 − slope at 4 ,1 . −4 Tangent line: y y 4x 2y x2 x x x2 x fx 1 1 2x x1 2x 12 x2 1 2 1 1 2x 2x 2 4 2 0 x x 5x x 2 3 3x 1 x 22 25 x 4 2 x 5x 22 3x 2 6 x 41 x 2x x 4 2 2 2 69. f x fx 71. f x gx gx 2. 6 2 2 xx x 2 12 fx 2 0 when x 0 or x f and g differ by a constant. Horizontal tangents are at 0, 0 and 2, 4 . x n sin x x n cos x x When n When n When n When n n 1 73. f x fx 75. Area nx n 1 At 2 31 t 2 2 2 2t 1 1 t 2 12 t 2t3 2 t1 2 sin x At sin x. 2 sin x . 3 sin x . 4 sin x . n sin x . x cos x n sin x x cos x x x cos x x2 x cos x x3 x cos x xn 1 1: f x 2: f x 3: f x 4: f x 3t1 1 t 2 12 6t 1 2 cm sec 2t For general n, f x x cos x S ection 2.3 The Product and Quotient Rules and Higher-Order Derivatives 71 77. C dC dx 100 100 200 x2 400 x3 10: 15: 20: x x x dC dx dC dx dC dx 30 , 1≤x 79. Pt Pt 500 1 500 500 2000 50 200 50 50 50 4t 50 t2 t2 4 4t 2t 50 t2 2 4t 2 t2 2 t2 t2 2 30 30 2 (a) When x (b) When x (c) When x $38.13. $10.37. P2 $3.80. 31.55 bacteria per hour As the order size increases, the cost per item decreases. 81. (a) sec x d sec x dx (b) csc x d csc x dx (c) cot x d cot x dx 1 cos x d 1 dx cos x 1 sin x d1 dx sin x cos x sin x d cos x dx sin x sin x sin x cos x cos x sin x 2 sin2 x cos2 x sin2 x x x x 2 x 1 3 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 sin x 0 sin x 1 cos x 2 cos x sin x sin x 1 sin x cos x sin x csc x cot x 1 sin2x csc2 x 83. f x fx fx 4x3 6x1 3x 2 2 85. f x fx x fx 1 11 x1 x1 2 12 3 x 1 1 2 87. f x fx fx 3 sin x 3 cos x 3 sin x 89. f x fx x2 2x 91. f x f 4 2x 1 2x 2 12 x 1 x 93. 4 3 2 y 95. f x fx f2 2g x 2g x 2g 2 2 2 hx hx h2 4 97. f x fx f2 gx hx hxg x hx h2g 2 h2 1 10 2 1 2 gxh x 2 g2h 2 2 1 x 1 2 3 4 0 34 f2 0 x 2 2. One such function is f x 72 Chapter 2 Differentiatio...
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