C H A P T E R 2 Differentiation

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© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. C H A P T E R 2 Differentiation Section 2.1... The Derivative and the Slope of a Graph ............................................67 Section 2.2 Some Rules for Differentiation............................................................82 Section 2.3 Rates of Change: Velocity and Marginals...........................................88 Section 2.4 The Product and Quotient Rules..........................................................95 Quiz Yourself .............................................................................................................104 Section 2.5 The Chain Rule...................................................................................106 Section 2.6 Higher-Order Derivatives...................................................................113 Section 2.7 Implicit Differentiation.......................................................................118 Section 2.8 Related Rates ......................................................................................126 Review Exercises ........................................................................................................130 Test Yourself .............................................................................................................143 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 67 C H A P T E R 2 Differentiation Section 2.1 The Derivative and the Slope of a Graph 1. 2. Skills Warm Up 1. (2, 1), (2, 4) 2 P Q x = 2. P(2, 2), Q = (−5, 2) 2 2 0 5 2 m − = = − − 2 0( 2) 2 y x y − = − = 3. P(2, 0), Q = (3, −1) 1 0 1 3 2 m − − = = − − 0 1( 2) 2 y x y x − = − − = − 4. P(3, 5), Q(−1, −7) 7 5 12 3 1 3 4 m − − − = = = − − − 5 3( 3) 3 4 y x y x − = − = − 5. ( )2 ( ) 0 0 0 2 2 lim lim lim 2 2 x x x xx x x x x x x x x x Δ → Δ → Δ → Δ + Δ Δ + Δ = Δ Δ = + Δ = 6. ( ) ( ) ( ) ( ) 2 2 3 0 2 2 0 2 2 0 2 3 3 lim 3 3 lim lim 3 3 3 x x x x x x x x x x x xx x x x xx x x Δ → Δ → Δ → Δ + Δ + Δ Δ Δ ⎡ + Δ + Δ ⎤ = ⎣ ⎦ Δ = + Δ + Δ = 7. 0 ( ) 2 lim 1 1 Δx→ x x x x = + Δ 8. ( ) ( ) ( ) ( ) 2 2 0 2 2 2 0 2 0 0 lim 2 lim 2 lim 2 lim 2 x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − Δ + Δ + Δ − = Δ Δ + Δ = Δ Δ + Δ = Δ = 9. f (x) = 3x Domain: (−∞, ∞) 10. ( ) 1 1 f x x = − Domain: (−∞, 1) ∪ (1, ∞) 11. ( ) 1 3 2 1 5 3 f x = x − 2x + x − 1 Domain: (−∞, ∞) 12. ( ) 3 f x 6x x x = + Domain: (−∞, 0) ∪ (0, ∞) x y x y 68 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3. 4. 5. 6. 7. The slope is m = 1. 8. The slope is 43 m = . 9. The slope is m = 0. 10. The slope is 1 4m = . 11. The slope is 1 3m = − . 12. The slope is m = −3. 13. 2005: m ≈ 119 2007: m ≈ 161 The slope is the rate of change of revenue at the given point in time. 14. 2006: m ≈ 250 2008: m ≈ 100 The slope is the rate of change of sales at that point in time. 15. 3: 9 7: 0 10: 10 t m t m t m = ≈ = ≈ = ≈ − The slope is the rate of change of the average temperature at the given point in time. 16. (a) At t1, f ′(t1) > g′(t1), so the runner given by f is running faster. (b) At t2 , g′(t2 ) > f ′(t2 ), so the runner given by g is running faster. The runner given by f has traveled farther. (c) At t3, the runners are at the same location, but the runner given by g is running faster. (d) The runner given by g will finish first because that runner finishes the distance at a lesser value of t. 17. ( ) ( ) ( ) sec 0 0 1 1 0 0 f x f m x x x + Δ − = Δ − − − = Δ = Δ = sec 0 0 lim lim 0 0 x x m m Δ → Δ → = = = 18. ( ) ( ) sec 2 2 6 6 0 0 f x f m x x x − + Δ − − = Δ − = Δ = Δ = sec 0 0 lim lim 0 0 x x m m Δ → Δ → = = = 19. ( ) ( ) ( ) sec 2 2 8 32 2 8 6 3 2 3 3 f x f m x x x x x x x + Δ − = Δ ⎡⎣ − + Δ ⎤⎦ − = Δ − − Δ − = Δ − Δ = Δ = − sec 0 0 lim lim 3 3 x x m m Δ → Δ → = = − = − x y y x y y Section 2.1 The Derivative and the Slope of a Graph 69 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 20. ( ) ( ) ( ) sec 1 1 6 1 3 9 6 6 3 9 6 6 f x f m x x x x x x x + Δ − = Δ ⎡⎣ + Δ + ⎤⎦ − = Δ + Δ + − = Δ Δ = Δ = sec 0 0 lim lim 6 6 x x m m Δ → Δ → = = = 21. ( ) ( ) ( ) ( ( ) ) ( ) ( ) ( ) sec 2 2 2 2 2 2 2 3 5 2 4 4 3 5 8 8 2 3 5 2 4 2 4 f x f m x x x x x x x x x x x x x + Δ − = Δ ⎡ + Δ − ⎤ − = ⎣ ⎦ Δ ⎡ + Δ + Δ − ⎤ − ⎢⎣ ⎥⎦ = Δ ⎡ + Δ + Δ − ⎤ − = ⎣ ⎦ Δ Δ + Δ = Δ = + Δ ( ( )) sec 0 0 lim lim 2 4 8 x x m m x Δ → Δ → = = + Δ = 22. ( ) ( ) ( ) ( ) ( ) ( ) ( ) sec 2 2 2 2 2 2 4 2 4 2 4 4 4 4 4 4 f x f m x x x x x x x x x x x x x + Δ − = Δ − + Δ − ⎡ − ⎤ = ⎣ ⎦ Δ − − Δ − Δ = Δ − Δ − Δ = Δ Δ − − Δ = Δ = − − Δ sec ( ) 0 0 lim lim 4 4 x x m m x Δ → Δ → = = − − Δ = − 23. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ) sec 3 3 2 3 2 3 2 2 2 2 2 2 2 2 8 12 6 2 6 11 6 11 6 11 6 f x f m x x x x x x x x x x x x x x x x x x x + Δ − = Δ + Δ − + Δ − ⎡ − ⎤ = ⎣ ⎦ Δ + Δ + Δ + Δ − − Δ − = Δ Δ + Δ + Δ = Δ Δ + Δ + Δ = Δ = + Δ + Δ ( ( )2 ) sec 0 0 lim lim 11 6 11 x x m m x x Δ → Δ → = = + Δ + Δ = 24. ( ) ( ) ( ) ( ) () () ( ) ( ) ( ) ( ) ( ( ) ) ( ) sec 3 3 2 3 2 3 2 2 1 1 1 21 1 2 1 1 3 3 2 2 3 5 3 5 3 5 3 f x f m x x x x x x x x x x x x x x x x x x x + Δ − = Δ + Δ + + Δ − ⎡ + ⎤ = ⎣ ⎦ Δ + Δ + Δ + Δ + + Δ − = Δ Δ + Δ + Δ = Δ Δ + Δ + Δ = Δ = + Δ + Δ ( ( )2 ) sec 0 0 lim lim 5 3 5 x x m m x x Δ → Δ → = = + Δ + Δ = 25. ( ) ( ) ( ) ( ) ( ) ( ) sec 4 4 2 4 2 4 2 4 4 2 4 4 2 4 4 4 4 16 2 4 4 16 4 16 2 4 4 4 2 4 4 4 2 4 4 f x f m x x x x x x x x x x x x x x x x x + Δ − = Δ + Δ − = Δ + Δ − + Δ + = ⋅ Δ + Δ + + Δ − = Δ +Δ + + Δ − = Δ +Δ + Δ = Δ +Δ + = + Δ + sec 0 0 lim lim 4 2 4 4 4 1 2 4 4 2 x x m m Δ → Δ → x = = + Δ + = = + 70 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 26. ( ) ( ) ( ) ( ) sec 8 8 8 1 8 1 9 3 9 3 9 3 9 3 9 9 9 3 9 3 1 9 3 f x f m x x x x x x x x x x x x x x x x + Δ − = Δ + Δ + − + = Δ + Δ − = Δ + Δ − + Δ + = ⋅ Δ + Δ + + Δ − = Δ + Δ + Δ = Δ + Δ + = + Δ + sec 0 0 lim lim 1 9 3 1 9 3 1 6 x x m m x Δ → Δ → = = + Δ + = + = 27. ( ) ( ) ( ) 0 0 0 0 lim lim 3 3 lim 0 lim 0 0 x x x x f x x f x f x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = Δ = Δ = = 28. ( ) ( ) ( ) ( ) 0 0 0 0 lim 2 2 lim lim 0 lim 0 0 x x x x f x x f x f x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ − − − = Δ = Δ = = 29. ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 lim 5 5 lim lim 5 5 5 lim 5 lim 5 5 x x x x x f x x f x f x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − + Δ − − = Δ − − Δ + = Δ − Δ = Δ = − = − 30. ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 lim 4 1 4 1 lim lim 4 4 1 4 1 lim 4 lim 4 4 x x x x x f x x f x f x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ + Δ + − − = Δ Δ = Δ = = 31. ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 1 1 3 3 1 1 1 3 3 3 1 3 lim 2 2 lim 2 2 lim lim lim 1 3 1 3 s s s s s g s s g s g s s s s s s s s s s s s Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ + Δ + − − = Δ Δ = Δ = = 32. ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 1 1 2 2 1 1 1 2 2 2 1 2 lim 6 6 lim 6 6 lim lim lim 1 2 1 2 t t t t t h t t h t h t t t t t t t t t t t t Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − + Δ − − = Δ − − Δ − + = Δ − Δ = Δ = − = − Section 2.1 The Derivative and the Slope of a Graph 71 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 33. ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ) 0 2 2 0 2 2 2 0 2 2 2 0 0 0 lim 4 5 4 5 lim 4 2 5 5 4 5 lim 4 8 4 5 5 4 5 lim 4 2 5 lim 4 lim 4 2 5 8 5 4 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡ + Δ − + Δ ⎤ − − = ⎣ ⎦ Δ ⎡ + Δ + Δ − − Δ ⎤ − + = ⎢⎣ ⎥⎦ Δ + Δ + Δ − − Δ − + = Δ Δ ⎛ + Δ − ⎞ ⎜ ⎟ = ⎝ ⎠ Δ = ⎛ + Δ − ⎞ = − ⎜ ⎟ ⎝ ⎠ 34. ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ) 0 2 2 0 2 2 2 0 2 2 2 0 0 0 lim 2 7 2 7 lim 2 2 7 7 2 7 lim 2 4 2 7 7 2 7 lim 2 2 7 lim 2 lim 2 2 7 4 7 2 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡ + Δ + + Δ ⎤ − + = ⎣ ⎦ Δ ⎡ + Δ + Δ + + Δ ⎤ − − = ⎢⎣ ⎥⎦ Δ + Δ + Δ + + Δ − − = Δ Δ ⎛ + Δ + ⎞ ⎜ ⎟ = ⎝ ⎠ Δ = ⎛ + Δ + ⎞ = + ⎜ ⎟ ⎝ ⎠ 35. ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 lim lim 1 1 lim 1 1 1 1 1 1 1 1 lim 1 1 lim 1 1 lim 1 1 1 1 2 1 t t t t t t h t t h t h t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = + Δ − − − = Δ + Δ − − − + Δ − + − = ⋅ Δ + Δ − + − + Δ − − − = Δ + Δ − + − Δ = Δ + Δ − + − = + Δ − + − = − 72 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 36. ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 lim lim 2 2 lim 2 2 2 2 2 2 2 2 lim 2 2 lim 2 2 lim 1 2 2 1 2 2 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ + Δ + − + + Δ + + + = ⋅ Δ + Δ + + + + Δ + − + = Δ +Δ + + + Δ = Δ +Δ + + + = + Δ + + + = + 37. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ( ) ) 0 3 3 0 3 2 2 3 3 0 2 2 3 0 2 2 0 2 2 0 2 lim 12 12 lim 3 3 12 12 12 lim 3 3 12 lim 3 3 12 lim lim 3 3 12 3 12 t t t t t t f t t f t f t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t tt t t Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − + Δ − − = Δ + Δ + Δ + Δ − − Δ − + = Δ Δ + Δ + Δ − Δ = Δ Δ + Δ + Δ − = Δ = + Δ + Δ − = − 38. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ( ) ) 0 3 2 3 2 0 3 2 2 3 2 2 3 2 0 2 2 3 2 0 2 2 0 2 2 0 2 lim lim 3 3 2 lim 3 3 2 lim 3 3 2 lim lim 3 3 2 3 2 t t t t t t f t t f t f t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t tt t t t t t Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + + Δ − + = Δ + Δ + Δ + Δ + + Δ + Δ − − = Δ Δ + Δ + Δ + Δ + Δ = Δ Δ + Δ + Δ + + Δ = Δ = + Δ + Δ + + Δ = + Section 2.1 The Derivative and the Slope of a Graph 73 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 39. ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) 0 0 0 0 0 0 2 lim 1 1 lim 2 2 1 2 1 2 lim 2 2 2 2 2 2 2 2 lim lim 2 2 lim 1 2 2 1 2 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = + Δ + + Δ + +Δ + ⋅ − ⋅ = + Δ + + + + Δ + Δ + − + Δ + + Δ + + = Δ −Δ = Δ + Δ + + − = + Δ + + = − + 40. ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) 0 0 0 0 0 0 2 lim 1 1 lim 1 1 1 1 1 1 lim 1 1 1 1 1 1 1 1 lim lim 1 1 lim 1 1 1 1 1 s s s s s s g s s g s g s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = + Δ − − Δ − +Δ − ⋅ − ⋅ = + Δ − − − + Δ − Δ − − + Δ − + Δ − − = Δ −Δ = Δ + Δ − − − = + Δ − − = − − 41. ( ) ( ) ( ) ( ) ( ( ) ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 0 1 1 2 2 1 1 2 2 lim lim 2 lim lim lim lim x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − = Δ + Δ + Δ − = Δ Δ + Δ = Δ Δ + Δ = Δ = + Δ = m = f ′(2) = 2 2 2( 2) 2 2 y x y x − = − = − 42. ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 0 lim lim 2 lim 2 lim 2 lim lim 2 2 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − + Δ − − = Δ − + Δ + Δ + = Δ − Δ − Δ = Δ Δ − − Δ = Δ = − −Δ = − m = f ′(−1) = −2(−1) = 2 ( 1) 2 ( 1) 2 1 y x y x − − = ⎡⎣ − − ⎤⎦ = + −6 6 −4 4 (2, 2) −3 3 −3 1 (−1, −1) 74 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 43. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 0 lim 1 1 lim 2 2 2 1 2 1 lim 2 2 lim 2 2 lim lim 2 2 2 2 x x x x x x f x x f x f x x x x x x x xx x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − − − = Δ + Δ − + Δ − Δ + − + − = Δ Δ + Δ − Δ = Δ Δ + Δ − = Δ = + Δ − = − m = f ′(−2) = 2(−2) − 2 = −6 9 6 ( 2) 6 3 y x y x − = − ⎡⎣ − − ⎤⎦ = − − 44. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 0 lim 2 1 2 1 lim 2 4 2 1 2 1 lim 4 2 lim 4 2 lim lim 4 2 4 x x x x x x f x x f x f x x x x x x x xx x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − − − = Δ + Δ + Δ − − + = Δ Δ + Δ = Δ Δ + Δ = Δ = + Δ = m = f ′(0) = 4(0) = 0 ( 1) 0( 0) 1 y x y − − = − = − 45. ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 lim 1 1 lim lim lim lim lim 1 1 2 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ + Δ − + Δ + = ⋅ Δ + Δ + + Δ − = Δ +Δ + Δ = Δ +Δ + = + Δ + = (4) 1 1 2 4 4 m = f ′ = = 3 1( 4) 4 1 2 4 y x y x − = − = + −12 12 −5 11 (−2, 9) −6 6 −3 5 (0, −1) −2 7 −1 5 (4, 3) Section 2.1 The Derivative and the Slope of a Graph 75 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 46. ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 0 lim lim 2 2 lim 2 2 2 2 2 2 2 2 lim 2 2 lim 2 2 lim 1 2 2 1 2 2 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ + Δ + − + + Δ + + + = ⋅ Δ + Δ + + + + Δ + − + = Δ +Δ + + + Δ = Δ +Δ + + + = + Δ + + + = + (7) 1 1 2 7 2 6 m = f ′ = = + 3 1( 7) 6 1 11 6 6 y x y x − = − = + 47. ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) 0 0 0 0 0 0 2 lim 1 1 lim 1 1 lim lim lim lim 1 1 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = + Δ Δ + Δ ⋅ − ⋅ = + Δ + Δ Δ − + Δ + Δ = Δ −Δ = Δ +Δ − = + Δ = − ( ) ( )2 1 1 1 1 m = f ′ = − = − 1 1( 1) 2 y x y x − = − − = − + −3 12 −3 7 (7, 3) −1 5 −1 3 (1, 1) 76 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 48. ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) 0 0 0 0 0 0 2 lim 1 1 lim 3 3 1 3 1 3 lim 3 3 3 3 3 3 3 3 lim lim 3 3 lim 1 1 3 3 3 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = + Δ − − Δ − +Δ − ⋅ − ⋅ = + Δ − − − + Δ − Δ − − + Δ − + Δ − − = Δ − Δ = + Δ − − Δ − = =− + Δ − − − ( ) ( ) ( ) ( ) 2 2 1 1 2 3 1 1 2 1 2 1 m f y x y x y x = ′ = − = − − − − = − − + = − + = − + 49. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 0 1 1 4 4 1 1 1 1 4 2 4 4 1 1 2 4 1 1 2 4 lim lim lim lim lim lim 1 1 2 4 1 2 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − + Δ − − = Δ − − Δ − Δ + = Δ − Δ − Δ = Δ Δ − − Δ = Δ = ⎛− − Δ ⎞ ⎜ ⎟ ⎝ ⎠ = − Since the slope of the given line is −1, ( ) 1 2 1 2 and 2 1. x x f − = − = = − At the point (2, −1), the tangent line parallel to x + y = 0 is ( 1) 1( 2) 1. y x y x − − = − − = − + 50. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 0 0 lim 7 7 lim 2 7 7 lim 2 lim lim 2 2 x x x x x f x x f x f x x x x x x x xx x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡ + Δ − ⎤ − − = ⎣ ⎦ Δ + Δ + Δ − − + = Δ Δ + Δ = Δ = + Δ = Since the slope of the given line is −2, ( ) 2 2 1 and 1 6. x x f = − = − − = − At the point (−1, −6), the tangent line parallel to 2x + y = 0 is ( 6) 2( ( 1)) 2 8. y x y x − − = − − − = − − −2 7 −3 3 (2, −1) Section 2.1 The Derivative and the Slope of a Graph 77 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 51. ( ) ( ) ( ) ( ) ( ) ( ( ) ( ) ) ( ) ( ) ( ( ) ) ( ( ) ) 0 3 3 0 3 2 2 3 3 0 3 2 2 3 3 0 2 2 0 2 2 2 0 1 1 3 3 1 1 3 3 1 1 1 3 3 3 1 3 1 3 lim lim 3 3 lim lim lim lim x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − + Δ − − = Δ − + Δ + Δ + Δ + = Δ − − Δ − Δ − Δ + = Δ Δ − − Δ − Δ = Δ = − − Δ − Δ = − Since the slope of the given line is −9, ( ) ( ) 2 2 9 9 3 and 3 9 and 3 9. x x x f f − = − = = ± = − − = At the point (3, −9), the tangent line parallel to 9x + y − 6 = 0 is ( 9) 9( 3) 9 18. y x y x − − = − − = − + At the point (−3, 9), the tangent line parallel to 9x + y − 6 = 0 is 9 9( ( 3)) 9 18. y x y x − = − − − = − − 52. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 0 lim lim 2 lim 2 lim 2 1 lim lim 2 1 2 1 x x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − + Δ − − = Δ + Δ + Δ − − Δ − + = Δ Δ + Δ − Δ = Δ Δ + Δ − = Δ = + Δ − = − Since the slope of the given line is 1 2− , ( ) 1 2 1 1 3 4 4 16 2 1 and . x x f − = − = = − At the point (1 3 ) 4 16 , − the tangent line parallel to x + 2y − 6 = 0 is ( 3 ) 1 ( 1 ) 16 2 4 1 1 2 16 . y x y x − − = − − = − − 53. y is differentiable for all x ≠ −3. At (−3, 0) the graph has a node. 54. y is differentiable for all x ≠ ±3. At (±3, 0) the graph has a cusp. 55. y is differentiable for all x ≠ 3. At (3, 0) the graph has a cusp. 56. y is differentiable for all x > 1. The derivative does not exist at endpoints. 57. y is differentiable for all x ≠ ±2. The function is not defined at x = ±2. 58. y is differentiable for all x ≠ 0. The function is discontinuous at x = 0. 78 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 59. Since f ′(x) = −3 for all x, f is a line of the form f (x) = −3x + b. f (0) = 2, so 2 = (−3)(0) + b, or b = 2. Thus, f (x) = −3x + 2. 60. Sample answer: Since f (−2) = f (4) = 0, (x + 2)(x − 4) = 0. A function with these zeros is f (x) = x2 − 2x − 8. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 Then lim 2 8 2 8 lim 2 2 2 8 2 8 lim 2 2 lim lim 2 2 2 2. x x x x x f x x f x f x x x x x x x x x x xx x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡ + Δ − + Δ − ⎤ − − − = ⎣ ⎦ Δ + Δ + Δ − − Δ − − + + = Δ Δ + Δ − Δ = Δ = + Δ − = − So f ′(1) = 2(1) − 2 = 0.Sketching f (x) shows that f ′(x) < 0 for x < 1 and f ′(0) > 0 for x > 1. 61. Analytically, the slope of ( ) 1 3 4 f x = x is ( ) ( ) ( ) ( ) ( ) ( ) 0 3 3 0 2 2 3 0 2 2 0 2 1 1 4 4 1 4 lim lim 3 3 lim lim 1 3 3 4 3 . 4 x x x x f x x f x m x x x x x x x x x x x x xx x x Δ → Δ → Δ → Δ → + Δ − = Δ + Δ − = Δ ⎡ Δ + Δ + Δ ⎤ = ⎣ ⎦ Δ = ⎡ + Δ + Δ ⎤ ⎣ ⎦ = x −2 32 − −1 1 2 − 0 1 2 1 32 2 f (x) −2 −0.8438 −0.25 −0.0313 0 0.0313 0.25 0.8438 2 f ′(x) 3 1.6875 0.75 0.1875 0 0.1875 0.75 1.6875 3 x y −4 −3 −2 −1 2 3 4 −2 −3 1 2 5 x y −3 −1 1 2 3 5 6 −2 −8 −9 1 −2 −2 2 2 Section 2.1 The Derivative and the Slope of a Graph 79 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 62. Analytically, the slope of ( ) 2 34 f x = x is ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 0 0 3 3 4 4 3 3 3 3 4 2 4 4 3 3 2 4 lim lim lim lim lim 3 3 3 . 2 4 2 x x x x x f x x f x m x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − = Δ + Δ − = Δ + Δ + Δ − = Δ Δ + Δ = Δ = ⎛ + Δ ⎞ = ⎜ ⎟ ⎝ ⎠ 63. Analytically, the slope of ( ) 1 3 2 f x = − x is ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 3 3 0 3 2 2 3 3 0 2 2 3 0 2 2 0 2 1 1 2 2 1 1 2 2 1 2 lim lim 3 3 lim 3 3 lim lim 1 3 3 2 3 . 2 x x x x x f x x f x m x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − = Δ − + Δ + = Δ − ⎡ + Δ + Δ + Δ ⎤ + = ⎣ ⎦ Δ − ⎡ Δ + Δ + Δ ⎤ = ⎣ ⎦ Δ = − ⎡ + Δ + Δ ⎤ ⎣ ⎦ = − x −2 32 − −1 1 2 − 0 1 2 1 32 2 f (x) 4 1.6875 0.5 0.0625 0 −0.0625 −0.5 −1.6875 −4 f ′(x) −6 −3.375 −1.5 −0.375 0 −0.375 −1.5 −3.375 −6 x 2 − 32 − −1 1 2 − 0 1 2 1 32 2 f (x) 3 1.6875 0.75 0.1875 0 0.1875 0.75 1.6875 3 f ′(x) −3 −2.25 −1.5 −0.75 0 0.75 1.5 2.25 2 −2 2 −3 3 −2 −2 2 2 80 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 64. Analytically, the slope of ( ) 2 32 f x = − x is ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 3 3 2 2 3 3 2 2 32 lim lim 2 lim 2 lim lim 3 2 2 3 . x x x x x f x x f x m x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − = Δ − + Δ − − = Δ − ⎡ + Δ + Δ ⎤ + = ⎣ ⎦ Δ − ⎡ Δ + Δ ⎤ = ⎣ ⎦ Δ = − + Δ = − 65. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 0 0 lim 4 4 lim 2 4 lim lim 2 4 2 4 x x x x f x x f x f x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − + Δ − − = Δ Δ + Δ − Δ = Δ = + Δ − = − The x-intercept of the derivative indicates a point of horizontal tangency for f. 66. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 0 0 2 0 0 2 6 2 6 lim lim 6 2 lim lim 6 2 6 2 x x x x f x x f x x x x x x x f x x x x xx x x x x x Δ → Δ → Δ → Δ → + Δ − + + Δ − + Δ − + − ′ = = Δ Δ Δ − Δ − Δ = = − − Δ = − Δ The x-intercept of the derivative indicates a point of horizontal tangency for f. x −2 32 − −1 1 2 − 0 1 2 1 32 2 f (x) −6 −3.375 −1.5 −0.375 0 −0.375 −1.5 −3.375 −6 f ′(x) 6 4.5 3 1.5 0 −1.5 −3 −4.5 −6 2 − 8 − 2 2 −6 −4 12 8 −7 14 −1 13 Section 2.1 The Derivative and the Slope of a Graph 81 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 67. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) 0 3 3 0 3 2 2 3 3 0 2 2 3 0 2 2 0 2 lim 3 3 lim 3 3 3 3 3 lim 3 3 3 lim lim 3 3 3 3 3 x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x x xx x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − + Δ − − = Δ + Δ + Δ + Δ − − Δ − + = Δ Δ + Δ + Δ − Δ = Δ = + Δ + Δ − = − The x-intercepts of the derivative indicate points of horizontal tangency for f. 68. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ( ) ) 0 3 2 3 2 0 3 2 2 3 2 2 3 2 0 2 2 0 2 lim 6 6 lim 3 3 6 2 6 lim lim 3 3 12 6 3 12 x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x xx x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − + Δ − − = Δ − Δ + Δ + Δ − + Δ + Δ − + = Δ = + Δ + Δ − − Δ = − The x-intercepts of the derivative indicate points of horizontal tangency for f. 69. True. The slope of the graph is given by f ′(x) = 2x, which is different for each different x value. 70. False. f (x) = x is continuous, but not differentiable at x = 0. 71. True. See page 122. 72. True. See page 115. 73. The graph of f (x) = x2 + 1 is smooth at (0, 1), but the graph of g(x) = x + 1 has a node at (0, 1). The function g is not differentiable at (0, 1). 6 −4 −6 4 8 −36 −4 4 −4 −1 4 6 82 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 2.2 Some Rules for Differentiation 1. 3 0 y y = ′ = 2. ( ) ( ) 8 0 f x f x = − ′ = 3. 5 5 4 y x y x = ′ = 4. ( ) 5 f x 1 x = ( ) 6 6 f x 5x 5 x ′ = − − = − 5. ( ) ( ) 3 2 3 9 h x x h x x = ′ = 6. ( ) ( ) 5 4 2 10 h x x h x x = ′ = 7. 3 2 2 3 2 y x y x = ′ = 8. ( ) ( ) 3 2 4 3 2 g t t gt t = ′ = 9. ( ) ( ) 4 4 fx x f x = ′ = Skills Warm Up 1. (a) 2x2 , x = 2 2(22 ) = 2(4) = 8 (b) ( )2 2x , x = 2 ( ) ⎡2 2 ⎤2 = 42 = 16 ⎣ ⎦ (c) 2x−2 , x = 2 ( ) 2 (1 ) 1 4 2 2 2 2 − = = 2. (a) ( )2 1 , 2 3 x x = ( ) 2 2 1 1 1 3 2 6 36 = = ⎡⎣ ⎤⎦ (b) 3 1, 2 4 x x = ( 3) ( ) 1 1 1 4 2 4 8 32 = = (c) ( ) 3 2 2 , 2 4 x x x − − = ( ) ( ) ( ) ( ) 3 3 2 2 2 3 2 2 4 2 1 4 2 4 2 4 4 64 − − − − ⎡⎣ ⎤⎦ = = = 3. 4(3)x3 + 2(2)x = 12x3 + 4x = 4x(3x2 + 1) 4. 1( ) 2 3 1 2 3 2 3 3 1 2 ( 3 2 ) 2 2 2 2 2 3 x − x = x − x = x x − 1 5. 3 4 3 4 1 1 4 4 x x ⎛⎜ ⎞⎟ − = ⎝ ⎠ 6. ( )2 12 2 3 2 1 2 2 3 2 3 2 1 3 2 1 1 1 3 2 3 3 1 1 3 x x x x x x x x x − ⎛⎜ ⎞⎟ − + − = − − + − ⎝ ⎠ = − + 7. ( ) 2 23 3 2 0 3 2 0 0 3 2 0 x x x x x x x + = + = = + = → = − 8. ( ) ( )( ) 3 2 0 1 0 1 1 0 0 1 0 1 1 0 1 x x x x x x x x x x x x − = − = + − = = + = → = − − = → = 9. ( )( ) 2 8 20 0 10 2 0 10 0 10 2 0 2 x x x x x x x x + − = + − = + = → = − − = → = 10. ( )( ) 2 10 24 0 12 2 0 12 0 12 2 0 2 x x x x x x x x − − = − + = − = → = + = → = − Section 2.2 Some Rules for Differentiation 83 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 10. ( ) ( ) 3 3 gx x g x = ′ = 11. 3 2 8 3 y x y x = − = − 12. 2 6 2 y t y t = − ′ = 13. ( ) ( ) 4 2 3 8 3 fx x x f x x = − ′ = − 14. ( ) ( ) 2 3 2 4 2 12 gx x x gx x x = + ′ = + 15. ( ) ( ) 3 2 2 4 6 2 f t t t f t t = − + − ′ = − + 16. 3 2 2 9 2 3 18 y x x y x x = − + ′ = − 17. ( ) ( ) 3 2 2 4 3 2 s t t t s t t = − + ′ = − 18. 3 2 2 2 3 1 6 2 3 y x x x y x x = − + − ′ = − + 19. ( ) ( ) 2 3 1 3 1 3 2 2 3 3 g x x g x x x − = ′ = = 20. ( ) ( ) 5 2 5 3 2 2 h x x h x x = ′ = 21. ( ) 4 3 4 1 3 16 1 3 3 3 4 4 y t y t t = ′ = = 22. ( ) ( ) 1 2 1 2 1 2 10 5 5 5 f x x f x x x x − = ′ = = = 23. 2 2 3 1 3 4 2 8 4 8 4 y x x y x x x x − − = + ′ = − + = − + 24. ( ) ( ) ( ) 1 2 2 4 1 4 1 4 s t t s t t t − − = + ′ = − = − Function Rewrite Differentiate Simplify 25. 4 2 7 y x = 2 4 7 y = x− 8 5 7 y x− ′ − = 5 8 7 y x ′ = − 26. 2 2 3 y x = 2 2 3 y = x− 4 3 3 y′ = − x− 3 4 3 y x ′ = − 27. ( )3 1 4 y x = 1 3 64 y = x− 3 4 64 y′ = − x− 4 3 64 y x ′ = − 28. ( )2 3 y x π = 2 9 y x π − = 2 3 9 y x′ π − = − 3 2 9 y x ′ π = − 29. ( ) 5 4 2 y x − = y = 128x5 y′ = 128(5)x4 y′ = 640x4 30. 3 y 4x x− = y = 4x4 y′ = 16x3 y′ = 16x3 31. y = 6 x y = 6x1 2 (1 ) 1 2 2 y′ = 6 x− y 3 x ′ = 32. 3 4 y = x 3 1 2 4 y = x 3 1 1 2 4 2 y′ = ⎛⎜ ⎞⎟x− ⎝ ⎠ 3 8 y x ′ = 33. 5 1 5 y x = 1 1 5 5 y = x− 1 1 6 5 5 5 y′ = ⎛⎜− ⎞⎟x− ⎝ ⎠ 5 6 1 25 y x ′ = − 34. 4 3 3 2 y x = 3 3 4 2 y = x− 3 3 7 4 2 4 y′ = ⎛⎜− ⎞⎟x− ⎝ ⎠ 4 7 9 8 y x ′ = − 84 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Function Rewrite Differentiate Simplify 35. y = 3x ( )1 2 y = 3x 1(3 ) 1 2 (3) 2 y x′ = − 3 2 y x x ′ = 36. y = 3 6x2 ( )1 3( )2 3 y = 6 x 2(6)1 3( ) 1 3 3 y x′ = − 2 3 6 2 3 y x x ′ = 37. 3 2 1 2 32 y x y x = ′ = At the point ( ) 3( )1 2 3 2 2 1, 1 , y′ = 1 = = m. 38. 1 2 2 1 y x y x x − − = ′ = = − At the point ( )2 34 3, 4 , 1 16 . 4 3 9 ⎛⎜ ⎞⎟ y′ = − = − = m ⎝ ⎠ 39. ( ) ( ) 2 2 f t t f t t = ′ = At the point (4, 16), f ′(4) = 2(4) = 8 = m. 40. ( ) ( ) 1 3 4 3 4 3 1 1 3 3 f x x f x x x − − = ′ = − = − At the point ( ) ( )4 3 8, 1 , 8 1 1 . 2 3 8 48 ⎛⎜ ⎞⎟ f ′ = − = − = m ⎝ ⎠ 41. ( ) ( ) 3 2 2 2 8 4 6 16 1 f x x x x f x x x = + − − ′ = + − At the point ( ) ( ) ( )2 ( ) −1, 3 , f ′ −1 = 6 −1 + 16 −1 − 1 = −11 = m. 42. ( ) ( ) 4 3 2 3 2 3 5 6 10 12 15 12 10 fx x x x x f x x x x = − + − ′ = − + − At the point ( ) ( ) ( )3 ( )2 ( ) 1, −6 , f ′ 1 = 12 1 − 15 1 + 12 1 − 10 = −1 = m. 43. ( ) ( ) ( ) ( ) 2 3 2 1 1 1 2 2 2 1 3 2 2 1 3 2 2 1 1 2 fx x x x x f x x f = − + = − − ′ = − − ′ = − − = − 44. ( ) ( ) ( ) ( ) ( )( ) 3 5 2 75 30 3 2 30 6 5 30 65 0 fx x x x f x x f = − = − + ′ = − + ′ = − + = 45. (a) 4 2 3 2 5 3 8 10 y x x y x x = − + − ′ = − + m = y′(1) = −8 + 10 = 2 The equation of the tangent line is 0 2( 1) 2 2. y x y x − = − = − (b) and (c) 46. (a) y′ = 3x2 + 1 m = y′(−1) = 3 + 1 = 4 The equation of the tangent line is ( 2) 4 ( 1) 2 4 4 4 2. y x y x y x − − = ⎡⎣ − − ⎤⎦ + = + = + (b) and (c) −4.7 4.7 −3.1 3.1 −5 4 −5 1 Section 2.2 Some Rules for Differentiation 85 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 47. (a) ( ) ( ) 3 5 1 3 1 5 2 3 4 5 2 3 4 5 1 1 1 1 3 5 3 5 f x x x x x f x x x x x − − = + = + ′ = + = + (1) 1 1 8 3 5 15 m = f ′ = + = The equation of the tangent line is 2 8 ( 1) 15 8 22. 15 15 y x y x − = − = + (b) and (c) 48. (a) ( ) ( ) ( ) 2 3 3 2 5 3 1 2 1 3 1 2 1 1 3 3 fx x x x x f x x m f − − = − = − ′ = − − = ′ − = − = − The equation of the tangent line is 1( ) 3 1 5 3 3 2 1 . y x y x − = − + = − + (b) and (c) 49. (a) 2 3 2 3 2 3 6 9 y xx x y x y x = ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ = − ′ = ( )2 m = y′ = 9 2 = 36 The equation of the tangent line is 18 36( 2) 36 54. y x y x − = − = − (b) and (c) 50. (a) ( )2 2 2 1 4 4 1 8 4 y x y x x y x = + = + + ′ = + m = y′ = 8(0) + 4 = 4 The equation of the tangent line is 1 4( 0) 4 1. y x y x − = − = + (b) and (c) 51. ( ) ( ) 2 1 2 2 3 2 3 4 3 2 4 6 2 4 6 f x x x x f x x x x x x x − − − − = − − ′ = + + = + + 52. ( ) ( ) 2 2 3 3 4 3 4 3 3 5 2 3 6 15 2 3 6 15 f x x x x x f x x x x x x x − − − − = − − + ′ = − + − = − + − 53. ( ) ( ) 2 2 4 4 5 5 2 2 2 2 2 2 8 2 2 8 f x x x x x x x f x x x x x − − = − − = − − ′ = − + = − + 54. ( ) ( ) 2 2 1 2 2 4 1 4 2 4 2 4 1 f x x x x x x x f x x x x x − − = + + = + + ′ = + − = + − 55. ( ) ( ) 4 5 1 5 1 5 4 1 4 1 5 5 f x x x f x x x − = + ′ = + = + 56. ( ) ( ) 1 3 2 3 2 3 1 1 1 3 3 f x x f x x x − = − ′ = = 57. ( ) ( ) ( ) 2 3 2 1 3 1 f x xx x x f x x = + = + ′ = + 58. ( ) ( )( ) ( ) 2 3 2 2 2 1 3 2 3 6 2 fx x x x x x x f x x x = + + = + + ′ = + + 59. ( ) ( ) ( ) 3 2 2 2 3 3 3 3 3 3 2 4 3 2 4 3 6 2 6 2 3 2 6 2 f x x x x x x x x f x x x x x − − − + = = − + − − ′ = − = − = = −4.7 4.7 −3.1 3.1 −5 4 −2 4 5 −50 −5 50 −4 2 −1 3 86 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 60. ( ) ( ) 2 1 2 2 2 2 2 3 1 2 3 2 2 1 2 1 f x x x x x x f x x x x x − − − + = = − + ′ − = − = − = 61. ( ) ( ) 3 2 1 2 2 3 2 3 2 3 3 4 3 2 5 4 3 2 5 4 2 10 4 2 10 4 2 10 f x x x x x x x x f x x x x x x x x − − − − − + + = = − + + ′ − − = − − = − − = 62. ( ) ( ) 3 2 2 1 2 2 6 3 2 1 6 3 2 12 3 12 3 1 f x x x x x x x x f x x x x x − − − + − + = =− + − + ′ = − + − = − + − 63. ( ) 3 2 6 4 6 2 3 2 0 when 0,2 y′ = − x + x = x − x = x = ± If 6 , 2 x = ± then 4 2 6 3 6 1 9 3 3 1 5. 2 2 4 2 4 y ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = −⎜⎜± ⎟⎟ + ⎜⎜± ⎟⎟ − = − + ⎜ ⎟ − = ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ The function has horizontal tangent lines at the points (0, 1), 6 , 5 , 2 4 ⎛ ⎞ − ⎜⎜− ⎟⎟ ⎝ ⎠ and 6 , 5 . 2 4 ⎛ ⎞ ⎜⎜ ⎟⎟ ⎝ ⎠ 64. y′ = 3x2 + 6x = 3x(x + 2) = 0 when x = 0, −2. The function has horizontal tangent lines at the points (0, 0) and (−2, 4). 65. y′ = x + 5 = 0 when x = −5. The function has a horizontal tangent line at the point ( 25) 2 −5, − . 66. y′ = 2x + 2 = 0when x = −1. The function has a horizontal tangent line at the point (−1, −1). 67. (a) (b) ( ) ( ) ( ) ( ) 3 2 1 1 3 f x g x x f g ′ = ′ = ′ = ′ = (c) Tangent line to f at x = 1: ( ) ( ) 1 1 1 3 1 3 2 f y x y x = − = − = − Tangent line to g at x = 1: ( ) ( ) 1 4 4 3 1 3 1 g y x y x = − = − = + (d) f ′ and g′ are the same. 4 2 −2 −4 −4 −2 2 4 x g f y 4 2 −2 −4 −4 −2 2 4 x g f y Section 2.2 Some Rules for Differentiation 87 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 68. (a) (b) ( ) ( ) ( ) ( ) 2 1 2 6 1 6 f x x f gx x g ′ = ′ = ′ = ′ = (c) Tangent line to f at x = 1: ( ) ( ) 1 1 1 2 1 2 1 f y x y x = − = − = − Tangent line to g at x = 1: ( ) ( ) 1 3 3 6 1 6 3 g y x y x = − = − = − (d) g′ is 3 times f ′. 69. If g(x) = f (x) + 6, then g′(x) = f ′(x) because the derivative of a constant is 0, g′(x) = f ′(x). 70. If g(x) = 2 f (x), then g′(x) = 2 f ′(x) because of the Constant Multiple Rule. 71. If g(x) = −5 f (x), then g′(x) = −5 f ′(x) because of the Constant Multiple Rule. 72. If g(x) = 3 f (x) − 1, then g′(x) = 3 f ′(x) because of the Constant Multiple Rule and the derivative of a constant is 0. 73. (a) 2005: m ≈ 119.2; 2007: m ≈ 161 (b) These results are close to the estimates in Exercise 13 in Section 2.1. (c) The slope of the graph at time t is the rate at which sales are increasing in millions of dollars per year. 74. (a) 2006: m ≈ 259.1; 2008: m ≈ 93.4 (b) These results are close to the estimates in Exercise 14 in Section 2.1. (c) The slope of the graph at time t is the rate at which sales are increasing in millions of dollars. 75. (a) More men and women seem to suffer from migraines between 30 and 40 years old. More females than males suffer from migraines. Fewer people whose income is greater than or equal to $30,000 suffer from migraines than people whose income is less than $10,000. (b) The derivatives are positive up to approximately 37 years old and negative after about 37 years of age. The percent of adults suffering from migraines increases up to about 37 years old, then decreases. The units of the derivative are percent of adults suffering from migraines per year. 76. (a) The attendance rate for football games, g′(t), is greater at game 1. (b) The attendance rate for basketball games, f ′(t), is greater than the rate for football games, g′(t), at game 3. (c) The attendance rate for basketball games, f ′(t), is greater than the rate for football games, g′(t), at game 4. In addition, the attendance rate for football games is decreasing at game 4. (d) At game 5, the attendance rate for football continues to increase, while the attendance rate for basketball continues to decrease. 77. C = 7.75x + 500 C′ = 7.75, which equals the variable cost. 78. ( ) 150 7000 500 150 7000 C x P R C P x x = + = − = − + P = 350x − 7000 P′ = 350, which equals the profit on each dinner sold. 79. ( ) ( ) 3 2 2 4.1 12 2.5 12.3 24 2.5 fx x x x f x x x = − + ′ = − + f has horizontal tangents at (0.110, 0.135) and (1.841, −10.486). x y −3 −2 −1 1 2 3 −1 2 3 4 5 g f x y −3 −2 −1 1 2 3 −1 2 3 4 5 g f 0 −12 3 12 f ′ f 88 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 80. ( ) ( ) 3 2 2 1.4 0.96 1.44 3 2.8 0.96 f x x x x f x x x = − − + ′ = − − f has horizontal tangents at (1.2, 0) and (−0.267, 1.577). 81. False. Let f (x) = x and g(x) = x + 1. Then f ′(x) = g′(x) = 1, but f (x) ≠ g(x). 82. True. c is a constant. Section 2.3 Rates of Change: Velocity and Marginals 1. (a) 1980–1985: 115 63 $10.4 billion yr 5 0 − = − (b) 1985–1990: 152 115 $7.4 billion yr 10 5 − = − (c) 1990–1995: 184 152 $6.4 billion yr 15 10 − = − (d) 1995–2000: 267 184 $16.6 billion yr 20 15 − = − (e) 2000–2005: 322 267 $11.0 billion yr 25 20 − = − (f ) 1980–2008: 398 63 $11.96 billion yr 28 0 − = − (g) 1990–2008: 398 152 $13.67 billion yr 28 10 − = − (h) 2000–2008: 398 267 $16.38 billion yr 28 20 − = − Skills Warm Up 1. 63 ( 105) 42 21 7 14 3 − − − = − = 2. 37 54 91 16 3 13 7 − − − = − = − 3. 24 33 9 3 9 6 3 − − = = − − 4. 40 16 24 12 18 8 10 5 − = = − 5. 4 2 2 7 8 2 y x x y x = − + ′ = − 6. 3 2 2 3 2 8 9 4 y t t y t t = − + − ′ = − + 7. 16 2 24 30 32 24 s t t s t = − + + ′ = − + 8. 16 2 54 70 32 54 y x x y x = − + + ′ = − + 9. ( ) ( ) 3 2 2 2 1 10 1 10 3 3 1 5 5 2 2 3 5 6 6 5 A r r r A r r A r r = − + + ′ = − + + ′ = − + + 10. ( ) ( ) 3 2 2 2 1 9 1 9 6 18 63 15 18 36 63 2 4 7 y x x x y x x y x x = − + − ′= − + ′ = − + 11. 2 12 5000 12 2 5000 12 2500 y x x y x y x = − ′ = − ′ = − 12. 3 2 138 74 10,000 74 3 10,000 y x x y x = + − ′ = − f ′ f −2 2 −5 5 Section 2.3 Rates of Change: Velocity and Marginals 89 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2. (a) Imports: 1980–1990: 495 245 $25 billion yr 10 0 − = − (b) Exports: 1980–1990: 394 226 $16.8 billion yr 10 0 − = − (c) Imports: 1990–2000: 1218 495 $72.3 billion yr 20 10 − = − (d) Exports: 1990–2000: 782 394 $38.8 billion yr 20 10 − = − (e) Imports: 2000–2009: 1560 1218 $38.0 billion yr 29 20 − = − (f ) Exports: 2000–2009: 1056 782 $30.4 billion yr 29 20 − = − (g) Imports: 1980–2009: 1560 245 $45.3 billion yr 29 0 − = − (h) Exports: 1980–2009: 1056 226 $28.6 billion yr 29 0 − = − 3. Average rate of change: (2) (1) 11 8 3 2 1 1 y f f t Δ − − = = = Δ − f ′(t) = 3 Instantaneous rates of change: f ′(1) = 3, f ′(2) = 3 4. Average rate of change: (2) (0) 0 2 1 2 0 2 y f f x Δ − − = = = − Δ − h′(x) = −1 Instantaneous rates of change: f ′(0) = −1, f ′(2) = −1 5. Average rate of change: ( ) ( ) ( ) 2 2 2 14 4 2 2 4 y h h x Δ − − − − = = = − Δ − − h′(x) = 2x − 4 Instantaneous rates of change: h′(−2) = −8, h′(2) = 0 6. Average rate of change: ( ) ( ) ( ) 3 1 10 6 4 3 1 4 y f f x Δ − − − − = = = − Δ − − f ′(x) = 2x − 6 Instantaneous rates of change: f ′(−1) = −8, f ′(3) = 0 7. Average rate of change: (8) (1) 48 3 45 8 1 7 7 y f f x Δ − − = = = Δ − f ′(x) = 4x1 3 Instantaneous rates of change: f '(1) = 4, f ′(8) = 8 8. Average rate of change: (4) (1) 8 1 7 4 1 3 3 y f f x Δ − − = = = Δ − ( ) 3 1 2 2 f ′ x = x Instantaneous rates of change: (1) 3, 2 f ′ = f ′(4) = 3 −10 11 −2 12 (1, 8) (2, 11) −2 4 −1 3 (0, 2) (2, 0) −14 16 −4 16 (−2, 14) (2, −2) −10 17 −11 7 (3, −10) (−1, 6) 0 10 0 54 (1, 3) (8, 48) −5 10 −1 9 (4, 8) (1, 1) 90 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 9. Average rate of change: (4) (1) (1 4) 1 1 4 1 3 4 y f f x Δ − − = = = − Δ − ( ) 2 f x 1 x ′ = − Instantaneous rates of change: f ′(1) = −1, (4) 1 16 f ′ = − 10. Average rate of change: (4) (1) 1 2 1 1 4 1 3 6 y f f x Δ − − = = = − Δ − ( ) 3 2 1 2 f x x ′ = − Instantaneous rates of change: (4) 1 , 16 f ′ = − (1) 1 2 f ′ = − 11. Average rate of change: (3) (1) 74 2 36 3 1 2 y g g x Δ − − = = = Δ − g′(x) = 4x3 − 2x Instantaneous rates of change: g′(1) = 2, g′(3) = 102 12. Average rate of change: ( ) ( ) ( ) ( ) 1 1 0 2 1 1 1 2 y g g x Δ − − − − = = = Δ − − g′(x) = 3x2 Instantaneous rates of change: g′(−1) = 3, g′(1) = 3 13. (a) 0 1400 467 3 − ≈ ≈ − The number of visitors to the park is decreasing at an average rate of 467 people per month from September to December. (b) Answers will vary. Sample answer: [4, 11] Both the instantaneous rate of change at t = 8 and the average rate of change on [4, 11] are about zero. 14. (a) The average rate of change is the greatest over [5, 6]. (b) Answers will vary. Sample answer: [2, 5] Both the instantaneous rate of change at t = 4 and the average rate of change on [2, 5] is about zero. 15. s = −16t2 + 30t + 250 Instantaneous: v(t) = s′(t) = −32t + 30 (a) Average: [0, 1]: (1) (0) 264 250 14 ft sec 1 0 1 s − s − = = − v(0) = s′(0) = 30 ft sec v(1) = s′(1) = −2 ft sec (b) Average: [1, 2]: (2) (1) 246 264 18 ft sec 2 1 1 s − s − = = − − v(1) = s′(1) = −2 ft sec v(2) = s′(2) = −34 ft sec (c) Average: [2, 3]: (3) (2) 196 246 50 ft sec 3 2 1 s − s − = = − − v(2) = s′(2) = −34 ft sec v(3) = s′(3) = −66 ft sec (d) Average: [3, 4]: (4) (3) 114 196 82 ft sec 4 3 1 s − s − = = − − v(3) = s′(3) = −66 ft sec v(4) = s′(4) = −98 ft sec 0 6 0 4 (1, 1) ) ) 4 4, 1 −1 5 −1 3 4, 1 2 ( ( (1, 1) 0 −10 4 90 (3, 74) (1, 2) −6 −4 6 4 (1, 0) (−1, −2) Section 2.3 Rates of Change: Velocity and Marginals 91 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 16. (a) ( ) 33 10 1 1 2 1 33 5 1 2 H v v v ′ = ⎡⎢ ⎛⎜ − ⎞⎟ − ⎤⎥ = ⎡⎢ − ⎤⎥ ⎣ ⎝ ⎠ ⎦ ⎣ ⎦ Rate of change of heat loss with respect to velocity. (b) ( ) 2 3 3 3 2 33 5 1 2 83.673 kcal m hr m sec 83.673 kcal sec m hr 83.673 kcal 1 m 3600 0.023 kcal m H ′ = ⎡ − ⎤ ⎢⎣ ⎥⎦ ≈ = ⋅ = ⋅ = ( ) 2 3 3 3 5 33 5 1 5 40.790 kcal m hr m sec 40.790 kcal sec m hr 40.790 kcal 1 m 3600 0.11 kcal m H ′ = ⎡ − ⎤ ⎢⎣ ⎥⎦ ≈ = ⋅ = ⋅ = 17. s = −16t2 + 555 (a) Average velocity (3) (2) 3 2 411 491 1 80 ft sec s − s = − − = = − (b) v = s′(t) = −32t, v(2) = −64 ft sec, v(3) = −96 ft sec (c) 2 2 2 16 555 0 16 555 555 16 5.89 seconds s t t t t = − + = = = ≈ (d) v(5.89) ≈ −188.5 ft sec 18. (a) ( ) ( ) ( ) 16 2 18 210 32 18 s t t t v t s t t = − − + = ′ = − − (b) [1, 2]: (2) (1) 110 176 66 ft sec 2 1 1 s − s − = = − − (c) v(1) = −50 ft sec v(2) = −82 ft sec (d) Set s(t) = 0. ( ) ( ) ( )( ) ( ) 2 2 16 18 210 0 18 18 4 16 210 18 13,764 3.10 sec 2 16 32 t t t − − + = − ± − − − ± = − = ≈ − − (e) v(3.10) = −117.2 ft sec 19. 205,000 9800 9800 C x dC dx = + = 20. 3 2 150,000 7 21 C x dC x dx = + = 21. 55,000 470 0.25 2, 0 940 470 0.5 C x x x dC x dx = + − ≤ ≤ = − 22. ( ) 1 2 100 9 3 100 0 3 1 150 2 C x dC x dx x − = + ⎡ ⎛ ⎞⎤ = ⎢ + ⎜ ⎟⎥ = ⎣ ⎝ ⎠⎦ 23. 50 0.5 2 50 R x x dR x dx = − = − 24. 30 2 30 2 R x x dR x dx = − = − 92 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 25. 3 2 2 6 8 200 18 16 200 R x x x dR x x dx = − + + = − + + 26. ( 3 2 ) 1 2 50 20 50 20 3 1000 75 2 R x x dR x x dx = − = ⎡ − ⎤ = − ⎢⎣ ⎥⎦ 27. 2 2 72 145 4 72 P x x dP x dx = − + − = − + 28. 0.25 2 2000 1,250,000 0.5 2000 P x x dP x dx = − + − = − + 29. 3 2 0.0013 12 0.0039 12 P x x dP x dx = + = + 30. 3 2 2 0.5 30 164.25 1000 1.5 60 164.25 P x x x dP x x dx = − + − − = − + − 31. C = 3.6 x + 500 (a) C(10) − C(9) ≈ $0.584 (b) ( ) ( ) 1.8 9 $0.60 per unit. C x x C ′ = ′ = (c) The answers are close. 32. R = 2x(900 + 32x − x2 ) (a) (15) (14) 2(15) 900 32(15) 152 2(14) 900 32(14) 142 34,650 32,256 $2394 R − R = ⎡⎣ + − ⎤⎦ − ⎡⎣ + − ⎤⎦ = − = (b) ( ) ( ) 2 3 2 1800 64 2 1800 128 6 14 $2416 R x x x Rx x x R = + − ′ = + − ′ = (c) The answers are close. 33. P = −0.04x2 + 25x − 1500 (a) (151) (150) 1362.96 1350 $12.96 151 150 1 P P P x Δ − − = = = Δ − (b) ( ) ( ) 0.08 25 150 $13 dP x P x dx P = − + = ′ ′ = (c) The results are close. 34. 2 36,000 2048 1 , 150 275 8 P x x x = + − ≤ ≤ 1 2 ( 3) 3 2048 1 1 2 2 8 1024 1 4 dP x x dx x x = ⎛⎜ − ⎞⎟ − − − ⎝ ⎠ = + (a) When x = 150, dP $83.61. dx ≈ (b) When x = 175, dP $77.41. dx ≈ (c) When x = 200, dP $72.41. dx ≈ (d) When x = 225, dP $68.27. dx ≈ (e) When x = 250, dP $64.76. dx ≈ (f ) When x = 275, dP $61.75. dx ≈ Section 2.3 Rates of Change: Velocity and Marginals 93 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 35. P = −15.56t2 + 802.1t + 117,001 (a) P(0) = 117,001.0 thousand people P(5) = 120,622.5 thousand people P(10) = 123,466.0 thousand people P(15) = 125,531.5 thousand people P(20) = 126,819.0 thousand people P(25) = 127,328.5 thousand people P(30) = 127,060.0 thousand people The population is growing from 1980 to 2005. It then begins to decline. (b) dP 31.12t 802.1 P (t) dt = − + = ′ (c) P′(0) = 802.1 thousand people per year P′(5) = 646.5 thousand people per year P′(10) = 490.9 thousand people per year P′(15) = 335.3 thousand people per year P′(20) = 179.7 thousand people per year P′(25) = 24.1 thousand people per year P′(30) = −131.5 thousand people per year The rate of growth is decreasing. 36. (a) (b) For t < 4, the slopes are positive, and the fever is increasing. For t > 4, the slopes are negative, and the fever is decreasing. (c) T(0) = 100.4°F T′(4) = 101°F T(8) = 100.4°F T(12) = 98.6°F (d) dT 0.075t 0.3; dt = − + the rate of change of temperature with respect to time (e) T′(0) = 0.3°F per hour T′(4) = 0°F per hour T′(8) = −0.3°F per hour T(12) = −0.6°F per hour 37. (a) TR = −10Q2 + 160Q (b) (TR)′ = MR = −20Q + 160 (c) 38. (a) R = xp = x(5 − 0.001x) = 5x − 0.001x2 (b) P = ( 2 ) ( ) 2 5 0.001 35 1.5 0.001 3.5 35 R C x x x x x − = − − + = − + − (c) 5 0.002 3.5 0.002 dR x dx dP x dx = − = − 39. (a) (400, 1.75), (500, 1.50) Slope 1.50 1.75 0.0025 500 400 − = = − − 1.75 0.0025( 400) 0.0025 2.75 p x p x − = − − = − + P = ( ) ( ) 2 0.0025 2.75 0.1 25 0.0025 2.65 25 R C xp c x x x x x − = − = − + − + = − + − (b) At x = 300, P has a positive slope. At x = 700, P has a negative slope. (c) P′(x) = −0.005x + 2.65 P′(300) = $1.15 per unit P′(700) = −$.85 per unit x 600 1200 1800 2400 3000 dR dx 3.8 2.6 1.4 0.2 −1.0 dP dx 2.3 1.1 −0.1 −1.3 −2.5 P 1705 2725 3025 2605 1465 Q 0 2 4 6 8 10 Model 160 120 80 40 0 −40 Table − 130 90 50 10 −30 15 98 0 103 1200 0 0 800 94 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 40. (36,000, 6), (33,000, 7) ( ) Slope 7 6 1 33,000 36,000 3000 6 1 36,000 3000 1 18 (demand function) 3000 p x p x − = =− − − = − − = − + (a) ( ) 1 2 18 0.2 85,000 17.8 85,000 3000 3000 P = R − C = xp − C = x⎛− x + ⎞ − x − = x + x − ⎜ ⎟ ⎝ ⎠ (b) (c) ( ) 17.8 1500 P x x ′ − = + At x = 18,000, P has a positive slope. P′(18,000) = $5.80 per ticket At x = 36,000, P has a negative slope. P′(36,000) = −$6.20 per ticket 41. (a) ( ) 15,000 mi 1 gal 2.95 dollars yr mi 1 gal C x x ⎛ ⎞⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ ( ) 44,250 dollars yr C x x = (b) 2 dollars 44,250 yr mpg dC dx x = − The marginal cost is the change of savings for a 1-mile per gallon increase in fuel efficiency. (c) (d) The driver who gets 15 miles per gallon would benefit more than the driver who gets 35 miles per gallon. The value of dC dx is a greater savings for x = 15 than for x = 35. 42. (a) f ′(2.959) is the rate of change of a gallon of gasoline when the price is $2.959 gallon. (b) In general, it should be negative. Demand tends to decrease as price increases. Answers will vary. 43. (a) Average rate of change from 1995 to 2009: 10,428.05 5117.12 379.35 dollars per year 19 5 p t Δ − = ≈ Δ − (b) Average rate of change from 1996 to 2000: 10,786.85 6448.26 $1084.65 10 6 p t Δ − = ≈ Δ − So, the instantaneous rate of change for 1998 is p′(8) ≈ $1084.65 yr. (c) Average rate of change from 1997 to1999: 11,497.12 7908.24 $1794.44 9 7 p t Δ − = ≈ Δ − So, the instantaneous rate of change for 1998 is p′(8) ≈ $1794.44 yr. (d) The average rate of change from 1997 to 1999 is a better estimate because the data is closer to the year in question. 44. Answers will vary. Sample answer: The rate of growth in the lag phase is relatively slow when compared with the rapid growth in the acceleration phase. The population grows slower in the deceleration phase, and there is no growth at equilibrium. These changes could be explained by food supply or seasonal growth. x 10 15 20 25 30 35 40 C 4425.00 2950.00 2212.50 1770.00 1475.00 1264.29 1106.25 dC dx −442.5 −196.67 −110.63 −70.80 −49.17 −36.12 −27.66 60,000 0 0 200,000 Section 2.4 The Product and Quotient Rules 95 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 2.4 The Product and Quotient Rules Skills Warm Up 1. ( )( ) ( )( ) ( ) 2 2 2 2 2 1 2 2 7 2 2 2 4 14 6 14 2 2 3 7 1 x x x x x x x x x x + + + = + + + = + + = + + 2. ( )( ) ( )( ) ( ) 3 2 2 2 4 2 4 2 4 2 2 2 8 4 2 3 16 8 8 12 24 20 4 6 5 x x x x x x x x x x x x x − + − = − + − = − = − 3. x(4)(x2 + 2)3(2x) + (x2 + 4)(1) = 8x2 (x2 + 2)3(x2 + 4) 4. ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 2 4 3 2 2 1 2 2 1 2 4 2 1 2 2 1 2 2 1 2 2 1 x x x x x x x x x x x x + + + = + + + = + ⎡ + + ⎤ ⎣ ⎦ 5. ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 7 5 5 6 2 10 35 10 12 2 7 2 7 23 2 7 x x x x x x x + − + + − − = + + = + 6. ( )( ) ( )( ) ( ) ( ) ( ) 2 2 3 2 3 2 2 2 2 2 2 2 2 4 2 1 2 2 8 4 2 2 4 4 8 4 4 x x x x xx x x x x x x x x x − + − + + − − − − = − − − − − = − 7. ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 1 2 2 2 4 2 1 1 2 2 2 1 2 1 1 x x x x x x x x x x x x x x + − + + − − = + + − − + = + − + − = + 8. ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 4 3 4 4 3 4 2 4 2 4 3 4 2 4 3 4 2 1 4 4 1 4 4 4 16 4 1 1 12 4 4 1 4 3 1 1 x x x x x x x x x x x x x x − − − − − + − = − − − + = − − + = − 96 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. f (x) = (2x − 3)(1 − 5x) ( ) (2 3)( 5) (1 5 )(2) 10 15 2 10 20 17 f x x x x x x ′ = − − + − = − + + − = − + 2. g(x) = (x − 4)(x + 2) ( ) ( 4)(1) ( 2)(1) 4 2 2 2 g x x x x x x ′ = − + + = − + + = − 3. f (x) = (6x − x2 )(4 + 3x) ( ) ( 2 )( ) ( )( ) 2 2 6 3 4 3 6 2 18 3 24 8 18 6 9 28 24 f x x x x x x x x x x x2 x ′ = − + + − = − + − + − = − + + Skills Warm Up —continued— 9. ( 1 )( ) ( )( 2 ) 1 ( 1 2 ) 2 2 3 2 2 2 2 3 1 2 2 2 2 3 3 4 3 3 4 3 3 4 3 3 x x x x x x x x x x x x x x x x − − − − − − + + − − + = + + − + + − = + − = + − − + = 10. ( )( ) ( )( ) ( ) ( ) 1 2 1 1 2 2 1 2 1 2 2 2 2 2 2 1 1 4 1 4 1 1 2 2 4 2 1 2 4 1 x x x x x x x x x x x x x x x x x x − − − − − − − − − − − ⎛ − − + ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟ − ⎝ − + ⎠⎝ ⎠ − + = − + − + = − 11. ( ) ( ) ( ) ( ) 3 2 4 6 1 2 62 1 12 1 11 f x x x f x x f = − + ′ = − ′ = − = − = 12. ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 2 8 3 2 8 2 32 22 8 3 4 4 8 0 f x x x x f x x x f = − + + ′ = − + + ′ = − + + = − + + = 13. ( ) ( ) ( ) 2 2 1 1 2 1 2 1 4 f x x f x x f = ′ = − ′ = − = − 14. ( ) ( ) ( ) ( ) 2 2 3 3 1 2 2 2 2 2 2 2 4 2 8 4 1 4 17 4 f x x x f x x x f = − ′ = + ′ = + = + = + = Section 2.4 The Product and Quotient Rules 97 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4. f (x) = (x2 + 1)(2x + 5) ( ) ( 2 )( ) ( )( ) 2 2 2 1 2 2 5 2 2 2 4 10 6 10 2 f x x x x x x x x x ′ = + + + = + + + = + + 5. f (x) = x(x2 + 3) ( ) ( ) ( 2 )( ) 2 2 2 2 31 2 3 3 3 f x x x x x x x ′ = + + = + + = + 6. f (x) = x2 (3x3 − 1) ( ) 2 ( 2 ) ( 3 )( ) 4 4 4 9 3 12 9 6 2 15 2 f x x x x x x x x x x ′ = + − = + − = − 7. h(x) 2 3 (x2 7) (2x 1 3)(x2 7) x = ⎛⎜ − ⎞⎟ + = − − + ⎝ ⎠ ( ) ( 1 )( ) ( 2 )( 2 ) 2 2 2 32 7 2 4 6 2 14 6 2 14 h x x x x x x x x x − − − ′ = − + + − = − − − = − + − 8. ( ) ( ) ( )( 2 ) 2 f x 3 x 4 5 3 x 4x 5 x = − ⎛⎜ − ⎞⎟ = − − − ⎝ ⎠ ( ) ( )( 3) ( 2 )( ) 3 2 2 3 2 3 8 4 5 1 24 8 4 5 24 4 5 f x x x x x x x x x − − − − − ′ = − − + − − = − + − + = − + + 9. g(x) = (x2 − 4x + 3)(x − 2) ( ) ( 2 )( ) ( )( ) 2 2 2 4 31 2 2 4 4 3 2 4 4 8 3 12 11 g x x x x x x x x x x x x ′ = − + + − − = − + + − − + = − + 10. g(x) = (x2 − 2x + 1)(x3 − 1) ( ) ( 2 )( 2 ) ( 3 )( ) 4 3 2 4 3 4 3 2 2 13 1 2 2 3 6 3 2 2 2 2 5 8 3 2 2 g x x x x x x x x x x x x x x x x ′ = − + + − − = − + + − − + = − + − + 11. ( ) 5 h x x x = − ( ) ( )( ) ( ) ( )2 ( )2 ( )2 5 1 1 5 5 5 5 5 x x x x h x x x x − − − − ′ = = = − − − − 12. ( ) 2 3 h x x x = + ( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 3 2 1 3 2 6 3 6 3 x x x h x x x x x x x x x + − ′ = + + − = + + = + 13. ( ) 2 2 3 3 1 f t t t − = + ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 3 1 4 2 3 3 3 1 12 4 6 9 3 1 6 4 9 3 1 t t t f t t t t t t t t t + − − ′ = + + − + = + + + = + 14. ( ) 1 1 f x x x + = − ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 1 1 1 1 1 1 1 1 2 1 x x f x x x x x x − − + ′ = − − − − = − = − − 15. ( ) 2 1 4 f t t t − = + ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 4 2 1 1 4 2 8 1 4 8 1 4 t t t f t t t t t t t t t + − − ′ = + + − + = + + + = + 16. ( ) 2 4 5 1 g x x x − = − ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 1 4 4 5 2 1 4 4 8 10 1 4 10 4 1 x x x g x x x x x x x x x − − − ′ = − − − + = − − + − = − 98 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 17. ( ) 2 6 5 2 1 f x x x x + + = − ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 2 6 6 5 2 2 1 4 12 2 6 2 12 10 2 1 2 2 16 2 1 x x x x f x x x x x x x x x x x − + − + + ′ = − + − − − − − = − − − = − 18. ( ) 4 2 1 2 f x x x x − + = + ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 8 1 4 1 1 2 8 16 2 4 1 2 4 16 3 2 x x x x f x x x x x x x x x x x + − − − + ′ = + − + − − + − = + + − = + 19. ( ) 6 2 1 3 1 f x x x + − = − ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 1 2 1 2 2 1 2 2 2 2 2 2 3 1 2 6 2 3 3 1 6 2 18 6 3 1 2 12 18 3 1 2 12 18 2 12 18 3 1 3 1 x x x f x x x x x x x x x x x x x x x x − − − − − − − − − − + ′ = − − + − − = − − − = − − − − − = = − − 20. ( ) 5 2 2 f x x x − − = + ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) 3 2 2 3 2 2 3 2 2 3 2 2 3 3 2 2 2 5 1 2 2 4 5 2 4 3 5 2 4 3 5 2 4 2 5 2 x x x f x x x x x x x x x x x x x x x x − − − − − − − + − − ′ = + + − + = + + − = + + − = + + − = + Function Rewrite Differentiate Simplify 21. ( ) 3 6 3 f x x x + = ( ) 1 3 2 3 f x = x + x f ′(x) = x2 + 2 f ′(x) = x2 + 2 22. ( ) 3 2 7 f x = x ( ) 3 2 7 f x == x ( ) 6 7 f ′ x = x ( ) 6 7 f ′ x = x 23. 2 2 3 y x x + = 1( 2 2 ) 3 y = x + x ' 1(2 2) 3 y = x + 2( 1) 3 y′ = x + 24. 4x3 2 y x = y = 4x1 2 , x ≠ 0 y′ = 2x−1 2 y 2 x ′ = 25. 3 7 3 y x = 7 3 3 y = x− y' = −7x−4 4 y 7 x ′ = − 26. 2 4 5 y x = 4 2 5 y = x− 8 3 5 y′ = − x− 3 8 5 y x ′ = − 27. 4 2 3 8 y x x x − = 1 3 2 3 1 2 , 0 2 8 y = x − x x ≠ 3 1 2 3 1 2 4 16 y′ = x − x− 3 3 4 16 y x x ′ = − 28. 5(3 2 5 ) 8 x x y x + = 15 25, 0 8 8 y = x + x ≠ 15, 0 8 y′ = x ≠ 15, 0 8 y′ = x ≠ Section 2.4 The Product and Quotient Rules 99 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Function Rewrite Differentiate Simplify 29. ( ) 2 4 3 2 1 y x x x − + = − 1( 3), 1 2 y = x − x ≠ 1(1), 1 2 y′ = x ≠ 1, 1 2 y′ = x ≠ 30. ( ) 2 4 4 2 y x x − = + 1( 2), 2 4 y = x − x ≠ − 1(1), 2 4 y′ = x ≠ − 1, 2 4 y′ = x ≠ − 31. ( ) ( 3 )( ) ( 2 )( 2 ) 4 3 2 4 3 2 4 3 2 3 4 3 3 3 2 3 5 4 3 12 9 6 9 9 9 15 10 12 3 18 15 f x x x x x x x x x x x x x x x x x x x ′ = − + + − + + = + − − + + + − − = + − − − Product Rule 32. ( ) ( 5 )( ) ( 4 )( 2 ) 6 5 6 5 4 6 5 4 1 8 7 5 4 7 3 8 7 8 7 20 35 15 28 42 15 8 7 h t t t t t t t t t t t t t t t t ′ = − − + − − = − − + + − − = − − − + Product Rule 33. ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 3 2 2 4 4 2 2 2 4 2 2 2 1 3 3 3 2 2 1 3 3 2 6 4 1 6 4 3 1 x x x x x f x x x x x x x x x x x − + − + + ′ = − − − − − = − − − − = − Quotient Rule 34. ( ) ( )( ) ( )( ) ( ) ( ) ( )( ) 5 2 5 2 2 4 5 3 2 2 2 2 2 2 3 1 3 5 3 3 2 5 3 2 6 3 3 fx x x x x x x f x x x x x x x x x x x x − − − − − = − = − ′ = − + − − = − − + = + Product Rule 35. ( ) ( )( ) 2 20 5 4 5, 4 4 4 x x x x f x x x x x − − − + = = = − ≠ − + + f ′(x) = 1, x ≠ −4 Simple Power Rule 36. ( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 2 2 1 , 2 5 6 2 3 3 3 0 1 1 , 2 3 1 , 2 3 h t t t t t t t t t t h t t t t t + + = = = ≠ − + + + + + + − ′ = ≠− + = − ≠ − + Quotient Rule 37. ( ) ( ) ( )( ) ( ) ( )( ) ( )( ) ( ) 3 2 3 3 3 2 3 2 2 3 2 1 2 12 1 2 16 2 16 12 2 1 g t t t t gt t t t t t t = − = − − ′ = − + − = − Product Rule 38. ( ) ( ) 2 3 1 3 1 3 2 3 1 3 1 3 2 3 1 1 3 1 1 3 3 4 1 3 3 f x x x x x x x x x − − ′ = + ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ = + + = + Product Rule 39. ( ) ( ) ( ) ( )( ) 2 2 1 2 1 2 2 1 2 3 2 1 2 3 2 1 2 1 2 2 1 2 1 2 3 2 3 2 1 2 1 5 2 2 2 5 2 5 2 2 2 5 2 2 3 5 3 2 5 2 2 2 g s s s s s s s s s s s s g s s s s s s s s s s s s s s − − − − − + − + = = − − − + ′ = − − + − = − − = − − = Quotient Rule 40. ( ) ( ) ( )( ) 1 2 1 2 1 2 1 2 1 2 1 2 1 2 3 2 1 2 1 1 2 2 1 1 1 1 1 2 2 f x x x x x x x x f x x x x x x x x − − − − + + = = − + ′ = − − = = − Quotient Rule 100 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 41. ( ) ( 2)(3 1) 3 2 5 2 4 2 4 2 x x x x f x x x − + − − = = + + ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 4 2 6 5 3 5 2 4 4 2 24 8 10 12 20 8 4 2 12 12 2 4 2 1 2 6 6 1 2 2 1 6 6 1 2 2 1 x x x x f x x x x x x x x x x x x x x x x + − − − − ′ = + − − − + + = + + − = + + − = + + − = + Quotient Rule 42. ( ) ( )( ) 2 1 2 7 2 5 7 2 1 2 1 x x x x f x x x + − − − = = + + ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 4 5 2 5 7 2 2 1 8 6 5 4 10 14 2 1 4 4 9 2 1 x x x x f x x x x x x x x x x + − − − − ′ = + − − − + + = + + + = + Quotient Rule 43. ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) 3 2 2 2 3 2 2 3 2 3 2 2 3 2 2 3 2 1 5 3 4 4 4 3 2 5 5 3 1 4 3 10 13 20 5 3 4 2 11 8 17 4 g x x x x x x x x x x x x x x x g x x x x x x x x x x x x x − − − − = + + = + + + − − − − − − ′ = + + − − − + + + = + + − − = + Quotient Rule 44. ( ) ( )( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( ) 3 3 2 3 2 2 4 3 2 4 3 2 4 3 2 3 4 5 1 3 4 4 5 3 4 2 4 4 59 4 6 12 8 16 9 36 41 16 20 15 48 33 32 20 f x x x x x x x x x f x x x x x x x x x x x x x x x x x x x = + − + = + − − ′ = + − + − − + = − + − + − − − − = − − − − Product Rule 45. f (x) = (5x + 2)(x2 + x) ( ) ( )( ) ( 2 )( ) 2 2 2 5 2 2 1 5 10 9 2 5 5 15 14 2 f x x x x x x x x x x x ′ = + + + + = + + + + = + + m = f ′(−1) = 3 0 3( ( 1)) 3 3 y x y x − = − − = + 46. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 4 2 3 3 1 2 1 4 4 2 4 2 4 2 24 h x x x x h x x x h = − = − + ′ = − ′ − = − − − = − 9 24( 2) 24 39 y x y x − = − + = − − 47. f (x) = x3(x2 − 4) ( ) 2 ( ) ( 2 )( 2 ) 4 4 2 4 2 2 43 2 3 12 5 12 f x x x x x x x x x x ′ = + − = + − = − m = f ′(1) = −7 ( 3) 7( 1) 7 4 y x y x − − = − − = − + 48. f (x) = x(x − 3) = x1 2 (x − 3) ( ) 1 2 ( ) ( )( 1 2 ) 1 2 1 2 1 2 1 2 1 2 1 2 1 3 2 2 3 3 2 2 f x x 1 x 3 x x x x x x − − − ′ = + − = + − = − ( ) 3( )1 2 3( ) 1 2 9 1 2 2 2 2 m f 9 9 9 4 = ′ = − − = − = 18 4( 9) 4 18 y x y x − = − = − 3 −5 −5 12 5 −10 −5 (−1, 0) 10 0 20 0 10 (9, 18) 5 −10 −5 (1, −3) 10 Section 2.4 The Product and Quotient Rules 101 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 49. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 3 1 1 1 3 4 x x f x x x f + − − ′ = = + + ′ = 1 3( 1) 2 4 3 5 4 4 y x y x + = − = − 50. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 3 1 1 2 3 x x f x x x f − − + − ′ = = − − ′ = − 5 3( 2) 3 11 y x y x − = − − = − + 51. ( ) ( )( ) 2 3 2 6 5 18 3 10 2 3 2 3 x x x x f x x x − + + − = = − − ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 3 36 3 18 3 10 2 2 3 72 102 9 36 6 20 2 3 36 108 11 2 3 x x x x f x x x x x x x x x x − + − + − ′ = − − − − − − = − − + = − ( ) ( ) ( ) ( ( ) ) 2 2 36 1 108 1 11 31 1 2 1 3 5 m f − − − + = ′ − = = − − ( 1) 31( ( 1)) 5 31 26 5 5 y x y x − − = − − = + 52. ( ) ( 2)( 2 ) 3 3 2 2 4 4 x x xx f x x x x x + + + + = = − − ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 3 2 2 3 2 2 3 2 2 3 2 2 4 3 6 2 3 2 1 4 3 12 6 24 2 8 3 2 4 2 9 24 8 4 x x x x x x f x x x x x x x x x x x x x x x − + + − + + ′ = − − + − + − − − − = − − − − = − ( ) ( ) ( ) ( ) ( ) 3 2 2 2 1 9 1 24 1 8 13 1 1 4 3 m f − − − = ′ = = − − ( 2) 13( 1) 3 13 7 3 3 y x y x − − = − − = + 53. ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 2 1 1 x x x x x f x x x − − − ′ = = − − f ′(x) = 0 when x2 − 2x = x(x − 2) = 0, which implies that x = 0 or x = 2. Thus, the horizontal tangent lines occur at (0, 0) and (2, 4). 54. ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 2 2 2 1 2 2 2 1 1 x x x x x f x x x + − ′ = = + + f ′(x) = 0 when 2x = 0, which implies that x = 0.Thus, the horizontal tangent line occurs at (0, 0). 55. ( ) ( )( ) ( ) ( ) ( ) 3 3 4 2 6 3 3 2 3 2 1 4 3 4 1 1 x x x xx f x x x x + − + ′ = = + + f ′(x) = 0 when x6 + 4x3 = x3(x3 + 4) = 0, which implies that x = 0 or x = 3 −4.Thus, the horizontal tangent lines occur at (0, 0) and ( 3 −4, −2.117). −1 −3 5 1 1, − ( ( 1 2 −7 8 −2 8 (2, 5) 5 −10 −5 (−1, −1) 10 −8 4 −4 4 (1, −2) 102 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 56. ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) 2 3 4 2 2 2 2 2 2 1 4 3 2 1 2 3 1 1 x x x x f x x x x x x + − + ′ = + + − = + f ′(x) = 0 when 2x(x2 + 3)(x2 − 1) = 0, which implies that x = 0 or x = ±1.Thus, the horizontal tangent lines occur at (0, 3), (1, 2), and (−1, 2). 57. ( ) ( ) ( ) 1 2 2 1 f x xx x x f x x = + = + ′ = + 58. ( ) ( ) ( ) ( ) 2 3 2 2 1 3 2 3 2 f x x x x x f x x x x x = + = + ′ = + = + 59. ( ) ( )( ) ( ) 3 2 1 1 3 1 f x x x x x x f x x = + − = − ′ = − 60. ( ) ( )( ) ( ) ( ) 2 4 2 3 2 1 1 4 2 2 2 1 f x x x x x x f x x x x x = + − = − ′ = − = − 61. ( ) ( )( ) ( )2 ( )2 275 1 3 5 1 5 1 3 3 5 3 275 275 5 1 5 1 x p p dx p p dp p p ⎛ ⎞ = ⎜ − ⎟ ⎝ + ⎠ ⎡ + − ⎤ ⎡ ⎤ = − ⎢ ⎥ = − ⎢ ⎥ ⎢⎣ + ⎥⎦ ⎢⎣ + ⎥⎦ When p = 4, ( )2 275 3 $1.87 21 dx dp ⎡ ⎤ = − ⎢ ⎥ ≈ − ⎢⎣ ⎥⎦ per unit. 62. ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 2 2 1 0 1 1 1 2 1 1 2 1 2 3 1 dx p p dp p p p p p p p + − = − − + = − − + − + − = + − − − = + When p = 3, 9 6 3 $1.13 16 dx dp − − − = ≈ − per unit. 63. ( )( ) ( )( ) ( ) ( ) 2 2 2 2 2 2 50 4 4 2 200 4 500 500 50 50 t t t t P t t ⎡ + − ⎤ ⎡ − ⎤ ′ = ⎢ ⎥ = ⎢ ⎥ ⎢ + ⎥ ⎢ + ⎥ ⎣ ⎦ ⎣ ⎦ When ( )2 2, 500 184 31.55 bacteria hour. 54 t P ⎡ ⎤ = ′ = ⎢ ⎥ ≈ ⎢⎣ ⎥⎦ 64. ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 50 2 1 1750 50 50 2 50 2 1750 2500 2 1748 50 2 874 25 2 dP t t dt t t t t t t + − + = ⎡⎣ + ⎤⎦ ⎡⎣ + − + ⎤⎦ = + − = + − = + (a) When t = 1, 874 3.88 percent day. 225 dP dt − = ≈ − (b) When t = 10, 874 3600 437 1800 0.24 percent day. dP dt − = = − ≈ − 65. ( ) 2 2 1 1 f t t t t − + = + ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 2 2 2 2 1 2 1 1 2 1 1 1 t t t t tt f t t t + − − − + − ′ = = + + (a) f ′(0.5) = −0.480 week (b) f ′(2) = 0.120 week (c) f ′(8) = 0.015 week Each rate in parts (a), (b), and (c) is the rate at which the level of oxygen in the pond is changing at that particular time. f f −3 −2 2 6 f ′ f −2 2 −1 3 f f −6 −2 2 11 f ′ f −2 2 −1 3 Section 2.4 The Product and Quotient Rules 103 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 66. 2 2 10 4 16 75 4 10 T t t t t ⎛ + + ⎞ = ⎜ ⎟ ⎝ + + ⎠ Initial temperature: T(0) = 75°F ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 4 10 8 16 4 16 75 2 4 700 2 10 4 10 4 10 t t t t t t t T t t t t t ⎛ + + + − + + + ⎞ − + ′ = ⎜ ⎟ = ⎜⎜ + + ⎟⎟ + + ⎝ ⎠ (a) T′(1) ≈ −9.33°F hr (b) T′(3) ≈ −3.64°F hr (c) T′(5) ≈ −1.62°F hr (d) T′(10) ≈ −0.37°F hr Each rate in parts (a), (b), (c), and (d) is the rate at which the temperature of the food in the refrigerator is changing at that particular time. 67. C = x3 − 15x2 + 87x − 73, 4 ≤ x ≤ 9 Marginal cost: dC 3x2 30x 87 dx = − + Average cost: C x2 15x 87 73 x x = − + − (a) (b) Point of intersection: 2 2 2 3 2 3 30 87 15 87 73 2 15 73 0 2 15 73 0 6.683 x x x x x x x x x x x − + = − + − − + = − + = ≈ When x = 6.683, C dC 20.50. x dx = ≈ Thus, the point of intersection is (6.683, 20.50). At this point average cost is at a minimum. 68. (a) As time passes, the percent of people aware of the product approaches approximately 95%. (b) As time passes, the rate of change of the percent of people aware of the product approaches zero. 69. 2 100 200 , 30 C x x 1 x x = ⎛ + ⎞ ⎜ + ⎟ ⎝ ⎠ ≥ ( ) ( ) ( ) ( ) 3 2 3 2 30 100 2 200 30 100 400 30 30 x x C x x x x − ⎡ + − ⎤ ′ = ⎢− + ⎥ ⎢⎣ + ⎥⎦ ⎡ ⎤ = ⎢− + ⎥ ⎢⎣ + ⎥⎦ (a) ( ) 3 2 10 100 400 30 38.125 10 40 C′ = ⎛⎜− + ⎞⎟ = − ⎝ ⎠ (b) C′(15) ≈ −10.37 (c) C′(20) ≈ −3.8 Increasing the order size reduces the cost per item. An order size of 2000 should be chosen since the cost per item is the smallest at x = 20. 70. (a) P = ax2 + bx + c When x = 10, P = 50: 50 = 100a + 10b + c. When x = 12, P = 60: 60 = 144a + 12b + c. When x = 14, P = 65: 65 = 196a + 14b + c. Solving this system, we have 5 8a = − , 75 4 b = , and c = −75. Thus, 5 2 75 8 4 P = − x + x − 75. (b) (c) Marginal profit: 5 75 4 4 P′ = − x + = 0 ⇒ x = 15 This is the maximum point on the graph of P, so selling 15 units will maximize the profit. 0 0 20 80 4 9 0 60 dC dx C x 104 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 71. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 4 0 fx gx hx f x gx hx f = + ′ = ′ + ′ ′ = − + = 72. ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 2 fx gx f x g x f = − ′ = − ′ ′ = − − = 73. f (x) = g(x)h(x) f ′(x) = g(x)h′(x) + h(x)g′(x) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 3 4 1 2 14 f ′ = g h′ + h g′ = + − − = 74. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 2 2 1 2 3 4 2 10 1 g x f x h x h x g x g x h x f x h x f = ′ − ′ ′ = ⎡⎣ ⎤⎦ − − − ′ = = − − 75. Answers will vary. Chapter 2 Quiz Yourself 1. f (x) = 5x + 3 ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 lim 5 3 5 3 lim lim 5 5 3 5 3 lim 5 lim 5 5 x x x x x f x x f x f x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡⎣ + Δ + ⎤⎦ − + = Δ + Δ + − − = Δ Δ = Δ = = At (−2, −7): m = 5 2. f (x) = x + 3 ( ) ( ) ( ) ( ) ( ) 0 0 0 0 lim lim 3 3 3 3 lim 3 3 lim 1 3 3 1 2 3 x x x x f x x f x f x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ + Δ + − + = Δ + Δ + + + = + Δ + + + = + At (1, 2): 1 1 2 1 3 4 m = = + 3. f (x) = x2 − 2x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 0 0 lim 2 2 lim 2 2 2 2 lim 2 2 lim lim 2 2 2 2 x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡ + Δ − + Δ ⎤ − − = ⎣ ⎦ Δ + Δ + Δ − − Δ − + = Δ Δ + Δ − = Δ = + Δ − = − At (3, 3): m = 2(3) = −2 = 4 4. ( ) ( ) 12 0 f x f x = ′ = 5. ( ) ( ) 19 9 19 f x x f x = + ′ = Chapter 2 Quiz Yourself 105 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 6. ( ) ( ) 5 3 2 6 fx x f x x = − ′ = − 7. ( ) ( ) 1 4 3 4 3 4 12 3 3 f x x f x x x − = ′ = = 8. ( ) ( ) 2 3 3 4 8 8 f x x f x x x − − = ′ = − = − 9. ( ) ( ) 1 2 1 2 2 2 1 fx x x f x x x − = = ′ = = 10. ( ) ( ) ( )( ) ( )( ) ( )2 ( )2 2 3 3 2 3 2 2 2 3 3 5 3 2 3 2 f x x x x x f x x x + = + + − + ′ = =− + + 11. ( ) ( )( ) ( ) ( )( ) ( )( ) ( ) 2 2 2 1 2 4 1 2 2 4 2 6 8 2 f x x x f x x x x f x x x = + − + ′ = + − + − + ′ = − + − 12. f (x) = (x2 + 3x + 4)(5x − 2) ( ) ( 2 )( ) ( )( ) 2 2 2 3 4 5 5 2 2 3 5 15 20 10 11 6 15 26 14 f x x x x x x x x x x x ′ = + + + − + = + + + + − = + + 13. ( ) 2 4 3 f x x x = + ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 3 4 4 2 3 4 12 8 3 4 12 3 4 3 3 x x x f x x x x x x x x x + − ′ = + + − = + − + = + − − = + 14. Average rate of change: (3) (0) 1 1 0 3 0 3 y f f x Δ − − = = = Δ − f ′(x) = 2x − 3 Instantaneous rates of change: f ′(0) = −3, f ′(3) = 3 15. Average rate of change: ( ) ( ) ( ) 1 1 6 4 1 1 1 2 y f f x Δ − − − = = = Δ − − f ′(x) = 6x2 + 2x − 1 Instantaneous rates of change: f ′(−1) = 3, f ′(1) = 7 16. Average rate of change: ( ) ( ) 1 1 5 2 10 4 1 5 2 3 20 y f f x Δ − − = = = − Δ − ( ) ( ) 1 2 1 1 2 2 1 2 fx x x f x x = = − ′ = − Instantaneous rates of change: (2) 1, 8 f ′ = − (5) 1 50 f ′ = − 0 3 −2 2 −1 1 3 7 2 5 −0.25 0.5 106 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 17. Average rate of change: (27) (8) 3 2 1 27 8 19 19 y f f x Δ − − = = = Δ − ( ) ( ) 3 1 3 2 3 2 3 1 1 3 3 f x x x f x x x − = = ′ = = Instantaneous rates of change: (8) 1 , 12 f ′ = (27) 1 27 f ′ = 18. P = −0.0125x2 + 16x − 600 (a) (176) (175) 1828.8 1817.1875 $11.6125 P − P = − = (b) dP 0.025x 16 dx = − + When x 175, dP 11.625 dx = = (c) The results are very similar. 19. ( ) ( ) 5 2 6 1 10 6 f x x x f x x = + − ′ = + At (−1, −2), m = −4. 2 4( 1) 4 6 y x y x + = − + = − − 20. f (x) = (x2 + 1)(4x − 3) ( ) ( 2 )( ) ( )( ) 2 2 2 1 4 4 3 2 4 4 8 6 12 6 4 f x x x x x x x x x ′ = + + − = + + − = − + ( ) ( )2 ( ) m = f ′ 1 = 12 1 − 6 1 + 4 = 10 2 10( 1) 10 8 y x y x − = − = − 21. S = −0.13556t3 + 1.8682t2 − 4.351t + 23.52 (a) dS S (t) 0.40668t2 3.7364t 4.351 dt = ′ = − + − (b) 2004: S′(4) ≈ $4.088 yr 2007: S′(7) ≈ $1.876 yr 2008: S′(8) ≈ −$0.487 yr Section 2.5 The Chain Rule Skills Warm Up 1. 5 (1 − 5x)2 = (1 − 5x)2 5 2. 4 (2x − 1)3 = (2x − 1)3 4 3. ( )2 1 2 2 1 4 1 4 1 x x − = + + 4. ( ) 1 3 3 1 6 6 x x − = − − 5. 1 2 ( ) 1 3 3 1 2 1 2 x x x x − = − − 6. ( ) ( ) ( ) ( ) 3 3 2 3 7 3 7 1 3 2 2 3 7 2 2 x x x x x x − − − = = − 7. ( ) ( ) ( )( ) 3 2 2 2 3 6 5 10 3 2 5 2 3 5 2 x x x x x x x x − + − = − + − = + − 8. ( ) ( ) ( )( ) 5 5 1 5 1 1 5 1 1 5 1 x x x x x x x x x − − + = − − − = − − 9. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 3 2 2 2 2 2 3 4 1 1 1 4 1 1 4 x xx x x x x x x + − + = + ⎡ − + ⎤ ⎣ ⎦ = + − − 10. ( ) ( ) ( )( ) ( )( )( ) 5 3 2 3 2 2 3 2 2 2 3 3 3 1 3 1 3 1 13 x x x x x x x x x x x x − + + − = − + − − + = − − + = − + + − 8 27 0 4 −5 4 −4 2 (−1, −2) 5 −10 −5 (1, 2) 10 Section 2.5 The Chain Rule 107 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. y = f (g(x)) u = g(x) y = f (u) 1. ( )4 y = 6x − 5 u = 6x − 5 y = u4 2. ( )y = x2 − 2x + 3 3 u = x2 − 2x + 3 y = u3 3. y = 5x − 2 u = 5x − 2 y = u 4. y = 1 − x2 u = 1 − x2 y = u 5. ( )1 y 3x 1 − = + u = 3x + 1 y = u−1 6. ( ) y x2 3 1 2 − = − u = x2 − 3 y = u−1 2 7. 2 4 dy u du du dx = = dy (2u)(4) dx = = 8(4x + 7) = 32x + 56 8. 3 2 6 dy u du du x dx = = ( ) ( )dy 3u2 6x 18x 3x2 2 2 dx = = − 9. 1 1 2 2 2 dy u du du x dx = − = − 1 1 2 ( 2 ) 2 dy u x dx = ⎛⎜ − ⎞⎟ − ⎝ ⎠ ( ) ( ) 2 1 2 2 3 3 x x x x − = − − = − − 10. 1 2 5 dy u du du dx = − = dy 5u 1 2 dx = − 5(5 9) 1 2 5 5 9 x x − = + = + 11. ( ) ( ) ( ) ( ) 1 3 3 1 3 3 4 1 3 3 3 3 4 2 3 20 2 2 20 2 3 25 2 20 2 3 4 10 1 3 5 2 dy u du du x dx dy u x dx x x x x x x − − − = = − = − = − − − = − 12. ( ) ( ) ( ) ( ) 2 2 2 2 3 2 2 2 2 3 2 2 3 4 3 4 2 3 4 3 4 2 dy u du du x x dx dy u x x dx x x x x x x x x − − − = − = + = − + = − + + + = − + 13. ( ) ( ) 3 1 3 2 2 1 ; 1 f x x x − = = − − (c) General Power Rule 14. ( ) 3 2 ; 1 f x x x = − (d) Quotient Rule 15. f (x) = 3 82 ; (b) Constant Rule. 16. f (x) = 3 x2 = x2 3; (a) Simple Power Rule 17. ( ) 2 f x x 2 x 2x 1; x + − = = + (a) Simple Power Rule 18. ( ) 1 2 3 ; 2 5 f x x x x = + − (d) Quotient Rule 19. ( ) 2 2( 2) 1; 2 f x x x − = = − − (a) General Power Rule 20. ( ) ( ) 2 1 2 5 5 1 ; 1 f x x x − = = + + (c) General Power Rule 21. ( )2 ( ) ( )2 y′ = 3 2x − 7 2 = 6 2x − 7 22. ( ) ( )2 ( ) ( )2 g′ x = 3 4 − 2x −2 = −6 4 − 2x 23. ( ) ( )f x = 5x − x2 3 2 ( ) ( ) ( ) ( ) 3 2 1 2 3 2 2 2 f ′ x = 5x − x 5 − 2x = 5 − 2x 5x − x 24. y′ = 2(2x3 + 1)(6x2 ) = 24x5 + 12x2 25. h′(x) = 2(6x − x3 )(6 − 3x2 ) = 6x(6 − x2 )(2 − x2 ) 26. ( ) ( ) ( ) ( )( ) f ′ x = 3 4x − x2 2 4 − 2x = 6 2 − x 4x − x2 2 27. ( ) ( )1 2 f t = t + 1 = t + 1 ( ) 1( 1) 1 2 (1) 1 2 2 1 f t t t ′ = + − = + 108 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 28. ( ) ( ) ( ) ( ) ( ) 1 2 1 2 5 3 5 3 1 5 3 3 3 2 2 5 3 g x x x g x x x − = − = − ′ = − − = − − 29. ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 2 2 5 2 2 5 2 1 2 5 2 4 5 4 5 2 2 2 5 2 st t t t t s t t t t t t t − = + + = + + ′ + = + + + = + + 30. ( ) ( ) ( ) ( ) 3 3 3 1 3 2 3 2 3 2 3 2 3 3 4 3 4 1 3 4 9 4 9 4 3 3 3 4 y x x x x y x x x x x x − = + = + ′ + = + + = + 31. ( ) ( ) ( ) ( ) 3 2 2 1 3 2 2 3 2 2 3 9 4 9 4 1 9 4 18 6 3 9 4 y x x y x x x x − = + = + ′= + = + 32. ( ) ( ) ( ) 2 2 1 2 2 1 2 2 2 4 24 2 1 4 2 2 2 4 y x x y x x x x − = − = − ′ ⎛ ⎞ − = ⎜ ⎟ − − = ⎝ ⎠ − 33. ( ) ( ) 3 f x 2 2 9x − = − ( ) ( )( ) ( ) ( ) 4 4 2 3 2 9 9 54 2 9 f x x x ′ = − − − − = − 34. ( ) ( )g x 3 x2 8x 3 2 − = + ( ) ( ) ( ) ( ) ( ) 2 5 2 2 5 2 3 9 4 3 8 2 8 2 8 x g x x x x x x ′ = ⎛− ⎞ + − + = − + ⎜ ⎟ ⎝ ⎠ + 35. ( ) ( )f x x2 25 1 2 − = + ( ) ( ) ( ) ( ) 2 3 2 2 3 2 1 25 2 2 25 f x x x x x ′ = ⎛− ⎞ + − = − ⎜ ⎟ ⎝ ⎠ + 36. ( ) y 4 x3 4 3 − = − ( ) ( ) ( ) 7 3 2 3 2 2 7 3 4 4 3 4 3 3 4 y x x x x ′ = ⎛− ⎞ − − − = ⎜ ⎟ ⎝ ⎠ − 37. ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 1 2 12 1 2 24 3 216 2 54 f x x x x x f f ′ = − = − ′ = = = 54 216( 2) 216 378 y x y x − = − = − 38. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 4 12 9 4 9 108 9 4 2 12 14 9 296,352 2 3 14 115,248 f x x x f f ′ = − = − ′ = = = = 115,248 296,352( 2) 296,352 477,456 y x y x − = − = − 39. ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 2 4 7 4 7 1 4 7 8 4 2 4 7 f x x x f x x x x x − = − = − ′ = − = − ( ) ( ) 2 8 3 2 3 f f ′ = = 3 8( 2) 3 8 7 3 3 y x y x − = − = − 40. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 2 1 2 2 2 1 2 2 1 2 2 1 2 2 2 2 2 5 5 1 5 2 5 1 2 5 5 5 5 2 5 5 f x x x x x f x x x x x x x x x x x x x − − − = + = + ′ = ⎡ + ⎤ + + ⎢⎣ ⎥⎦ = + + + = + ⎡ + + ⎤ ⎣ ⎦ + = + ( ) ( ) ( ) 2 13 3 2 6 6 13 2 3 13 8 3 3 f f y x y x ′ = = − = − = − −2 −400 4 200 (2, 54) − 1 0 5 200,000 (2, 115,248) −1 −4 4 10 (2, 3) − 1 − 1 4 10 Section 2.5 The Chain Rule 109 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 41. ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 2 2 1 2 1 1 2 1 2 2 2 1 2 1 1 1 fx x x x x f x x x x x x x x x − = − + = − + ′ = − + − − = − + − = − ( ) ( ) 2 1 2 1 f f ′ = = 1 1( 2) 1 y x y x − = − = − 42. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 5 3 2 5 3 5 3 2 3 2 4 3 6 4 3 4 3 4 2 8 1 2 8 32 4 2 8 1 4 f x x x x x f f − − ′ = − − − = − ′ = = = − − − = − = 1 1( 2) 4 4 1 3 4 4 y x y x − = − − = − + 43. ( ) ( ) 2 32 2 2 1 3 4 2 1 f x x x x x ′ − − = + f has a horizontal tangent when f ′ = 0. 44. ( ) ( )3 2 2 2 1 f x x x ′ = + f ′ is never 0. 45. ( ) 3 2 1 2 1 f x x x ′ = − + f ′ is never 0. 46. ( ) 2 5 2 2 f x x x ′ − = f has a horizontal tangent when f ′ = 0. 47. ( ) y 4 x2 1 − = − ( )( ) ( ) ( ) 2 2 2 2 1 4 2 2 4 y x x x x − ′ = − − − = − General Power Rule 48. ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 2 2 2 1 3 1 3 1 1 3 1 2 3 2 3 3 1 s t t t t t s t t t t t t t − − = = + − + − ′ = − + − + + = − + − General Power Rule 49. ( ) ( ) ( ) ( ) 2 2 3 3 4 4 2 2 8 2 8 2 y t t y t t − − = − = − + + ′ = + = + General Power Rule 50. ( ) ( ) f x 3x x3 4 2 − = − ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 2 3 2 3 3 3 3 2 3 3 3 3 3 3 3 3 2 4 3 4 3 18 4 3 4 3 4 6 4 3 5 4 4 f x x x x x x x x x x x x x − − − − − ′ = ⎡ − − ⎤ + − ⎢⎣ ⎥⎦ = − − + − = − − ⎡ − − ⎤ ⎣ ⎦ − + = − Product Rule 51. f (x) = (2x − 1)(9 − 3x2 ) ( ) ( )( ) ( )( ) ( ) 2 2 2 2 2 2 1 6 9 3 2 12 6 18 6 18 6 18 6 3 2 3 f x x x x x x x x x x x ′ = − − + − = − + + − = + − = − − − Product Rule −2 −3 4 3 (2, 1) − 2 − 2 4 2 2 , 1 4 ( ( f f −2 −1 5 2 − 1 − 1 3 f f ′ 3 − 5 − 3 4 f f ′ 4 4 −1 −4 4 f f ′ 110 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 52. y = x3(7x + 2) = 7x4 + 2x3 y′ = 28x3 + 6x2 = 2x2 (14x + 3) Simple Power Rule 53. ( ) ( ) ( ) 1 2 3 2 3 2 1 2 2 1 2 1 2 2 2 y x x y x x − − = = + + ′ = − + = − + General Power Rule 54. ( ) ( ) ( ) ( ) ( ) ( ) 3 1 3 3 3 4 3 2 3 2 3 4 3 3 3 1 1 3 1 1 3 3 3 1 g x x x g x x x x x − − = = − − ′ = ⎛− ⎞ − = − ⎜ ⎟ ⎝ ⎠ − General Power Rule 55. ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 2 2 3 3 9 3 3 9 1 3 9 9 3 9 9 3 12 9 27 3 4 3 f x x x x x x x x x x x ′ = − + − = − ⎡⎣ + − ⎤⎦ = − − = − − Product and General Power Rules 56. ( ) ( ) ( ) ( ) ( ) ( )( ) 3 2 3 2 5 4 3 4 3 2 2 2 2 4 8 16 8 16 5 32 48 5 32 48 5 12 4 f x x x x x x x x x f x x x x x x x x x x = − = − + = − + ′ = − + = − + = − − Simple Power Rule 57. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 2 3 2 3 12 3 2 2 3 2 2 3 2 3 3 1 2 3 y x x x x y x x x x x x x x − − = + = + ′ = ⎡ + ⎤ + + ⎢⎣ ⎥⎦ = + ⎡⎣ + + ⎦⎤ + = + Product and General Power Rules 58. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 1 1 1 2 1 1 1 2 1 1 2 1 2 1 1 3 2 2 3 2 2 1 y t t tt y t t t t t t t t t t t t t − − − − = + = + ′ = ⎡ + ⎤ + + ⎢⎣ ⎥⎦ = + + + = + ⎡⎣ + + ⎤⎦ = + + + = + Product and General Power Rules 59. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 2 2 1 2 4 2 2 4 2 2 2 5 8 2 2 y t t t t y t t tt t t tt t tt t t t t − − = − = − ′ = ⎡ − ⎤ + − ⎢⎣ ⎥⎦ = − ⎡⎣ + − ⎤⎦ + − = − − = − Product and General Power Rules 60. ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 2 1 2 2 1 2 1 2 1 2 2 2 2 2 1 2 2 1 2 2 2 2 2 2 4 2 2 2 2 4 2 2 2 5 2 2 y xx x x y x x x x x x x x x x x x x x x x x x x − = − = − ′ = ⎡ − ⎤ + − ⎛ ⎞ ⎣ ⎦ ⎜ ⎟ ⎝ ⎠ − = − + − + − = − ⎡⎣ + − ⎤⎦ = − − = Product and General Power Rules 61. ( )( ) ( )( ) ( ) ( )( ) ( ) 2 2 2 2 2 2 2 2 3 6 5 1 6 5 1 5 6 5 2 2 1 1 2 6 5 5 12 5 1 y x x x x x x y x x x x x x ⎛ − ⎞ = ⎜ ⎟ ⎝ − ⎠ − ⎡ − − − − ⎤ ′ = ⎛ ⎞⎢ ⎥ ⎜ − ⎟⎢ ⎥ ⎝ ⎠ − ⎣ ⎦ − − + = − Quotient and General Power Rules Section 2.5 The Chain Rule 111 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 62. ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 6 3 3 5 6 2 6 5 5 4 4 4 64 3 3 3 6 33 1 64 3 6 3 3 6 64 192 3 3 y x x x x x x x x y x x x x x x x x ⎛ ⎞ ⎡ ⎤ = ⎜ ⎟ = ⎢ ⎥ ⎝ − ⎠ ⎢⎣ − ⎥⎦ ⎡ − − − − ⎤ ′ = ⎢ ⎥ ⎢⎣ − ⎥⎦ ⎧⎪ ⎡⎣ − + ⎤⎦⎪⎫ ⎡ − ⎤ = ⎨ ⎬ = ⎢ ⎥ ⎩⎪ − ⎪⎭ ⎢⎣ − ⎥⎦ Quotient Rule 63. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 36 36 3 3 72 3 1 72 3 0 72 8 27 3 f t t t f t t t f − − = = − − ′ = − − − = − ′ = = 4 8( 0) 3 8 4 3 y t y t − = − = + 64. ( )y = 4x3 + 3 2 = 16x6 + 24x3 + 9 y′ = 96x5 + 72x2 = 24x2 (4x3 + 3) ( ) ( )( ( )3 ) m = y′ −1 = 24 −1 4 −1 + 3 = −24 1 24( ( 1)) 24 23 y x y x − = − − − = − − 65. ( ) ( )f x = 3x3 + 4x 1 5 ( ) ( ) ( ) ( ) 3 4 5 2 2 3 4 5 13 4 9 4 5 9 4 5 3 4 f x x x x x x x − ′ = + + + = + (2) 1 2 m = f ′ = 2 1( 2) 2 1 1 2 y x y x − = − = + 66. ( ) ( ) 2 1 2 2 1 3 4 3 4 s x x x x x − = = − + − + ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 2 3 2 3 2 1 3 4 2 3 2 3 2 2 3 4 3 3 6 3 2 4 16 s x x x x x x x s − ′ = − − + − − = − + ′ − = = − 1 3 ( 3) 2 16 3 17 16 16 y x y x − = − − = − + 67. ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 1 2 2 1 2 1 2 1 2 2 1 2 2 2 1 2 2 5 2 9 2 2 9 2 9 2 9 1 2 2 2 2 1 9 2 2 2 2 2 1 9 2 2 2 2 1 9 2 4 2 2 2 5 4 9 2 2 4 2 f t t t t t f t t t t t t t tt t t t t t t t t t t t t t t − − − − − = − + = − + ′ = − ⎡ + ⎤ + + ⎢⎣ ⎥⎦ = − + + + = + ⎡ − + + ⎤ ⎢⎣ ⎥⎦ = + ⎡ − + + ⎤ ⎢⎣ ⎥⎦ = + ⎛ + − ⎞ ⎜ ⎟ ⎝ ⎠ + − = + f ′(−1) = −6 ( 8) 6 ( 1) 6 14 y t y t − − = − ⎡⎣ − − ⎤⎦ = − − 68. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 3 2 3 2 2 2 1 1 1 2 2 1 1 2 1 2 1 2 1 1 y x x x x x x x y x x x x x x − = = + + + − ⎛ ⎞ + ⎜ ⎟ ′ = ⎝ ⎠ + + − + = = + + ( ) ( ) 3 5 8 3 5 3 8 5 9 8 8 y y x y x ′ = − = − = + −4 −2 4 12 (0, 4) −2 −2 1 12 (−1, 1) 3 0 −3 (2, 2) 4 −1 5 −1 3 3 , 1 2 ( ( −4 −16 4 2 (−1, −8) 7 −1 −1 5 (3, 3) 112 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 69. ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 3 2 3 2 1 2 1 1 2 3 2 3 2 3 1 1 2 3 2 2 3 2 3 1 2 3 4 2 3 f x x x x x x x x f x x x x x x x − + + = = − − − − + − ′ = − − − + = − − = − (2) 1 3 2 1 f ′ − = = − 3 2( 2) 2 7 y x y x − = − − = − + 70. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 2 3 2 2 1 2 2 2 3 2 2 1 2 2 3 2 2 2 2 3 2 1 2 25 25 25 2 25 1 25 25 25 25 25 25 y x x x x y x x x x x x x x x x x − − − − − − = = + + ′ = ⎡− + ⎤ + + ⎢⎣ ⎥⎦ = − + + + = + ⎡− + + ⎤ ⎣ ⎦ = + ( ) ( ) 0 1 5 0 1 0 5 1 5 y y x y x ′ = − = − = 71. ( ) 59 59 1000 60 1 1 5000 1 12 12 12 A′ = ⎛ + r ⎞ ⎛ ⎞ = ⎛ + r ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (a) ( ) 59 0.08 50 1 0.08 12 $74.00 per percentage point A′ = ⎛⎜ + ⎞⎟ ⎝ ⎠ ≈ (b) ( ) 59 0.10 50 1 0.10 12 $81.59 per percentage point A′ = ⎛⎜ + ⎞⎟ ⎝ ⎠ ≈ (c) ( ) 59 0.12 50 1 0.12 12 $89.94 per percentage point A′ = ⎛⎜ + ⎞⎟ ⎝ ⎠ ≈ 72. ( ) N 400 1 3 t2 2 2 = ⎡ − + − ⎤ ⎢⎣ ⎥⎦ ( ) ( )( )( ) ( ) ( ) 2 3 2 3 400 3 2 2 2 4800 2 dN Nt t t dt t t = ′ = ⎡ − − + − ⎤ ⎢⎣ ⎥⎦ = + (a) N′(0) = 0 bacteria day (b) N′(1) ≈ 177.8 bacteria day (c) N′(2) ≈ 44.4 bacteria day (d) N′(3) ≈ 10.8 bacteria day (e) N′(4) ≈ 3.3 bacteria day (f ) The rate of change of the population is decreasing as time passes. 73. (a) 1 V k t = + When t = 0, V = 10,000. 10,000 10,000 0 1 10,000 1 k k V t = ⇒ = + = + (b) ( ) ( ) ( ) ( ) 1 2 3 2 3 2 10,000 1 5000 1 1 5000 1 V t dV t dt t − − = + = − + = − + When t = 1, ( )3 2 5000 2500 $1767.77 per year. 2 2 dV dt = − = − ≈ − (c) When t = 3, ( )3 2 5000 $625.00 per year. 4 dV dt = − = − −4 −2 8 6 (2, 3) − 3 − 1 3 1 Section 2.6 Higher-Order Derivatives 113 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 74. (a) From the graph, the tangent line at t = 4 is steeper than the tangent line at t = 1. So, the rate of change after 4 hours is greater. (b) The cost function is a composite function of x units, which is a function of the number of hours, which is not a linear function. 75. (a) ( )r = 2.8557t4 − 72.792t3 + 676.14t2 − 2706t + 4096 1 2 ( ) ( ) ( ) 4 3 2 1 2 3 2 3 2 4 3 2 1 2.8557 72.792 676.14 2706 4096 11.4228 218.376 1352.28 2706 2 11.4228 218.376 1352.28 2706 2 2.8557 72.792 676.14 2706 4096 dr r t t t t t t t t dt t t t t t t t − = ′ = − + − + ⋅ − + − − + − = − + − + Chain Rule (b) (c) t = 3 yr (d) t ≈ 4.52 yr, t ≈ 6.36 yr, t ≈ 8.24 yr Section 2.6 Higher-Order Derivatives Skills Warm Up 1. ( ) 16 2 24 0 16 24 0 0 16 24 0 1.5 t t t t t t t − + = − + = = − + = → = 2. ( ) ( )( ) 2 2 16 80 224 0 16 5 14 0 16 7 2 0 7 0 7 2 0 2 t t t t t t t t t t − + + = − − − = − − + = − = → = + = → = − 3. ( ) ( )( ) 2 2 16 128 320 0 16 8 20 0 16 10 2 0 10 0 10 2 0 2 t t t t t t t t t t − + + = − − − = − − + = − = → = + = → = − 4. −16t2 + 9t + 1440 = 0 ( )( ) ( ) 9 92 4 16 1440 2 16 9 92241 32 9 3 10249 32 t − ± − − = − − ± = − ± = t ≈ −9.21 and t ≈ 9.77 5. ( ) ( ) ( ) 2 2 2 2 2 2 7 2 2 2 7 2 4 14 6 14 y x x dy x xx dx x x x x x = + = + + = + + = + 3 9 −6 6 114 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. ( ) ( ) 2 0 f x f x ′ = − ′′ = 2. ( ) ( ) 4 0 f x f x ′ = ′′ = 3. ( ) ( ) 2 7 2 f x x f x ′ = + ′′ = 4. ( ) ( ) 6 4 6 f x x f x ′ = + ′′ = 5. ( ) ( ) 2 8 2 2 8 g t t t g t t ′ = − + ′′ = − 6. f (x) = −2x5 + 3x4 + 8x f ′(x) = −10x4 + 12x3 + 8 f ′′(x) = −40x3 + 36x2 7. ( ) ( ) ( ) 2 2 3 4 4 3 3 4 4 3 2 9 9 2 2 ft t t f t t f t t t − − − = = ′ = − ′′ = = 8. g(t) = 8t−1 g′(t) = −8t−2 ( ) 3 3 g t 16t 16 t ′′ = − = 9. ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 2 2 2 2 9 2 2 18 2 18 2 2 2 2 18 18 2 4 2 18 2 5 2 f x x x x x f x x x x x x x x x x ′ = − − = − − ′′ = − − − + − − = − ⎡ − − ⎤ ⎣ ⎦ = − − 10. ( )( ) ( ) ( )( ) 2 2 4 3 2 5 4 3 2 4 3 2 4 3 5 2 5 24 60 10 25 24 300 1200 1500 120 1200 3600 3000 y x x x x x x x x x x x y x x x x ′ = + + = + + + = + + + ′′= + + + 11. ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 3 3 1 1 1 1 1 2 2 1 1 4 1 1 4 1 x x f x x x x f x x x − − − − + ′ = − = − = − − − ′′ = − = − Skills Warm Up —continued— 6. ( )( ) ( )( ) ( )( ) 2 2 2 2 3 2 3 2 3 2 3 2 5 3 4 2 3 2 5 4 12 4 10 6 15 8 18 10 15 y x x x dy x x x x x dx x x x x x x x x = + − = + + + − = + + − + − = + − − 7. ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 7 2 7 2 2 2 7 4 14 2 2 7 2 14 2 7 2 7 2 7 y x x dy x x x dx x x x x x x x x x x x = + + − = + + − = + + = + + = + 8. ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 2 3 2 2 2 2 2 2 3 2 5 2 52 3 3 4 2 5 4 6 10 15 4 12 2 5 6 10 15 2 5 y x x x dy x x x x x dx x x x x x x x x x x + = − − + − + = − + − − − − = − − − − = − 9. f (x) = x2 − 4 Domain: (−∞, ∞) Range: [−4, ∞) 10. f (x) = x − 7 Domain: [7, ∞) Range: [0, ∞) Section 2.6 Higher-Order Derivatives 115 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 12. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 4 4 4 4 2 2 8 2 24 2 24 2 g t t t g t t g t t t − − − = − = − + + ′ = + ′′ = − + = − + 13. ( ) ( ) ( ) 4 3 3 2 2 5 12 20 36 60 72 f x x x f x x x f x x x ′ = − ′′ = − ′′′ = − 14. ( ) ( ) ( ) ( ) 3 2 2 4 6 12 12 24 12 12 2 1 f x x x f x x x f x x x ′ = − ′′ = − ′′′ = − = − 15. ( ) ( ) ( ) ( ) ( ) ( ) 3 3 2 4 3 2 3 2 2 5 4 5 12 48 64 5 60 240 320 20 180 480 320 60 360 480 120 360 f x x x x x x x x x x x f x x x x f x x x f x x = + = + + + = + + + ′ = + + + ′′ = + + ′′′ = + 16. ( ) ( ) ( ) ( ) ( ) 3 3 2 11 8 5 2 10 7 4 9 6 3 4 6 3 12 216 1296 2592 132 1728 6480 5184 1320 12,096 25,920 5184 f x x x x x x x f x x x x x f x x x x ′ = − = − + − ′′ = − + − ′′′ = − + − 17. ( ) ( ) ( ) ( ) 2 2 3 4 5 5 3 3 16 16 3 8 9 8 9 9 2 2 fx x x f x x f x x f x x x − − − − = = ′ = − ′′ = ′′′ = − = − 18. f (x) = −2x−1 f ′(x) = 2x−2 f ′′(x) = −4x−3 ( ) 4 4 f x 12x 12 x ′′′ = − = 19. ( ) ( ) ( ) ( ) 3 2 20 20 60 20 2 60 4 20 260 g t t t g t t g ′ = + ′′ = + ′′ = + = 20. ( ) ( ) ( ) ( ) 9 2 2 2 5 2 fx x fx x f x f = − ′ = − ′′ = − ′′− = − 21. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 3 2 5 2 5 2 5 2 4 4 1 4 2 1 4 4 3 4 3 8 8 4 5 3 1 8 9 648 f x x x f x x f x x f x x x f − − − = − = − ′ = − − ′′ = − − ′′′ − = − − = − ′′′ − − = = − 22. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 3 2 3 2 5 2 5 2 2 3 2 3 12 3 2 2 3 2 12 3 2 2 3 2 3 2 3 2 3 2 2 3 1 3 2 32 f t t t f t t t f t t t f t t t f − − − − − = + = + ′ = + = + ′′ = − + = − + ′′′ = + =+ ′′′⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 23. ( ) ( )f x = x3 − 2x 3 = x9 − 6x7 + 12x5 − 8x3 f ′(x) = 9x8 − 42x6 + 60x4 − 24x2 f ′′(x) = 72x7 − 252x5 + 240x3 − 48x f ′′(1) = 12 24. ( ) ( )g x = x2 + 3x 4 = x8 + 12x7 + 54x6 + 108x5 + 81x4 g′(x) = 8x7 + 84x6 + 324x5 + 540x4 + 324x3 g′′(x) = 56x6 + 504x5 + 1620x4 + 2160x3 + 972x2 g′′(−1) = −16 25. f ′′(x) = 4x 26. f ′′′(x) = 60x2 − 72x = 12x(5x − 6) 27. ( ) ( ) ( )( ) ( ) 1 2 4 1 2 2 1 2 1 2 1 1 1 2 1 f x x x f x x x − ′′′ = − = − = ⎛ ⎞ − = ⎜ ⎟ − ⎝ ⎠ 116 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 28. f ′′′(x) = 4x−4 f (4) (x) = −16x−5 (5) ( ) 6 6 f x 80x 80 x = − = 29. ( )( ) ( )( ) ( )( ) 5 2 3 6 2 2 12 4 4 12 4 f x x x x x f x x = + = + = + 30. ( ) ( )( ) ( )( ) 4 5 4 7 4 0 f x x f x f x ′′′ = + = = 31. ( ) ( ) 3 2 18 27 6 18 f x x x f x x ′ = − + ′′ = − f ′′(x) = 0 ⇒ 6 18 3 x x = = 32. ( ) ( )( )( )( ) ( )( ) ( ) ( ) 2 2 4 2 3 2 2 2 3 3 4 9 13 36 4 26 12 26 f x x x x x x x x x f x x x f x x = + − + − = − − = − + ′ = − ′′ = − f ′′(x) = 0 ⇒ 12 2 26 13 78 6 6 x x = = ± = ± 33. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 1 2 1 2 2 1 2 2 2 2 2 1 2 1 1 1 1 2 1 1 2 1 f x x x x x f x x x x x x x x − = − = − ′ = − + − = + − − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 2 1 2 2 2 1 2 2 1 2 2 2 3 2 2 3 2 2 1 2 2 3 2 3 2 1 2 1 1 2 2 1 1 2 1 2 1 2 1 1 1 1 2 3 1 x x x x x fx x x x x x x x x x x x x x x − − − − ⎛ ⎞ − ⎜ ⎟ ′′ = ⎝ ⎠ + − − − − − = + ⋅ − − − − = − f ′′(x) = 0 ⇒ 2x3 − 3x = (2 2 3) 0 3 6 2 2 x x x − = = ± = ± x = 0 is not in the domain of f. 34. ( ) ( )( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 3 2 2 2 2 3 2 2 3 3 1 2 3 3 3 3 3 3 2 3 2 3 2 2 3 23 3 2 9 3 2 9 3 x x x x f x x x x x f x x x x x x x x x x x x x x x x − − − − + − − ′ = = = − + + + ′′ = − ⎡− + ⎤ + + − ⎢⎣ ⎥⎦ = − + ⎡ − + + ⎤ ⎣ ⎦ − − = + − = + f ′′(x) = 0 ⇒ 2 ( 2 9) 0 0, 3 x x x − = = ± Section 2.6 Higher-Order Derivatives 117 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 35. (a) ( ) ( ) ( ) ( ) ( ) ( ) 16 2 144 32 144 32 s t t t v t s t t a t v t s t = − + = ′ = − + = ′ = ′′ = − (b) ( ) ( ) ( ) 2 3 288 ft 3 48 ftsec 3 32 ftsec s v a = = = − (c) v(t) = 0 32 144 0 32 144 4.5 sec t t t − + = − = − = s(4.5) = 324 ft (d) s(t) = 0 ( ) 16 2 144 0 16 9 0 0 sec 9 sec t t t t t t − + = − − = = = v(9) = −32(9) + 144 = −144 ft sec This is the same speed as the initial velocity. 36. (a) ( ) ( ) ( ) ( ) ( ) 16 2 1250 32 32 s t t v t s t t a t v t = − + = ′ = − = ′ = − (b) s(t) = 0 when 16t2 = 1250, or t = 78.125 ≈ 8.8 sec. (c) v(8.8) ≈ −282.8 ft sec 37. ( )( ) ( )( ) ( ) ( ) 2 2 2 2 10 90 90 1 900 10 10 d s t t dt t t + − = = + + As time increases, the acceleration decreases. After 1 minute, the automobile is traveling at about 77.14 feet per second. 38. ( ) ( ) ( ) ( ) ( ) 8.25 2 66 16.50 66 16.50 s t t t v t s t t a t s t = − + = ′ = − + = ′′ = − It takes 4 seconds for the car to stop, at which time it has traveled 132 feet. 39. f (x) = x2 − 6x + 6 f ′(x) = 2x − 6 f ′′(x) = 2 (a) (b) The degree decreased by 1 for each successive derivative. (c) f (x) = 3x2 − 9x f ′(x) = 6x − 9 f ′′(x) = 6 (d) The degree decreases by 1 for each successive derivative. 40. Graph A is the position function. Graph B is the velocity function. Graph C is the acceleration function. Explanations will vary. Sample explanation: The position function appears to be a third-degree function, while the velocity is a second-degree function, and the acceleration is a linear function. t 0 1 2 3 4 5 s(t) 0 57.75 99 123.75 132 123.75 v(t) 66 49.50 33 16.50 0 −16.50 a(t) −16.50 −16.50 −16.50 −16.50 −16.50 −16.50 t 0 10 20 30 40 50 60 ds dt 0 45 60 67.5 72 75 77.14 2 2 d s dt 9 2.25 1 0.56 0.36 0.25 0.18 10 −3 −5 f f ″ f ′ 7 −4 −10 4 f f ′ f ′′ 10 118 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 41. (a) y = −68.991t3 + 1208.34t2 − 5445.4t + 10,145 (b) y′ = −206.973t2 + 2416.68t − 5445.4 y′′ = −413.946t + 2416.68 (c) Over the interval 5 ≤ t ≤ 8, y′ > 0, therefore y is increasing over 5 ≤ t ≤ 8 or from 2005 to 2008. (d) y′′ = 0 413.946 2416.68 0 413.946 2416.68 5.84 or 2005 t t t − + = − = − ≈ 42. Let y = xf (x). Then, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 . y xf x f x y xf x f x f x xf x f x y xf x f x f x xf x f x ′ = ′ + ′′ = ′′ + ′ + ′ = ′′ + ′ ′′′ = ′′′ + ′′ + ′′ = ′′′ + ′′ In general ( ) ( ) ( ) ( ) ( ) ( 1) ( ). y n = ⎡xf x ⎤ n = xf n x + nf n− x ⎣ ⎦ 43. False. The Product Rule is ⎡⎣ f (x)g(x)⎤⎦′ = f ′(x)g(x) + f (x)g′(x). 44. True. h′(c) = f ′(c)g(c) + f (c)g′(c) = 0 45. Answers will vary. Section 2.7 Implicit Differentiation Skills Warm Up 1. 2 2 2 2 2 2 2 x y x x y x y x x y x x − = − = − = − = − 2. 4 1 3 4 3 3 4 x y y x y x = − = − − = 3. ( ) 6 6 6 6 6 6 6 6 1, 6 xy x y xy y x y x x y x x y x − + = + = + + = + + = + = ≠ − 4. ( ) 2 2 2 2 2 2 2 2 12 3 4 3 4 12 3 4 12 4 12 4, 3 3 y x xy y x y x y x x y x x x + = + − = − − = − − = = − ≠ ± − 5. 2 2 2 2 2 5 5 5 x y y x y x + = = − = ± − 6. 2 2 2 2 2 2 2 2 6 6 6 6 6 x y x y x y x y x y = ± − = − − = − − = ± − = 7. ( ) 2 2 3 4, 2, 1 3 x y − ( ) ( ) 2 ( ) 2 3 2 4 3 4 4 8 3 1 3 3 − − = = 8. ( ) 2 2, 0, 3 1 x y − − − ( ) 02 2 2 1 1 3 4 2 − − = = − − − 9. ( ) 2 5 , 1,2 3 12 5 x y y − − + ( ) ( 2 ) ( ) ( ) 5 1 5 5 5 3 2 12 2 5 3 4 24 5 7 7 − − − = = = − + − + − Section 2.7 Implicit Differentiation 119 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. 4 0 xy xdy y dx xdy y dx dy y dx x = + = = − = − 2. 3 2 8 6 8 8 6 6 8 x y x x dy dx dy x dx dy x dx − = − = − = − = − 3. 2 1 2 2 2 y x y dy x dx dy x dx y = − = − = − 4. ( ) 3 3 2 2 2 2 2 2 4 2 3 12 2 12 2 2 6 1 3 3 y x x y dy x dx dy x x dx y y = + = + + + = = 5. 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 0 2 2 2 2 2 2 1 x y x x dy y xy dx x y dy xy dx dy xy dx x y xy x y − = + − = = − − = − = 6. ( ) 2 2 2 2 4 10 2 4 4 0 2 4 4 4 2 4 xy xy y xy dy y xdy dx dx xy x dy y y dx dy y y dx xy x + = + + + = + =− − + = − + 7. ( ) 4 2 2 8 0 8 8 y xy y dy xdy y dx dx y x dy y dx dy y dx y x − = − − = − = = − 8. ( ) 3 2 3 2 2 2 2 3 3 2 2 2 2 2 6 2 0 6 2 2 2 2 6 xy x y y xy dy xy x dy dx dx xy x dy xy y dx dy xy y dx xy x − = + − − = − = − − = − 9. ( ) ( ) 2 2 1 1 0 xy y y x xy y y x y x y x y y dy dx − = − − = − − = − − = − = 10. 2 1 5 2 5 6 1 6 1 6 x y x y x y x y y x y x dy dx + = − + = − = − = − = − 11. ( ) ( ) 2 2 2 5 3 2 5 15 2 1 10 1 10 2 1 2 5 1 y x y y x y dy y dy dx dx y dy dx dy dx y − = − − = − − = − = − = − − 120 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 12. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 9 9 8 4 2 2 9 8 9 2 9 72 2 9 2 9 72 9 36 y x y y y dy y y dy dx dx x y y dy y y dx x y y dy dx x y dy x y dx y dy x y dx y = − − ⎛ ⎞ − ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠= − − − = − − = − − = − − = 13. 2 2 16 2 2 0 2 2 x y x y dy dx y dy x dx dy x dx y + = + = = − = − At (0, 4), 0 0. 4 dy dx = − = 14. 2 2 25 2 2 0 2 2 x y x y dy dx y dy x dx dy x dx y − = − = − = − = At (5, 0), dy dx is undefined. 15. ( ) 4 0 1 1 y xy dy x dy y dx dx dy x y dx dy y dx x + = + + = + = − = − + At ( 5, 1), 1. 4 dy dx − − = − 16. 16 2 25 2 400 32 50 0 50 32 16 25 x y x y dy dx y dy x dx dy x dx y + = + = = − = − At (5, 4), 4. 5 dy dx = − 17. ( ) 3 2 2 2 2 4 3 2 0 2 3 3 2 x xy y x x dy y y dy dx dx dy y x y x dx dy y x dx y x − + = − − + = − = − − = − At (0, −2), 1. 2 dy dx = 18. ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 0 2 2 2 2 x y y x x dy xy y y dy x dx dx dy x xy xy y dx dy y x y dx x x y + = − + + + = + = − − + = + At (2, −1), y′ is undefined. 19. ( ) 3 3 3 2 2 3 3 2 2 3 2 3 3 2 3 3 1 3 1 1 3 1 3 3 1 x y y x x y dy x y dy dx dx dy xy xy dx dy x y dx x y − = + − = − = − − = − At (0, 0), dy 1. dx = − Section 2.7 Implicit Differentiation 121 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 20. ( ) ( ) ( ) 3 3 2 2 2 2 2 2 2 2 2 2 2 2 6 3 3 6 3 3 6 6 3 6 6 3 6 3 3 6 3 2 3 2 2 2 x y xy x y dy xdy y dx dx x y dy x dy y dx dx y x dy y x dx dy y x dx y x dy y x dx y x dy y x dx y x + = + = ⎛ + ⎞ ⎜ ⎟ ⎝ ⎠ + = + − = − − = − − = − − = − At 4, 8 , 3 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 4. 5 dy dx = 21. 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 9 1 1 0 2 2 0 x y x y dy dx x y dy dx dy x y dx y x − − − − − − + = + = + = − = =− At (16, 25), 5. 4 dy dx = − 22. 2 3 2 3 1 3 1 3 1 3 1 3 3 1 3 1 3 5 2 2 0 3 3 x y x y dy dx dy x y y dx y x x − − − − + = + = − = =− =− At (8, 1), 1. 2 dy dx = − 23. ( ) ( ) 1 2 1 2 2 2 1 1 1 2 2 2 2 1 2 2 1 2 2 2 2 2 2 4 2 2 4 2 2 5 5 8 xy x y x y x y x y dy y x dy dx dx x dy dy y y dx dx x y dy x x y dx x x y y xy y x xy x y y x x y x y x y − − = − = − ⎛ ⎞ + ⎛ ⎞ = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + = − − = ⋅ + − = + − − = + − − = − At (4, 1), 1. 4 dy dx = 24. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 1 3 3 3 3 3 3 2 2 2 2 2 x y x y x y dy x y dy dx dx x y x y dy x y dy dx dx x y dy y dy x x y dx dx dy x y y x x xy y dx dy xy y y x y dx x xy x x y + = + + ⎛ + ⎞ = + ⎜ ⎟ ⎝ ⎠ + + + = + + − = − + ⎡ + − ⎤ = − + + ⎣ ⎦ − + + = =− + + At (−1, 1), dy 1. dx = − 25. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 3 2 3 3 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 4 4 2 4 2 2 2 2 2 2 2 y x y x y x y dy x y y dy x dx dx xy y dy x y dy y dy x dx dx dx dy y xy x xy dx dy x y dx y y x dy x y dx y y x + = ⎛ + ⎞ + + ⎛ ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + + + = + = − − = + − = + At (1, 1), 1. 3 dy dx = 122 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 26. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 3 2 2 3 2 2 3 2 3 2 3 2 2 3 2 2 2 2 3 2 8 2 2 2 8 16 4 4 4 4 8 16 4 4 8 16 4 4 4 4 4 2 4 2 x y xy x y x y dy x dy y x dx dx x x y dy xy y dy x dy xy dx dx dx dy xy y x xy x xy dx dy xy x xy dx x y y x dy x y x y dx x y y x + = + ⎛ + ⎞ = + ⎜ ⎟ ⎝ ⎠ + + + = + + − = − − − − = + − − − = + − At (2, 2), dy 0. dx = 27. 3 2 2 5 0 6 2 0 3 x y x dy dx dy x dx − + = − = = At (1, 4), dy 3. dx = 28. 4 2 2 1 0 8 2 0 8 4 2 x y x dy dx dy x x dx + − = + = = − = − dy ( 1) 4( 1) 4 dx − = − − = 29. 2 2 4 2 2 0 x y x y dy dx dy x dx y + = + = = − At (0, 2), dy 0. dx = 30. 4 2 9 2 36 8 18 0 4 9 x y x dy dx dy x dx y + = + = = − At 5, 4 , 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ( ) 4 5 5. 9 4 3 3 dy dx = − = − 31. ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 2 2 3 2 2 3 3 2 2 3 2 2 2 2 2 4 4 4 3 1 2 4 2 12 3 4 2 12 2 4 2 6 2 4 6 4 x y x y x x dy x x x y dx x y dy x x x dx x y dy x x dx x dy x x dx y x dy x x dx y x − = = − − − − = − − + = − − = − − = − − − = − − At (2, 7), dy 2. dx = 32. 2 3 2 2 0 2 3 0 2 3 x y x y dy dx dy x dx y − = − = = At (−1, 1), 2. 3 dy dx = − Section 2.7 Implicit Differentiation 123 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 33. Implicitly: 1 2 0 1 2 y dy dx dy dx y − = = Explicitly: ( ) ( ) ( ) ( ) 1 2 1 2 1 1 1 1 1 2 1 2 1 1 2 1 1 2 y x x dy x dx x x y − = ± − = ± − = ± − = ± − = ± − = At (2, −1), 1. 2 dy dx = − 34. Implicitly: 8 2 0 4 y dy x dx dy x dx y − = = Explicitly: ( ) ( ) ( ) 2 2 1 2 2 1 2 2 2 1 7 2 1 7 2 1 7 2 4 2 7 4 1 7 2 4 y x x dy x x dx x x x x x y − = ± + = ± + = ± + = ± + = ⎛± + ⎞ ⎜ ⎟ ⎝ ⎠ = At (3, 2), 3. 8 dy dx = 35. 2 2 100 2 2 0 x y x y dy dx dy x dx y + = + = = − At (8, 6): ( ) 4 3 6 4 8 3 4 50 3 3 m y x y x = − − = − − = − + At (−6, 8): 3 4 m = 8 3( 6) 4 3 25 4 2 y x y x − = + = + 36. 2 2 9 2 2 0 x y x y dy dx dy x dx y + = + = = − At (0, 3): ( ) 0 3 0 0 3 m y x y = − = − = At (2, 5): ( ) 2 2 5 5 5 5 2 5 2 5 2 5 9 5 5 5 m y x y x = − = − − = − − = − + y y = − y = x − 1 (2, −1) x − 1 2 1 −1 −2 2 3 4 x y y = − y = x 2 + 7 x 2 + 7 2 2 (3, 2) 2 2 −2 −4 −2 −4 4 4 x 24 −16 −24 16 (8, 6) (−6, 8) −9 −6 9 (0, 3) ( 2, 2 ) 6 5 124 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 37. 2 3 2 2 5 2 15 15 2 y x y dy x dx dy x dx y = = = At (1, 5): ( ) 15 3 5 2 5 2 5 3 5 1 2 3 5 5 2 2 m y x y x = = − = − = − At (1, − 5): 15 3 5 2 5 2 m − = = − 5 3 5 ( 1) 2 3 5 5 2 2 y x y x + = − − = − + 38. 4 2 5 4 4 2 0 4 2 2 4 2 xy x xdy y x dx dy y x y x dx x x + = + + = + + = − = − At (1, 1): ( ) 3 2 1 3 1 2 3 5 2 2 m y x y x = − − = − − = − + At (5, −1): ( ) 3 10 1 3 5 10 3 1 10 2 m y x y x = − + = − − = − + 39. 3 3 2 2 2 2 2 2 8 3 3 0 3 3 x y x y dy dx y dy x dx dy x dx y + = + = = − = − At (0, 2): ( ) 0 2 0 0 2 m dy dx y x y = = − = − = At (2, 0): m dy dx = is undefined. The tangent line is x = 2. 40. ( ) ( ) ( )() ( )( ) ( ) ( ) ( ) 3 3 3 3 3 3 2 2 2 2 2 2 6 1 6 1 1 6 1 1 6 1 6 11 1 6 3 6 1 3 6 1 6 6 6 1 7 3 6 1 x y xy y xy x y x x y x x dy x x y dx x y dy x x dx x dy dx y x + = − − = − − − = − + + = − − − + = − − − − = − = − − At (−1, 0): m dy dx = is undefined. The tangent line is x = −1. At (0, −1): 7 3 m dy dx = = − ( 1) 7( 0) 3 7 1 3 y x y x − − = − − = − − −30 −5 5 30 (1 , 5 ) (1 , − 5 ) 0 −2 8 2 (1, 1) (5, −1) −8 −5 8 (0, 2) (2, 0) 5 3 −2 −3 (−1, 0) (0, −1) 2 Section 2.7 Implicit Differentiation 125 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 41. ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 2 8 4 4 8 4 8 8 8 4 4 8 1 4 2 16 4 x y y x y y y x y x x dy x x dx dy x dx x − − − = − + = + = = = + + = − + = − + At (−2, 1): ( ) (( ) )2 2 16 2 32 1 2 4 64 2 m dy dx − = = − = = − + 1 1( ( 2)) 2 1 2 2 y x y x − = − − = + At 6, 1 : 5 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ( ) ( ) 2 2 16 6 96 3 6 4 1600 50 m dy dx = = − = − = − ⎡ + ⎤ ⎣ ⎦ 1 3 ( 6) 5 50 1 3 9 5 50 25 3 14 50 25 y x y x y x − = − − − = − + = − + 42. ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 3 2 2 2 2 2 4 4 3 1 2 4 2 6 2 4 6 4 y x x dy x x x y dx x dy x x y dx x dy x x dx y x = − − − − = − − = − − = − At (2, 2): ( ) 2 2 2 2 2 2 m y x y x = − = − = − At ( ) ( ) 2, 2 : 2 2 2 2 2 2 m y x y x − = − + = − − = − + 43. 3 2 , 0 0.00001 0.1 p x x x = ≥ + ( ) ( ) 3 2 2 2 2 2 2 0.00001 0.1 2 0.00003 0.1 2 0.00003 0.1 2 2 0.00003 0.1 x x p x dx dx dp dp p x dx dp p dx dp p x + = + = − + =− = − + 44. 2 4 , 0 0.000001 0.05 1 p x x x = ≥ + + ( ) ( ) 2 2 2 2 0.000001 0.05 1 4 0.000002 0.05 4 0.000002 0.05 4 4 0.000002 0.05 x x p x dx dx dp dp p x dx dp p dx dp p x + + = + = − + = − = − + 45. ( ) ( ) 2 2 2 2 200 , 0 200 2 2 200 2 2 2 2 1 4 4 2 1 p x x x xp x x p p dx dx dp dp p dx xp dp dx xp dp p − = < ≤ = − ⎛ ⎞ + ⎜ ⎟ = − ⎝ ⎠ + = − = − + 46. ( ) ( ) 2 2 2 2 500 , 0 500 2 2 500 2 2 2 2 1 4 4 2 1 p x x x xp x x p p dx dx dp dp dx p xp dp dx xp dp p − = < ≤ = − ⎛ ⎞ + ⎜ ⎟ = − ⎝ ⎠ + = − = − + 0 −3 4 (2, 2) (2, −2) 3 8 −5 −8 (−2, 1) (6, 1 ) 5 5 126 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 47. (a) ( ) 0.75 0.25 0.75 0.75 0.25 0.25 0.75 0.25 0.75 0.25 100 135,540 100 0.25 75 0 25 75 3 x y x y dy y x dx x dy y y dx x dy y dx x − − = ⎛ ⎞ + = ⎜ ⎟ ⎝ ⎠ ⋅ = − = − When x = 1500 and y = 1000, dy 2. dx = − (b) If more labor is used, then less capital is available. If more capital is used, then less labor is available. 48. (a) As price increases, the demand decreases. (b) For x > 0, the rate of change of demand, x, with respect to the price, p, is never decreasing; that is, for x > 0, dx dp is always increasing. 49. 2 3 2 2 2 1952.4 13.0345 168.469 465.66 2 39.1035 337.938 465.66 39.1035 337.983 465.66 2 y t t t y dy t t dt dy t t dt y − = − + = − + − + = (a) The number of cases of HIV AIDS decreases from 2004 through 2007, then begins to increase. (b) Using the graph, the number of reported cases was decreasing at the greatest rate during 2005. (c) The result is the same. Section 2.8 Related Rates Skills Warm Up 1. A = π r2 2. 3 43 V = π r 3. SA = 6s2 4. V = s3 5. 1 2 3 V = π r h 6. 1 2 A = bh 7. [ ] 2 2 2 2 9 9 2 2 0 2 2 2 2 x y d x y d dx dx x y dy dx y dy x dx dy x dx y x y + = ⎡⎣ + ⎤⎦ = + = = − − = − = 8. [ ] 2 2 3 6 3 6 3 3 2 0 3 2 3 2 3 3 xy x d xy x d dx dx y xdy x dx xdy x y dx dy x y dx x − = ⎡⎣ − ⎤⎦ = + − = = − − = 9. ( ) ( ) 2 2 2 12 2 12 2 2 0 2 2 2 2 2 2 x y xy d x y xy d dx dx x dy y xdy dx dx dy xdy y x dx dx dy x y x dx dy y x dx x + + = ⎡⎣ + + ⎤⎦ = + + + = + = − − + = − − − − = + t 4 5 6 7 8 y 44.11 41.06 38.46 37.50 39.21 y′ −2.95 −3.00 −2.01 0.22 3.38 0 0 2,000 100,000 9 37 3 45 Section 2.8 Related Rates 127 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. y = x, 1 1 2 1 , 2 2 dy x dx dx dt dt x dt = − = dx 2 x dy dt dt = (a) When x = 4 and dx 3, dt = 1 (3) 3. 2 4 4 dy dt = ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ (b) When x = 25 and dy 2, dt = dx 2 25(2) 20. dt = = 2. y = 2(x2 − 3x) = 2x2 − 6x, dy 4x dx 6 dx, dt dt dt = − 1 4 6 dx dy dt x dt = − (a) When x = 3 and dx 2, dt = dy 4(3)(2) 6(2) 12. dt = − = (b) When x = 1 and dy 5, dt = 1 (5) 5. 4 6 2 dx dt = = − − 3. xy = 4, x dy y dx 0, dt dt + = dy y dx, dt x dt = ⎛− ⎞ ⎜ ⎟ ⎝ ⎠ dx x dy dt y dt ⎛ ⎞ = ⎜− ⎟ ⎝ ⎠ (a) When x = 8, 1, 2 y = and dx 10, dt = 1 2(10) 5. 8 8 dy dt =− =− (b) When x = 1, y = 4, and dy 6, dt = − 1( 6) 3. 4 2 dx dt = − − = 4. x2 + y2 = 25, 2xdx 2y dy 0, dt dt + = dy x dx, dt y dt = − dx y dy dt x dt = − (a) When x = 3, y = 4, and dx 8, dt = 3(8) 6. 4 dy dt = − = − (b) When x = 4, y = 3, and dy 2, dt = − 3( 2) 3. 4 2 dx dt = − − = 5. A = π r2 , dr 3, dt = dA 2 r dr 6 r dt dt = π = π (a) When r = 6, dA 2 (6)(3) 36 in.2 min. dt = π = π (b) When r = 24, dA 2 (24)(3) 144 in.2 min. dt = π = π 6. 4 3, 3 V = π r dr 3, dt = dV 4 r2 dr 12 r2 dt dt = π = π (a) When r = 6, dV 12 (6)2 432 in.3 min. dt = π = π (b) When r = 24, dV 12 (24)2 6912 in.3 min. dt = π = π Skills Warm Up —continued— 10. [ ] ( ) 2 2 2 2 2 2 2 2 1 2 2 2 2 1 2 2 1 1 2 2 x xy y xy d x xy y d xy dx dx y xy dy y dy y x dy dx dx dx xy dy y dy xdy y y dx dx dx dy xy y x y y dx dy y y dx xy y x + − = ⎡⎣ + − ⎤⎦ = + + − = + − − = − − − − = − − − − = − − 128 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 7. A = π r2 , dA 2 r dr dt dt = π If dr dt is constant, then dA dt is not constant; dA dt is proportional to r. 8. 4 3, 3 V = π r dV 4 r2 dr dt dt = π If dr dt is constant, dV dt is not constant since it is proportional to the square of r. 9. 4 3, 3 V = π r dV 10, dt = dV 4 r2 dr , dt dt = π 2 1 4 dr dV dt π r dt = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (a) When r = 1, ( ) ( ) 2 1 10 5 ft min. 4 1 2 dr dt π π = = (b) When r = 2, ( ) ( ) 2 1 10 5 ft min. 4 2 8 dr dt π π = = 10. 1 2 1 2 (3 ) 3 3 3 V = π r h = π r r = π r dV 3 r2 dr 6 r2 dt dt = π = π (a) When r = 6, dV 6 (6)2 216 in.3 min. dt = π = π (b) When r = 24, dV 6 (24)2 3456 in.3 min. dt = π = π 11. (a) 0.75 0.75(150) 112.5 dollars per week dC dx dt dt = = = (b) ( ) ( )( ) 250 1 5 250 150 1 1000 150 5 7500 dollars per week dR dx x dx dt dt dt = − = − = (c) 7500 112.5 7387.5 dollars per week P R C dP dR dC dt dt dt = − = − = − = 12. (a) dC 1.05dx 1.05(250) 262.5 dollars week dt dt = = = (b) 2 2(5000) ( ) 500 500 250 25 25 25,000 dollars week dR x dx dt dt ⎛ ⎞ ⎛ ⎞ = ⎜ − ⎟ = ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ = (c) 25,000 262.5 24,737.5 dollars week P R C dP dR dC dt dt dt = − = − = − = 13. V = x3, dx 3, dt = dV 3x2 dx dt dt = (a) When x = 1, dV 3(1)2 (3) 9 cm3 sec. dt = = (b) When x = 10, dV 3(10)2 (3) 900 cm3 sec. dt = = 14. A = 6x2 , dx 3, dt = dA 12x dx 36x dt dt = = (a) When x = 1, dA 36(1) 36 cm2 sec. dt = = (b) When x = 10, dA 36(10) 360 cm2 sec. dt = = 15. 2 2 y x dy xdx dt dt = = (a) When x = −3 and dx 3 in. sec: dt = dy 2( 3)(3) 18 in. sec dt = − = − (b) When x = 0 and dx 3 in. sec: dt = dy 2(0)(3) 0 in. sec dt = = (c) When x = 1 and dx 3 in. sec: dt = dy 2(1)(3) 6 in. sec dt = = (d) When x = 3 and dx 3 in. sec: dt = dy 2(3)(3) 18 in. sec dt = = Section 2.8 Related Rates 129 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 16. ( ) ( )( ) ( ) ( ) 2 1 2 2 2 2 1 1 1 2 2 1 y x dy x x dx dt dt dy x dx dt x dt − − = + = − + = − + (a) When x = −2 and dx 2 in. sec: dt = ( ) ( ) ( ) 2 2 2 2 8 2 in.sec 1 2 25 dy dt − = − = ⎡ + − ⎤ ⎣ ⎦ (b) When x = 0 and dx 2 in. sec: dt = ( ) ( ) ( ) 2 2 2 0 2 0 in.sec 1 0 dy dt = − = ⎡ + ⎤ ⎣ ⎦ (c) When x = 6 and dx 2 in. sec: dt = ( ) ( ) ( ) 2 2 2 6 24 2 in. sec 1 6 1369 dy dt = − = − + (d) When x = 10 and dx 2 in. sec: dt = ( ) ( ) ( ) 2 2 2 10 40 2 in. sec 1 10 10,201 dy dt = − = − + 17. Let x be the distance from the boat to the dock and y be the length of the rope. 122 2 2 4 2 2 x y dy dt xdx y dy dt dt dx y dy dt x dt + = = − = = When y = 13, x = 5 and 13( 4) 10.4 ft sec. 5 dx dt = − = − As x → 0, dx dt increases. 18. (a) 15 15 15 6 6 9 15 5 3 y y x y y x y x y x = ⇒ − = − = = Find dy dt if dx 5 ft sec dt = when x = 10 ft. ( ) 5 3 5 5 25 ft sec 3 3 dy dx dt dt dy dt = = = (b) Find d (y x) dt − if dx 5 ft sec dt = and 25 ft sec 3 dy dt = when x = 10 ft. ( ) 25 5 10 ft sec 3 3 d y x dy dx dt dt dt − = − = − = 19. 2 62 2 2 2 x s x dx s ds dt dt dx s ds dt x dt + = = = When s = 10, x = 8 and ds 240: dt = 10( 240) 300 mi hr. 8 dx dt = − = 20. s2 = 902 + x2 , x = 26, dx 30 dt = − 2s ds 2x dx dt dt ds x dx dt s dt = = When x = 26, ( ) 2 2 26 30 8.33 ft sec. 90 26 ds dt = − ≈ − + y x 15 ft 6 ft 130 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 21. (a) L2 = x2 + y2 , dx 450, dt = − dy 600, dt = − and dL x(dx dt) y(dy dt) dt L + = When x = 150 and y = 200, L = 250 and 150( 450) 200( 600) 750 mph. 250 dL dt − + − = =− (b) 250 1 hr 20 min 750 3 t = = = 22. 2250 50 0.35 2 50 0.70 S x x dS dx xdx dt dt dt = + + = + 50(125) 0.70(1500)(125) $137,500 per week dS dt = + = 23. V = π r2h, h = 0.08, V = 0.08π r2 , dV 0.16 r dr dt dt = π When r = 150 and 1, 2 dr dt = 0.16 (150) 1 12 37.70 ft3 min. 2 dV dt = π ⎛⎜ ⎞⎟ = π = ⎝ ⎠ 24. ( ) ( 2 ) 2 2 2 50 0.01 4000 40 0.02 50 0.01 4000 40 0.02 0.01 10 4000 P R C xp C x x x x x x x x x x = − = − = − − + − = − − − + = + − dP 0.02x dx 10 dx dt dt dt = + When x = 800 and dx 25, dt = dP 0.02(800)(25) (10)(25) $650 week. dt = + = 25. ( ) ( ) 2 6000 25 2400 5200 25 3600 5200 P R C xp C x x x x x = − = − = − − + = − + − 50 3600 1 3600 50 dP xdx dx dt dt dt dx dP dt x dt = − + = − When x = 44 and dP 5600, dt = ( )( ) 1 5600 4 units per week. 3600 50 44 dx dt = = − 26. (a) For supply, if dx dt is negative, then dp dt is negative. For demand, if dx dt is negative, then dp dt is positive. (b) For supply, if dp dt is positive, then dx dt is positive. For demand, if dp dt is positive, then dx dt is negative. Review Exercises for Chapter 2 1. Slope 4 2 2 − ≈ = − 2. Slope 4 2 2 ≈ = 3. Slope ≈ 0 4. Slope 2 1 4 2 − ≈ = − 5. Answers will vary. Sample answer: t = 4; slope ≈ $290 million yr; Sales were increasing by about $290 million yr in 2004. t = 7; slope ≈ $320 million yr; Sales were increasing by about $320 million yr in 2007. 100 200 100 200 y L x x y 0 0 1,600 50,000 Review Exercises for Chapter 2 131 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 6. Answers will vary. Sample answer: t = 5: slope ≈ 25 million acres yr; The number of acres of farmland in the United States is increasing by about 25 million acres yr in 2005. t = 8: slope ≈ 50 million acres yr; The number of acres of farmland in the United States is increasing by about 50 million acres yr in 2008. 7. Answers will vary. Sample answer: t = 1: m ≈ 65 hundred thousand visitors month; The number of visitors to the national park is increasing at about 65,000,000 month in January. t = 8: m ≈ 0 visitors month; The number of visitors to the national park is neither increasing nor decreasing in August. t = 12: m ≈ −1000 hundred thousand month; The number of visitors to the national park is decreasing at about 1,000,000,000 visitors per month in December. 8. (a) At t1, the slope of g(t) is greater than the slope of f (t), so the rafter whose progress is given by g(t) is traveling faster. (b) At t2 , the slope of f (t) is greater than the slope of g(t), so the rafter whose progress is given by f (t) is traveling faster. (c) At t3, the slope of f (t) is greater than the slope of g(t), so the rafter whose progress is given by f (t) is traveling faster. (d) The rafter whose progress is given by f (t) finishes first. The value of t where f (t) = 9 is smaller than the value of t where g(t) = 9. 9. ( ) ( ) ( ) ( ) ( ) 0 0 0 lim 3 5 3 5 lim lim 3 3 x x x f x x f x f x x x x x x x x Δ → Δ → Δ → + Δ − ′ = Δ − + Δ − − − − = Δ − Δ = = − Δ f ′(−2) = −3 10. ( ) ( ) ( ) ( ) ( ) 0 0 0 lim 7 3 7 3 lim lim 7 7 x x x f x x f x f x x x x x x x x Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ Δ = = Δ f ′(−1) = 7 11. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 lim 4 4 lim 2 4 lim 2 4 lim lim 2 4 2 4 x x x x x f x x f x f x x x x x x x x x x xx x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − + Δ − − = Δ + Δ + Δ − Δ − = Δ Δ + Δ − Δ = Δ = + Δ − = − f ′(1) = 2(1) − 4 = −2 12. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 2 0 0 lim 10 10 lim 2 lim 2 lim lim 2 2 x x x x x f x x f x f x x x x x x x xx x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ + Δ + Δ − = Δ Δ + Δ = Δ = + Δ = f ′(2) = 4 13. ( ) ( ) ( ) ( ) ( ) 0 0 0 0 lim lim 9 9 9 9 9 9 9 9 lim 9 9 lim 1 1 9 9 2 9 x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + + Δ + + + = ⋅ Δ + Δ + + + + Δ + − + = Δ ⎡ + Δ + + + ⎤ ⎣ ⎦ = = + Δ + + + + ( 5) 1 4 f ′ − = 132 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 14. ( ) ( ) ( ) ( ) ( ) 0 0 0 0 lim lim 1 1 1 1 1 1 1 1 lim 1 1 lim 1 1 1 1 2 1 x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − = Δ + Δ − − − + Δ − + − = ⋅ Δ + Δ − + − + Δ − − − = Δ ⎡ + Δ − + − ⎤ ⎣ ⎦ = = + Δ − + − − (10) 1 6 f ′ = 15. ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 0 0 0 0 2 lim 1 1 lim 5 5 5 5 lim 5 5 lim 1 1 5 5 5 x x x x f x x f x f x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = + Δ − − Δ − − + Δ − = Δ + Δ − − − = =− + Δ − − − f ′(6) = −1 16. ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 0 0 0 0 2 lim 1 1 lim 4 4 4 4 lim 4 4 lim 1 4 4 1 4 x x x x f x x f x f x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = + Δ + + Δ + − + Δ + = Δ ⎡⎣ + Δ + + ⎤⎦ − = + Δ + + = − + f ′(−3) = −1 17. f (x) = 9x + 1 ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 lim 9 1 9 1 lim lim 9 9 1 9 1 lim 9 lim 9 9 x x x x x f x x f x f x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎣⎡ + Δ + ⎤⎦ − + = Δ + Δ + − − = Δ Δ = Δ = = 18. f (x) = 1 − 4x ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 lim 1 4 1 4 lim lim 1 4 4 1 4 lim 4 lim 4 4 x x x x x f x x f x f x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡⎣ − + Δ ⎤⎦ − − = Δ − − Δ − + = Δ − Δ = Δ = − = − 19. ( ) 1 2 2 2 f x = − x + x ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 0 0 1 1 2 2 1 1 1 2 2 2 1 2 1 2 lim 2 2 lim 2 2 2 lim 2 lim lim 2 2 x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡− + Δ + + Δ ⎤ − − + = ⎣ ⎦ Δ − − Δ − Δ + + Δ + − = Δ Δ − − Δ + = Δ = − − Δ + = − + Review Exercises for Chapter 2 133 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 20. f (x) = 4 − x2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 2 2 0 0 0 lim 4 4 lim 4 2 4 lim 2 lim lim 2 2 x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡ − + Δ ⎤ − − = ⎣ ⎦ Δ − − Δ − Δ − + = Δ Δ − − Δ = Δ = − −Δ = − 21. f (x) = x − 5 ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 lim 5 5 5 5 lim 5 5 5 5 lim 5 5 lim 1 1 5 5 2 5 x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − − − + Δ − + − = ⋅ Δ + Δ − + − + Δ − − − = Δ +Δ − + − = = + Δ − + − − 22. f (x) = x + 3 ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 lim 3 3 lim lim lim lim 1 1 2 x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ ⎡ + Δ + ⎤ − + = ⎣ ⎦ Δ + Δ − + Δ + = ⋅ Δ + Δ + + Δ − = Δ + Δ + = = + Δ + 23. f (x) 5 x = ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 0 2 lim 5 5 lim 5 5 lim lim 5 lim 5 5 x x x x x f x x f x f x x x x x x x x x x x x x x x x x x xx x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − + Δ = Δ − + Δ + Δ = Δ − Δ = Δ ⎡⎣ + Δ ⎤⎦ = − = − + Δ 134 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 24. ( ) 1 4 f x x = + ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) 0 0 0 0 0 2 lim 1 1 lim 4 4 4 4 4 4 lim lim 4 4 lim 1 1 4 4 4 x x x x x f x x f x f x x x x x x x x x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → Δ → + Δ − ′ = Δ − = + Δ + + Δ + − + Δ + + + Δ + = Δ −Δ = Δ ⎡⎣ + + Δ + ⎤⎦ = − = − + + Δ + + 25. y is not differentiable at x = 1, a discontinuity. 26. y is not differentiable at x = 0, a node. 27. y is not differentiable at x = 0, a discontinuity. 28. y is not differentiable at x = −1, a cusp. 29. 6 0 y y = − ′ = 30. ( ) ( ) 5 0 f x f x = ′ = 31. ( ) ( ) 3 3 2 f x x f x x = ′ = 32. ( ) ( ) ( ) ( ) 4 4 5 5 1 4 4 h x x h x x h x x h x x − − − = = ′ = − ′ − = 33. ( ) ( ) 4 2 8 fx x f x x = ′ = 34. ( ) ( ) 5 4 6 30 gt t gt t = ′ = 35. ( ) ( ) 4 3 2 5 8 5 f x x f x x = ′ = 36. 2 3 1 3 1 3 3 2 2 y x y x y x − = ′ = ′ = 37. ( ) ( ) 4 2 3 2 3 8 6 gx x x gx x x = + ′ = + 38. ( ) ( ) 6 2 4 12 4 fx x x f x x = − ′ = − 39. 2 6 7 2 6 y x x y x = + − ′ = + 40. 4 3 3 2 2 3 8 9 1 y x x x y x x = − + ′ = − + 41. ( ) ( ) ( ) ( ) ( ) 1 2 3 2 3 2 2 ; 4, 1 4 4 0.125 f x x f x x f − − − = ′ = − ′ = − = − 42. ( ) 1 2 2 1 2 2 3 3; 1, 6 2 2 3 3 2 3 3 2 2 1 3 6 2 2 y x y x y x x y − − = + ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = + ′ = − = − ′⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ 43. () ( ) ( ) ( ) ( ) ( ) 3 2 2 2 4 6 8; 1, 9 3 8 6 1 3 1 8 1 6 5 g x x x x g x x x g = − − + − ′ = − − ′ − = − − − − = 44. ( ) ( ) ( ) ( ) ( ) 4 3 2 3 2 3 2 2 5 6 ; 1, 2 8 15 12 1 1 81 151 121 1 4 y x x x x y x x x y = − + − ′ = − + − ′ = − + − = 45. ( ) ( ) 4 3 2 5 f x x f ′ = − ′ = 3 5( 2) 5 7 y x y x − = − = − 46. ( ) ( ) 44 3 10 1 34 7 34 1 34 27 y x x y y x y x ′ = − ′ − = − − = − + = − − −3 5 −2 10 (2, 3) −2 −2 0 20 (−1, 7) Review Exercises for Chapter 2 135 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 47. ( ) ( ) ( ) 1 2 1 2 1 2 3 2 3 2 1 1 1 2 2 1 1 2 2 1 1 fx x x x x f x x x x x f − − − = − = − ′ = + = + ′ = 0 1( 1) 1 y x y x − = − = − 48. ( ) ( ) 2 4 4 4 f x x f ′ = − − ′ − = 4 4( 4) 4 12 y x y x + = + = + 49. ( ) 4 3 2 3 2 0.7500 13.278 74.50 440.2 523 3 39.834 149 440.2 S t t t t ds s t t t t dt = − + − + + = ′ = − + − + (a) 2004: s′(4) = 289.544 2007: s′(7) = 320.066 (b) Results should be similar. (c) The slope shows the rate at which sales were increasing or decreasing in that particular year or value of t. 50. ( ) 3 2 2 0.0991 1.512 4.01 933.9 0.2973 3.024 4.01 L t t t dL L t t t dt = + + + = ′ = + + (a) 2005: L′(5) = 26.5625 2008: L′(8) = 47.2292 (b) The results should be similar. (c) The slope shows the rate at which the number of acres of farmland in millions is increasing or decreasing in that particular year or value of t. 51. Average rate of change: ( ) ( ) ( ) ( ) 1 3 7 9 4 1 3 4 f − f − − − = = − − Instantaneous rate of change: ( ) ( ) ( ) 4 1 4 3 4 f t f f ′ = ′ = ′ − = 52. Average rate of change: (8) (1) 4 1 3 8 1 7 7 f − f − = = − Instantaneous rate of change: ( ) ( ) ( ) 1 3 2 3 8 1 3 1 2 3 f x x f f ′ = ′ = ′ = 53. Average rate of change: ( ) ( ) ( ) 1 0 0 4 4 1 0 1 f − f − − = = − Instantaneous rate of change: ( ) ( ) ( ) 2 3 1 5 0 3 f x x f f ′ = + ′ = ′ = −1 −2 3 2 (1, 0) 4 −6 −8 2 (−4, −4) 1 −9 −3 9 3 −1 −3 3 5 −7 −5 7 136 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 54. Average rate of change: ( ) ( ) ( ) 2 2 10 ( 10) 5 2 2 4 f − f − − − = = − − Instantaneous rate of change: ( ) ( ) ( ) 3 2 1 2 13 2 13 f x x f f ′ = + ′ = ′ − = 55. s(t) = −16t2 + 24t + 300 (a) Average velocity (2) (1) 284 308 24 ft sec 2 1 1 s − s − = = = − − (b) ( ) ( ) ( ) ( ) 32 24 1 8 ftsec 3 72 ftsec v t s t t v v = ′ = − + = − = − (c) ( ) ( ) ( ) ( )( ) ( ) 2 2 2 0 16 24 300 0 16 24 300 0 24 24 4 16 300 24 19,776 2 16 32 5.14 sec s t t t t t t t = − + + = − − = − − ± − − − ± = = ≈ (d) v(5.14) = −140.48 ft sec 56. (a) ( ) ( ) ( ) 16 2 276 32 s t t v t s t t = − + = ′ = − (b) Average velocity (2) 2(0) 2 0 212 276 2 32 ft sec s − = − − = = − (c) ( ) ( ) ( ) 32 2 64 ft sec 3 96 ft sec v t t v v = − = − = − (d) ( ) 2 2 2 0 16 276 0 16 276 276 16 4.15 sec s t t t t t = − + = = = ≈ (e) v(4.15) = −132.8 ft sec 57. 2500 320 320 C x dC dx = + = 58. 3 2 3 24,000 9 C x dC x dx = + = 59. ( )( ) 1 2 1 2 370 2.55 370 2.25 1 2.55 1.275 2 C x x dC x dx x − = + = + = = 60. 1 3 3 5.25 2 3.5 3 dC x dx x = ⎛⎜ − ⎞⎟ = ⎝ ⎠ 61. 150 0.6 2 150 1.2 R x x dR x dx = − = − −6 6 −4 4 Review Exercises for Chapter 2 137 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 62. 150 3 2 4 R = x − x 150 3 2 dR x dx = − 63. 3 2 2 4 2 100 12 4 100 R x x x dR x x dx = − + + = − + + 64. 1 2 1 2 4 10 4 5 R x x dR dx x = + = + 65. 3 2 2 0.0002 6 2000 0.0006 12 1 P x x x dP x x dx = − + − − = − + − 66. 3 2 2 1 4000 120 144,000 15 1 8000 120 5 P x x x dP x x dx = − + − − = − + − 67. P = −0.05x2 + 20x − 1000 (a) Find P for 100 x 101. x Δ ≤ ≤ Δ (101) (100) 509.95 500 $9.95 101 100 P − P = − = − (b) Find dP dx when x = 100. dP 0.1x 20 P (x) dx = − + = ′ When x = 100, dP P (100) $10. dx = ′ = (c) Parts (a) and (b) differ by only $0.05. 68. P = −0.007t2 + 2.78t + 123.6 (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 123.6 5 137.325 10 150.7 15 163.725 20 176.4 25 188.725 30 200.7 P P P P P P P = = = = = = = These are the populations in millions for Brazil from 1980 to 2010. (b) dP 0.014t 2.78 P (t) dt = − + = ′ (c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2.78 5 2.71 10 2.64 15 2.57 20 2.5 25 2.43 30 2.36 P P P P P P P ′ = ′ = ′ = ′ = ′ = ′ = ′ = These are the rates at which the population of Brazil is changing in millions per year from 1980 to 2010. 69. ( ) ( ) ( ) ( ) 3 2 3 5 2 4 2 2 5 3 5 3 15 15 15 1 f x x x x x f x x x x x = − = − ′ = − = − Simple Power Rule 70. y = (3x2 + 7)(x2 − 2x) ( 2 )( ) ( )( 2 ) 3 2 3 2 3 2 3 7 2 2 6 2 6 14 6 14 6 12 12 18 14 14 y x x xx x x x x x x x x x ′ = + − + − = + − − + − = − + − Product Rule 71. y = (4x − 3)(x3 − 2x2 ) ( )( 2 ) ( 3 2 ) 3 2 3 2 3 2 4 3 3 4 4 2 12 25 12 4 8 16 33 12 y x x x x x x x x x x x x x ′ = − − + − = − + + − = − + Product Rule 72. ( ) ( )( ) ( )( ) ( )( ) 2 2 2 2 2 2 3 1 2 1 2 2 4 1 3 4 3 4 2 3 3 2 8 12 2 3 2 6 8 12 3 s t t t t t t s t t t t t t t t t t t t − − − − − − − − = ⎛ − ⎞ − = − − ⎜ ⎟ ⎝ ⎠ ′ = − − + − = − − + + − = − − Product Rule 73. ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 3 3 1 1 3 3 3 g x x x x x g x x g x x = + + − ′ = + ′ = + Quotient Rule 138 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 74. ( ) 2 5 3 1 f x x x − = + ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 3 1 5 2 5 3 3 1 15 5 6 15 3 1 11 3 1 x x f x x f x x x x f x x + − − − ′ = + ′ − − − + = + ′ = − + Quotient Rule 75. ( ) 2 6 5 1` f x x x − = + ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 1 6 6 5 2 1 6 10 6 1 2 3 5 3 1 x x x f x x x x x x x x + − − ′ = + + − = + + − = + Quotient Rule 76. ( ) 2 2 1 1 f x x x x + − = − ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 3 2 3 2 2 2 2 2 2 2 2 2 1 2 1 1 2 1 2 2 1 2 2 2 1 1 1 1 1 x x x x x f x x x x x x x x x x x x x − + − + − ′ = − + − − − − + = − − − + = = − − − Quotient Rule 77. ( ) ( )f x = 5x2 + 2 3 ( ) ( ) ( ) ( ) 2 2 2 2 3 5 2 10 30 5 2 fx x x x x ′ = + = + General Power Rule 78. ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 1 3 2 2 3 2 2 3 1 1 1 1 2 2 3 3 1 f x x x f x x x x x − = − = − ′ = − = − General Power Rule 79. ( ) ( ) ( ) ( ) ( ) 1 2 3 2 3 2 2 2 1 1 2 1 1 2 1 1 h x x x h x x x − − = = + + ′ = ⎛− ⎞ + ⎜ ⎟ ⎝ ⎠ = − + General Power Rule 80. ( ) ( ) ( ) ( ) ( ) 6 3 6 3 1 2 6 3 1 2 5 2 5 2 6 3 12 9 12 9 1 12 9 6 36 2 3 18 12 9 g x x x x x g x x x x x x x x x − = − + = − + ′ = − + − − = − + General Power Rule 81. ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 1 2 2 1 2 2 1 2 2 1 2 2 2 2 2 1 1 1 1 2 1 1 2 1 1 2 1 1 g x x x x x g x x x x x x x x x x − − = + = + ′ = ⎡ + ⎤ + + ⎢⎣ ⎥⎦ = + ⎡ + + ⎤ ⎣ ⎦ + = + Product and General Power Rules 82. ( ) ( )3 1 g t t t = − ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 6 3 2 6 4 4 1 1 31 1 1 1 31 1 1 3 2 1 1 1 t t t g t t t t t t t t t t t − − − − ′ = − − + − = − − + + = = − − Quotient Rule 83. ( ) ( )f x = x 1 − 4x2 2 ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 1 4 8 1 4 16 1 4 1 4 1 4 16 1 4 1 4 1 20 f x x x x x x x x x x x x x ′ = − − + − = − − + − = − ⎡− + − ⎤ ⎣ ⎦ = − − Product and General Power Rules Review Exercises for Chapter 2 139 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 84. ( ) ( ) ( ) ( ) ( ) 5 2 2 1 5 2 1 4 2 4 2 2 1 5 2 5 1 2 1 f x x x x x f x x x x x x x x x − − − = ⎛ + ⎞ = + ⎜ ⎟ ⎝ ⎠ ′ = + − = ⎛ + ⎞ ⎛ − ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ General Power Rule 85. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 6 3 6 2 5 3 5 2 5 2 2 3 2 3 3 2 3 2 6 2 3 6 2 3 2 3 18 2 3 1 h x x x x x h x x x x x x x x x x x x = ⎡⎣ + ⎤⎦ = + ′ = ⎡ + ⎤ + + ⎣ ⎦ = + ⎡⎣ + + ⎤⎦ = + + Product and General Power Rules 86. ( ) ( )( ) 2 f x = ⎣⎡ x − 2 x + 4 ⎦⎤ ( ) ( )( ) ( ) ( ) ( )( )( ) ( )( )( ) 2 2 4 2 4 2 2 4 2 2 4 2 4 1 f x x x x x x x x x x x ′ = ⎡⎣ − + ⎤⎦⎡⎣ − + + ⎤⎦ = − + + = − + + Product and General Power Rules 87. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 5 2 4 5 4 4 1 5 1 2 1 1 5 2 1 1 7 2 f x x x f x x x x x x x x x x x x = − ′ = − + − = − ⎡⎣ + − ⎤⎦ = − − Product and General Power Rules 88. ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 5 2 3 2 3 2 2 2 5 2 2 2 3 2 2 2 2 2 3 2 2 5 2 1 1 2 3 1 1 5 3 1 1 8 3 f s s s f s s s s s s s s s s s s s = − ′ = − + − = − ⎡ + − ⎤ ⎣ ⎦ = − − Product and General Power Rules 89. ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 1 2 2 2 2 12 1 2 4 1 2 3 3 3 1 3 1 1 3 1 3 1 3 12 3 1 3 3 1 2 1 3 3 1 3 3 1 1 3 32 3 16 1 3 3 9 5 2 3 11 3 t t h t t t t t t t h t t t t t t t t t − − + + = = − − − + − + − − ′ = − + ⎡⎣ − + + ⎤⎦ = − + = + − Quotient and General Power Rules 90. ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) 2 2 2 2 2 2 2 2 4 2 2 3 2 2 3 2 2 3 3 1 1 1 2 3 1 3 3 1 2 12 1 3 1 6 1 3 14 1 3 1 6 4 6 1 2 3 1 3 2 3 1 x g x x x x x x x g x x x x x x x x x x x x x x x + = + + + − + + ′ = + + ⎡ + − + ⎤ = ⎣ ⎦ + + − − + = + + + − = − + Quotient and General Power Rules 140 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 91. ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 2 2 2 1300 1300 2 25 2 25 1300 2 25 2 2 2600 1 2 25 T t t t t T t t t t t t t − − = = + + + + ′ = − + + + + = − + + (a) (1) 325 6.63 F hr 49 T′ = − ≈ − ° (3) 13 6.5 F hr 2 T′ = − ≈ − ° (5) 13 4.33 F hr 3 T′ = − ≈ − ° (10) 1144 1.36 F hr 841 T′ = − ≈ − ° (b) The rate of decrease is approaching zero. 92. When L = 12, ( ) ( ) ( ) ( ) 4 2 12 4 2 3 4 2 16 16 4 3 4 . 2 V L D D D dV D dD = − = − = − = − (a) When 8, 3 (8 4) 6 board ft in. 2 D dV dD = = ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ (b) When D = 16, 3 (16 4) 18 board ft in. 2 dV dD = ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ (c) When D = 24, 3 (24 4) 30 board ft in. 2 dV dD = ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ (d) When D = 36, 3 (36 4) 48 board ft in. 2 dV dD = ⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ 93. ( ) ( ) ( ) 3 2 7 1 6 7 6 f x x x f x x f x = + + ′ = + ′′ = 94. ( ) ( ) ( ) ( ) 4 2 3 2 2 5 6 2 20 12 2 60 12 12 5 1 f x x x x f x x x f x x x ′ = − + ′′ = − + ′′′ = − = − 95. ( ) ( )( ) ( )( ) 4 4 5 5 6 6 6 24 120 120 f x x f x x f x x x − − − ′′′ = − = = − = − 96. ( ) ( ) ( ) ( ) ( )( ) 1 2 1 2 3 2 5 2 4 7 2 7 2 1 2 1 4 3 8 15 15 16 16 f x x x f x x f x x f x x f x x x − − − − = = ′ = ′′ = − ′′′ = = − = − 97. ( ) ( ) 5 2 3 2 7 35 2 f x x f x x ′ = ′′ = 98. ( ) ( ) ( ) 2 2 1 2 3 3 3 3 2 3 2 6 2 6 f x x x x x f x x x f x x x − − − = + = + ′ = − ′′ = + = + 99. ( ) ( ) 3 13 2 3 2 3 6 6 2 2 f x x x f x x x − ′′ = = ′′′ = = 100. ( ) ( )( ) ( )( ) 4 4 3 3 4 3 4 5 2 5 2 5 20 2 20 2 80 6 240 24 240 24 f x x x x x f x x x f x x x x x − − − ′′′ = − = − = + = − = − 101. (a) s(t) = −16t2 + 5t + 30 (b) s(t) = 0 = −16t2 + 5t + 30 Using the Quadratic Formula, t ≈ 1.534 seconds. (c) ( ) ( ) ( ) 32 5 1.534 44.09 ft sec v t s t t v = ′ = − + ≈ − (d) a(t) = v′(t) = −32 ft sec2 102. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 4 4 1 1 2 1 2 1 2 1 6 1 6 1 s t t t t v t s t t t a t v t t t − − − = = + + + ′ = = − + = −+ ′ = = + =+ 0 0 24 60 Review Exercises for Chapter 2 141 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 103. ( ) ( ) 2 3 2 2 2 2 3 10 2 3 3 3 0 3 3 2 3 2 3 2 3 3 3 3 x xy y x xdy y y dy dx dx dy x y x y dx dy x y x y dx x y x y + + = + + + = + = − − − − + = = − + + 104. ( ) 2 9 2 0 2 9 9 2 0 9 2 2 9 2 9 2 9 9 2 9 2 x xy y x y x dy y dy dx dx x y dy x y dx dy x y x y dx x y x y + + = + + + = + =− − − − + = = − + + 105. ( ) 2 2 8 9 1 0 2 2 8 9 0 2 9 2 8 2 8 2 9 y x x y y dy x dy dx dx y dy x dx dy x dx y − + − − = − + − = − = − − = − 106. ( ) 2 2 6 2 5 0 2 2 6 2 0 2 6 2 2 2 2 1 2 6 3 y x y x y dy x dy dx dx dy y x dx dy x x dx y y + − − − = + − − = − = − − − = = − − 107. ( ) 2 2 1 2 1 2 1 1 1 2 1 y x y y dy dy dx dx y dy dy dx dx y dy dx dy dx y = − = − + = + = = + At (2, 1), 1. 3 dy dx = 1 1( 2) 3 1 1 3 3 y x y x − = − = + 108. 1 3 1 2 2 3 1 2 1 2 2 3 2 3 10 2 3 0 3 2 4 9 x y x y dy dx dy y dx x − − + = + = = − At (8, 4), 2. 9 dy dx = − 4 2( 8) 9 2 52 9 9 y x y x − = − − = − + 109. ( ) 2 2 2 2 2 2 2 2 y x xy y dy xdy y dx dx y x dy y dx dy y dx y x − = − = + − = + + = − At (1, 2), 4. 3 dy dx = 2 4( 1) 3 4 2 3 3 y x y x − = − = + 142 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 110. ( ) 3 2 2 2 2 2 2 2 2 2 2 2 2 3 1 3 2 4 6 3 0 3 2 6 4 3 4 3 3 2 6 y xy xy y dy x dy xy xy dy y dx dx dx dy y x xy xy y dx dy xy y dx y x xy − + = − − − + + = − + = − − = − + At (0, −1), dy 1. dx = − 1 1( 0) 1 y x y x + = − − = − − 111. 2 2 A r dA r dr dt dt π π = = (a) Find dA dt when r = 3 in. and dr 2 in. min. dt = ( )( ) 2 2 2 3 2 12 in. min 37.7 in. min dA dt = π = π ≈ (b) Find dA dt when r = 10 in. and dr 2 in. min. dt = ( )( ) 2 2 2 10 2 40 in. min 125.7 in. min dA dt = π = π ≈ 112. 1 2 1 1 2 2 1 2 y x dy x dx dt dt dy dx dt x dt − = = = (a) Find dy dt when 1 4 x = and dx 3 cm sec. dt = ( ) 1 4 1 3 3 cmsec 2 dy dt = = (b) Find dy dt when x = 1 and dx 3 cm sec. dt = 1 (3) 3 cm sec 2 1 2 dy dt = = (c) Find dy dt when x = 4 and dx 3 cm sec. dt = 1 (3) 3 cm sec 2 4 4 dy dt = = 113. Let b be the horizontal distance of the water and h be the depth of the water at the deep end. Then b = 8h for 0 ≤ h ≤ 5. ( ) ( ) ( ) 1 20 10 10 8 80 2 2 160 1 1 10 1 160 160 16 V bh bh hh h dV hdh dt dt dh dV dt h dt h h = = = = = = = = When h = 4, ( ) 1 1 ft min. 16 4 64 dh dt = = 114. (a) ( ) ( ) 2 211 0.002 30 1,500,000 181 0.002 1,500,000 P R C xp C x x x x x = − = − = − − + = − − (c) (b) ( ) ( ) 181 0.004 80,000 $ 139 per unit P x x P ′ = − ′ = − The maximum profit occurs for x = 45,250, which means p = 211 − 0.002(45,250) = $120.50. 0 0 100,000 5,000,000 Chapter 2 Test Yourself 143 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 2 Test Yourself 1. ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 0 2 0 0 lim 1 1 lim 2 lim lim 2 2 x x x x f x x f x f x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ + − + = Δ Δ + Δ = Δ = + Δ = At (2, 5): m = 2(2) = 4 2. ( ) ( ) ( ) ( ) 0 0 0 0 lim 2 2 lim lim lim 1 1 2 x x x x f x x f x f x x x x x x x x x x x x x x Δ → Δ → Δ → Δ → + Δ − ′ = Δ + Δ − − − = Δ + Δ − = Δ = + Δ + = At (4, 0): 1 1 2 4 4 m = = 3. ( ) ( ) 3 2 2 3 2 f t t t f t t = + ′ = + 4. ( ) ( ) 4 2 8 1 8 8 f x x x f x x = − + ′ = − 5. ( ) ( ) 3 2 3 1 2 3 2 2 f x x f x x x = ′ = = 6. ( ) ( )( ) ( ) ( ) 2 3 2 2 3 2 5 6 3 10 6 fx x x x f x x x x f x x x = + + = + + ′ = + + (Or use the Product Rule.) 7. ( ) ( ) 3 4 4 3 9 9 f x x f x x x − − = − ′ = = 8. ( ) ( ) ( ) 1 2 3 2 1 2 1 2 5 5 5 3 5 3 2 2 2 2 f x x x x x f x x x x x − = + = + ′ = + = + 9. ( ) ( ) ( ) ( )( ) 2 2 2 3 3 4 2 3 4 6 36 48 f x x fx x x x x = + ′ = + = + 10. ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 11 2 2 2 1 1 2 fx x x f x x x − = − = − ′ = − − = − − 11. ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 2 2 2 2 2 5 1 3 5 1 5 5 1 5 1 15 5 1 5 1 10 1 x f x x x x x f x x x x x x x x x − = − − − ′ = − ⎡⎣ − − ⎤⎦ = − + = 12. ( ) ( ) ( ) 2 2 1 1 1 1 1 1 2 1 f x x x f x x f = − ′ = + ′ = + = 0 2( 1) 2 2 y x y x − = − = − 13. S = −1.3241t3 + 26.562t3 − 155.81t + 314.3 (a) S t Δ Δ for 5 ≤ t ≤ 8 (8) (5) 89.8488 33.7875 8 5 3 $18.6871 billion yr S − S − = − = (b) ( ) ( ) ( ) 3.9723 2 53.124 155.81 5 $10.5025 billion yr 8 $14.9548 billion yr S t t t S S ′ = − + − ′ = ′ = (c) The annual sales of CVS Caremark from 2005 to 2008 increased on average by about $18.69 billion/year, and the instantaneous rates of change for 2005 and 2008 are $10.50 billion/year and $14.95 billion/year, respectively. −6 6 −4 4 (1, 0) 144 Chapter 2 Differentiation © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 14. P = 1700 − 0.016x, C = 715,000 + 240x Profit = Revenue − Cost (a) Revenue: ( ) 2 1700 0.016 1700 0.016 R xp R x x R x x = = − = − ( 2 ) ( ) 2 1700 0.016 715,000 240 0.016 1460 715,000 P R C P x x x P x x = − = − − + = − + − (b) ( ) ( ) 0.032 1460 700 $1437.60 dP x P x dx P = − + = ′ ′ = 15. ( ) ( ) ( ) ( ) 2 2 3 1 4 3 4 0 f x x x f x x f x f x = + + ′ = + ′′ = ′′′ = 16. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 3 2 3 2 3 3 1 3 1 1 3 2 2 1 1 3 1 1 3 2 2 4 f x x x f x x x f x x x − − − − = − = − ′ = − − = − − ′′ = − ⎛− ⎞ − − = − − ⎜ ⎟ ⎝ ⎠ ( ) ( ) ( ) ( ) ( ) 5 2 5 2 5 2 1 3 3 1 4 2 3 3 8 3 8 3 f x x x x − − ′′′ = − ⎛− ⎞ − − ⎜ ⎟ ⎝ ⎠ = − − = − − 17. ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 2 1 2 2 1 2 2 1 4 2 1 4 2 1 f x x x x x f x x x x − + = − − − + ′ = − = − = − − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 4 4 8 2 1 2 16 2 1 48 2 1 2 96 2 1 f x x x f x x x − − − ′′ = − = − ′′′ = − − = − − 18. ( ) ( ) ( ) ( ) ( ) ( ) 16 2 30 75 32 30 32 s t t t v t s t t a t v t s t = − + + = ′ = − + = ′ = ′′ = − At t = 2: ( ) ( ) ( ) 2 2 71 ft 2 34 ftsec 2 32 ftsec s v a = = − = − 19. 6 1 0 1 1 x xy xdy y dx xdy y dx dy y dx x + = + + = = − − + = − 20. ( ) 2 2 2 1 0 2 2 2 0 2 2 2 1 1 y x y y dy dy dx dx dy y dx dy dx y + − + = + − = − = − = − − 21. 2 2 2 4 2 4 0 2 x y x y dy dx dy x dx y − = − = = 22. 2 3 2 20 60 V rh r dV r dr dt dt π π π = = = (a) dV 60 (0.5)2 (0.25) 3.75 cm3 /min dt = π = π (b) dV 60 (1)2 (0.25) 15 cm3 /min dt = π = π [Show More]

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