CHAPTER 2 Differentiation

CHAPTER Differentiation Section. The Derivative and the Slope of a Graph............. 9 Section. Some Rules for Differentiation.................. 56 Section. Rates of Change: Velocit and Marginals............. 59 Section. The Product and Quotient Rules.................. 6 Section.5 The Chain Rule.......................... 66 Section.6 Higher-Order Derivatives..................... 69 Section.7 Implicit Differentiation....................... 7 Section.8 Related Rates............................ 75 Review Eercises................................. 77 Practice Test................................... 8

CHAPTER Differentiation Section. The Derivative and the Slope of a Graph Solutions to Odd-Numbered Eercises. The tangent line at, has a positive slope. The. The tangent line at, has a positive slope. The tangent line at, has a negative slope. tangent line at, has zero slope. 5. The slope is m. 7. The slope is m 0. 9. The slope is m.. For 997, t 7 and m 50. For 000, t 0 and m 00. For 00, t and m 0.. For t : m 65 5. For t 8: m 0 For t : m 000 f f f f f f 0 0 f f lim 0 0 7. f f f 5 f 5 5 5 f f 5 5 f f lim 5 0 9. f f f f f f f f lim 0 9

50 Chapter Differentiation. ht t ht t ht t ht t t t ht t ht t t t t t t t t t t t t t t t t t t t ht t ht lim t 0 t t. f t t t t t t f t t f t t t tt t t f t t f t t ft t t f t t f t lim t t 0 t t t t tt t t t t tt t 5. f f f f f f f f lim 0 7. f f f f f f lim f 0 At,, the slope of the tangent line is m. The figure shows the graph of f and the tangent line. f 6 f 6 6 6 (, )

Section. The Derivative and the Slope of a Graph 5 9. f f f f 0 f f 0 f f lim 0 f 0 (0, ) At 0,, the slope of the tangent line is 0.. f f f f f f f lim f 0 (, ) At,, the slope of the tangent line is m. The figure shows the graph of f and the tangent line.. f f f f f f f lim 0 At, 6, the slope of the tangent line is f. 5. f f lim 0 At,, the slope of the tangent line is f. f f f f f

5 Chapter Differentiation 7. f f f f f f f lim 0 At the point,, the slope of the tangent line is m. The equation of the tangent line is. (, ) 9. f f f f f f f lim 0 At the point, 9, the slope of the tangent line is m 6. The equation of the tangent line is 9 6 6. (, 9) 0 8 6. f f f (, ) f f f f lim 0 At the point,, f The equation of the tangent line is..

Section. The Derivative and the Slope of a Graph 5. f f f (, ) f f f f lim 0 At the point,, the slope of the tangent line is m The equation of the tangent line is. 5. f f f f f f f lim 0 (slope of tangent line) Since the slope of the given line is, we have and f. Therefore, at the point,, the tangent line parallel to 0 is.

5 Chapter Differentiation 7. f f f f f f f lim 0 (slope of tangent line) Since the slope of the given line is 6, we have 6 ± and f and f. At the point,, the tangent line is 6 6 8. At the point,, the tangent line is 6 6 8. 9. is not differentiable when. At, 0, the graph has a node. is differentiable for all. 5. is differentiable everwhere ecept at. At, 0, the graph has a cusp. 5. f is differentiable on the open interval,. 55. f is differentiable everwhere ecept at 0, which is a nonremovable discontinuit. 57. f 0 f 0.88 0.5 0.0 0 0.0 0.5 0.88 f.6875 0.75 0.875 0 0.875 0.75.6875 Analticall, the slope of f is f f m lim 0 lim 0 lim 0 lim 0.

Section. The Derivative and the Slope of a Graph 55 59. f 0 f.6875 0.5 0.065 0 0.065 0.5.6875 f 6.75.5 0.75 0 0.75.5.75 6 Analticall, the slope of f is f f m lim 0 lim 0 lim 0 lim 0 lim 0. 6. f f f lim 0 lim 0 lim 0 lim 0 The -intercept of the derivative indicates a point of horizontal tangenc for f. 6 8 6. f lim 0 f f lim 0 lim 0 lim 0 lim 0 The -intercepts of the derivative, ±, indicate points of horizontal tangenc for f. 6 6

56 Chapter Differentiation 65. f Other answers possible. f() = 67. The graph of f is smooth at 0,, but the graph of g has a sharp point at 0,. The function g is not differentiable at 0,. 6 69. False. f is continuous, but not differentiable at 0. 7. True Section. Some Rules for Differentiation. (a) At,,. (b) At,,.. (a) At,,. (b) At,,. 5. 07. f 9. f g g. ft 6t. st t 5. t 6 t 7. f 9. f 8 8. Function Rewrite Differentiate Simplif. 6 6 6 5. 7. f f f

Section. Some Rules for Differentiation 57 9. f. f f 8 At 0,,.. f 5. f f 6 6 f 8 5 8 5 7. f 9. f 8 f f 6 6. f. f 5 5 f 6 6 6 f 0 0 0 5 5. f 5 7. 5 f 5 5 55 8 0 At, 0, the slope is m 8 0. The equation of the tangent line is 0. 9. f 5 5. f 5 5 5 5 f 5 8 5 8 5 8 5 5 6 0 0, ± If ± 6, then ±6 6 6 9 5. The function has horizontal tangent lines at the points 0,, 6, 5, and 6, 5.

58 Chapter Differentiation 5 0 5. when 5. The function has a horizontal tangent line at the point 5, 5. 55. (a) 57. (a) h f h f 0 f g f h f h (b) h f h f h f 6 (b) f g h (c) h f f g h h f (c) h f (d) h f g f h h 0 f f h f 6 59. S 8.7097t.097t 885.75t 0,8.7t 68,95. t 7 corresponds to 997, 7 t. (a) St.8788t 0.78t 977.50t 0,8.7 998: S8 0. 00: S 79. 00: S 56. (b) The results are close to the estimates. (c) The units of S are millions of dollars per ear per ear. 6. C 0.60 50 6. R.00 P R C.00 0.60 50 0.0 50 dp 0.0 d Therefore, the derivative is constant and is equal to the profit on each can bar sold. f 0 f There are horizontal tangents at 0., 0. and.8, 0.9. 65. False. Let f and g. Then f g, but f g.

Section. Rates of Change: Velocit and Marginals 59 Section. Rates of Change: Velocit and Marginals. (a) 980 985: 5 60 $ billion per ear 5 (b) 985 990: (c) 990 995: (e) 980 00: 80 50 $6 billion per ear 5 70 60 $9.5 billion per ear (d) 995 000: (f) 990 00: Answers will var. 50 5 5 60 80 5 70 50 $7 billion per ear $6 billion per ear $0 billion per ear. ft 5. Average rate of change: Instantaneous rates of change: f f 9 t f, f h Average rate of change: Instantaneous rates of change: h h h 8, h 0 (, ) (, ) 8 (, 9) t 6 6 8 0 (, ) 7. f 9. Average rate of change: f f g g g 7 Average rate of change: 6 Instantaneous rates of change: g, g 0 90 Instantaneous rates of change: f, f 6 0 0 (, ) (, ). (a) 0 00 67. The number of visitors to the park is decreasing at an average rate of 67 people per month from September to December. (b) At t 8, the instantaneous rate of change is about 0. Answers will var. For eample,,.

60 Chapter Differentiation. Et 7 9 6t t 9 t t (a) E E0 0 7 0 0 7 (b) E E 7 7 7 E0 E 9 E 9 E (c) E E 7 5 7 (d) E E 07 7 7 E E 0 E 0 E 5 9 5. s 6t 555 (a) Average velocit (b) v st t, v 6 ftsec, v 96 ftsec (c) s 6t 555 0 6t 555 t 555 6 t 5.89 seconds (d) v5.89 88.5 ftsec 5 5 9 80 ftsec 7. dc.7 d 9. dc 70 0.5 d. dr 50 d. dr d 8 6 00 5. dp 7 d 7. dp 0.0005. d 9. C.6 500. p 0.05 0 000 (a) C0 C9 $0.58 (a) p5 p50 879.95 875 $.95 (b) C.8 C9 $0.6 per unit. (b) p 0. 0 p50 0.50 0 $5.00 per unit. (c) The answers are ver close. (c) The results are nearl the same.

Section. Rates of Change: Velocit and Marginals 6. T 0.075t 0.t 00. 5. (a) 0 0 98 (b) For t <, the slopes are positve and the fever is going up. For t >, the slopes are negative and the fever is diminshing. (c) T0 00.F T 0F T8 00.F T 98.6F dt (d) 0.075t 0. (e) T0 0.F per hour T 0F per hour 5 T8 0.F per hour T 0.6F per hour p 5 0.00, C 5.5 (a) (b) P R C 5 0.00 5.5 (c) R p 5 0.00 5 0.00 0.00.5 5 R 5 0.00 P.5 0.00 600 00 800 00 000 drd dpd.8.6. 0... 0...0.5 P 705 75 05 605 65 500 00 7. (a) The slope is Hence 0.50 0.75 00. (b) ( p is price, is number of glasses sold.) 50 00 00p 0.75 P R C p C 00p 700 p 700. 00 700 0.05 0 00 0.005.7 0 9. 6,000, 6, 0,000, 7 Slope (a) (b) 0,000 7 6 0,000 6,000 6000 p 6 6,000 6000 p 6000 P R C p 0.0 85,000 (demand function) 6000 0. 85,000.8 85,000 6000 0 0 P 0.005.7 P00 > 0 Profit increasing, slope positive P00 < 0 Profit decreasing, slope negative (c) P00 0.7 P00 0. 700 (c) 0 00,000 P.8 000 70,000 P8,000 5.8 dollars, positive P6,000 0.0 dollars, negative

6 Chapter Differentiation. p 50, < 8000, C 0.5 500. (a) 790 to 900: P R C p C 50 0.5 500 50 0.5 500 P 5 0.5 (a) P900 $0. per unit (b) P600 $0. per unit (c) P500 $0 per unit (d) P600 $0.08 per unit P500 0 indicates that 500 is the optimal value of. Hence, p 50 50500 $.00. (b) 900 to 000: (c) 790 to 000: 80 miles. mir 0 ears 50 miles.5 mir 00 ears 80 miles.85 mir 0 ears 5. N fp 7. (a) f.79 is the rate of change of a gallon of gasoline when the price is $.79/gallon. (b) In general, it should be negative. Demand tends to decrease as price increases. Answers will var. f, f f has a horizontal tangent at 0. f f 9. The population in each phase is increasing. During the acceleration phase the growth is the greatest. Therefore, the slopes of the tangent lines are greater than the slopes of the tangent lines during the lag phase and the deceleration phase. Possible reasons for the changing rates could be seasonal growth or food supplies. Section. The Product and Quotient Rules. f 9 5. f f 6 f0 0 5. g 7. g 6 6 8 8 5 h 5 5 5 h6 5 9. ft t t. g 5 ft t t t t g 5 5 5 6t t 9 t g6 6 5 f 69 9 9 75 00

Section. The Product and Quotient Rules 6. f t t t ft t t t t 8t t t f 8 5 5 5. 7. 9. Function 7 8 Rewrite, 0 7 8 Differentiate, 0 7, 6 Simplif, 0 7 6.,,. f 5 9 6 9 9 9 5 0 8 5 5. gt t t 6 t 7. f 56 gt t 5 t t t f 5 6 6 5 6 6 9. f. f f 5, f,. f 5. gs s s 5 s s s 5s f gs s s 5 s s s 5 s 7. g 5 g 5 5 0 0 5 8 7

( 6 Chapter Differentiation 9. f. f f0 5 5 0 5 f f f (0, ) 5 6 5 (,. f 5 5 f 5 (0, 5) 0 6 0 f0 6 5 6 0 6 5 5. f f 0 when 0, therefore 0 or. Thus, the horizontal tangent lines occur at 0, 0 and,. 7. f 6 f 0 when 6 0. Thus, the horizontal tangent lines occur at 0, 0 and,.7. 9. f 5. f 6 f f f f f f 6

Section. The Product and Quotient Rules 65 5. 75 When p, p 5p 55. d dp 75 5p p5 5p 75 d dp 75 $.87 per unit. 5p ft t t t t t t t t (a) (b) f0.5 0.8 per week f 0. per week (c) f8 0.05 per week 57. P 500 50 t tt 8 When t, P 500 5.55 bacteriahour. 50 t 500 00 t 50 t 59. (a) (b) (c) 6,000,000 k k p, 5 6 k 000 6,000,000 p p 6,000,000 p 000 C 50 0,000 P R C p C 000 50 0,000 000 50 0,000 6. (a) (b) f f f d d f 7000 0 0 80 Let 6 and p $500. (c) f d d f 0 Since (a) and (b) are increasing functions, onl (c) could represent a demand function.

66 Chapter Differentiation 6. (a) C0 00 00 0 0 0 8.5 (b) C5 0.7 (c) C 00 00 0, C 00 00 C0.8 Increasing the order size reduces the cost per item. 0 0 00 00 0 0 65. P.7 0.t 0.07t 0,000 0.06t 0.0t 0,000,700 0t 7t 0,000 060t t Pt 0,000 060t t 0 6t,700 0t 7t 060 t 0,000 060t t 50t 860t 0,900 8t 55t 500 P5 0.09 P7 0.07 P9 0.99 P 0.0577 These derivatives give the rate of change of price at ear t. Section.5 The Chain Rule. f g 6 5 u g u 6 5 f u u 5. 5 u 5 u f g u g f u. u u 7. u u 9. f is most efficientl done b the General Power Rule (c).. f 8 is most efficientl done b the Constant Rule (b).. (a) First rewrite f, then use the simple 5. (c) Rewrite as f and use the General power rule. Power Rule. 7. 7 6 7 9. g 6. h 6 6 66. f 9 9

Section.5 The Chain Rule 67 5. f t t t 7. ft t t st t 5t t 5t st t 5t t 5 t 5 t 5t 9. 9 9. 9 8 6 9 f 9 f 9 9 7 9. h 7 7 5. f 00 f 6 (, 5) f 5 5 6 6 78 00 7. f 7 7 9. Point:, f, f 7 8 When, the slope is f 8, and the equation of the tangent line is 8 8 7 7 f f f 0 (, ) (, ). f. f f f f has a horizontal tangent when f 0. f is never 0. f f f 5 5 f

68 Chapter Differentiation 5. 7. t 8t 8 t 9. f 5. f 6 gt t gt t t t t 5. f 9 9 55. 9 9 9 9 9 7 57. t t 59. t t tt t t tt t tt t t5t 8 t f f 8 6. f 6. 6 5 f 6 5 5 6 5 6 55 5 65. ft 6 6 t t 67. f t t 9t ft 7 t 7 t ft t 9 t t t f0 7 7 8 8 t 0 f 8 6 8 6 6 8 t (0, ) (, 8) 6

Section.6 Higher-Order Derivatives 69 69. f f f 7 6 (, ) 7. A 000 r 60 A 00060 r 59 5000 r 59 (a) A0.08 50 0.08 per percentage point 59 $7.00 (b) A0.0 50 0.0 per percentage point 59 $8.59 (c) A0. 50 0. per percentage point 59 $89.9 8 7. N 00 t 00 00t dn 00t t 800t t The rate of growth of N is decreasing. 75. (a) V k t When t 0, V 0,000 k 0,000. Therefore, V 0,000 t. t 0 dn 0 77.78. 0.8.9 (b) V 0,000 t When t, dv 0,000 $.8 per ear. (c) When t, dv 0,000 $5.97 per ear. 77. False. B the chain rule, Section.6 Higher-Order Derivatives. f. f 75. gt t 8t f 0 f gt t 8 7. f t t 9. ft t ft 9 t 9 t f f 9 8 f 8 8 8 8 5. f. f 8 8 6 6

70 Chapter Differentiation 5. f 5 7. f 0 6 f 60 7 f 5 5 8 6 5 60 0 0 f 0 80 80 0 f 60 60 80 f 0 60 9. f 6. f 8 gt 0t 0t gt 60t 0 g 60 0 60 f 9 8 f 9 5 9 5. f 5. f f f 8 5 f5 895 68 8 5 f f 9 8 f 6 8 8 f 7 8 f 7 8 6 7. f 9. f f 0. f 5. f 6 f 8 7 f 6 8 0 f 0 when. 5. f 5 7 60 f 8 7 f 6 8 f 0 when 6 8 0.

Section.6 Higher-Order Derivatives 7 7. f f f f 0 0 ± ±6 (Note: 0 is not in the domain of f.) 9. f f 9 9 f 0 when 9 0 0, ±.. (a) st 6t t (b) vt st t (c) at vt vt 0 t when t s.5 feet.5 sec. (d) st 0 when 6t t, or t 0, 9 sec. v9 ft/sec, which is the same speed as the initial velocit.. d s t 090 90t 900 t 0 t 0 t 0 0 0 0 0 50 60 ds d s 0 5 60 67.5 7 75 77. 9.5 0.56 0.6 0.5 0.8 As time increases, the acceleration decreases. After minute, the automobile has traveled about 77. feet. 5. f 6 6 f 6 f The degrees of the successive derivatives decrease b. 5 f 7 f f 0 7. The degree of f is, and the degrees of the successive derivatives decrease b. f f f

7 Chapter Differentiation 9. (a) 75 5 (b) 0.088t.t 7.06t 58.7 0.8t.886t 7.06 0.968t.886 (c) > 0 on 5 t which shows that the price is increasing. (d) is greatest at t 9.85, or 999. 0. (e) To find maimum, set Answers will var. 5. False. The Product Rule is f g fg f g. 5. True. hc fcgc f cgc 0 55. True 57. Let Then, f. f f f f f f f f f f f f. In general n f n f n nf n. Section.7 Implicit Differentiation. 5. 5 5 0 5 5 5. 7. 0 8 0 8 8

Section.7 Implicit Differentiation 7 9. 5. 5 5 0 d d 0 d d 5. Note: simplifies to and hence. 9 0 At 0, 7, 0 7 0. 5. 7. 0 d d d d At 5,, d. 0 (Product Rule) d d At 0,, d. d d 9. d d d At 0, 0, d.. d 9 d 0 d 0 At 6, 5, d 5. d. 5 5. d 0 At 8,, d. d 5 0 6 d 0 d At,,. d 7. 9. 9 6 0 8 8 0 9 At 0,, 0. At 5,, 5 9 5.

7 Chapter Differentiation. Implicitl: 0. d At,, d. d Eplicitl: ±5 d ± 5 ± 5 ±5 Implicitl: 0 Eplicitl: ± ± ± ± ± (, ) 6 = 5 6 6 At,,. 6 = 5 5. 69 7. d 0 At 5, : At, 5: d m 5 5 5 5 69 m 5 5 5 69 5 5 5 5 5 At, 5 : At, 5 : 5 5 5 5 0 5 5 5 0 5 5 5 5 5 0 5 5 5 5 5 5 5 5 0 5 5 (, 5) 6 (5, ) 0 (, 5) 5 5 6 0 (, 5)

Section.8 Related Rates 75 9. 5 8. 0 At 0,, 0 and (tangent line). At, 0, is undefined and (tangent line). 8 5 (0, ) (, 0) 8 p 0.006 0.0 0, 0 dp d d 0.0 0.0 dp dp dp 0.0 0.0 d dp d dp 0.0 0.0. p 00 p d dp p d dp p d dp p p 00, 0 < 00 d dp p p 5. (a) (b) 00 0.75 0.50.75 d 0.5 75 0.5 0 When 00,000 500 and 00 0.75 0.5 5,50 5 0.75 750.5 0.75 d 0.5 000, d. d 0 0 000 If more labor is used, then less capital is available. If more capital is used, then less labor is available. Section.8 Related Rates., (a) When and (b) When 6 and, we have d d d d 8, 8 6. 6 6 d 8 d 55 d 8, d 8 55 85.., (a) When (b) When d 0, d, d 8, d, and 0, 8 0 5 8. d,, and 6 6,.

76 Chapter Differentiation 5. A dr da dr r,7., r (a) When da r 6, 6 in min. (b) When r, da 96 in min. A r, da dr r dr da da If is constant, then is not constant; is proportional to r. dr 9. V dv dv dr dv 0, r r,., r (a) C 5,000 0.75 dr (a) When 5 r, 0 dr (b) When 5 r, 0 ftmin. ftmin. (b) (c) dc 0.75d 0.7550.5 dollars per week R 50 0 dr 50d 5 d 5050 5 00050 P R C dp dr dc 7500.5 7500 dollars per week 787.5 dollars per week dv. V d d,, 5., (a) When (b) When dv, 9 cm sec. dv 0, 0 900 cm sec. (a) When (b) When (c) When (d) When d, d, cmmin. 0, 0 0 cmmin., cmmin., cmmin. 7. Let be the distance from the ground to the top of the ladder and let be the distance from the house to the base of the ladder. 5, (a) When (b) When (c) When d 0, d 7 7, 576, 7 ftsec. 5 5, 00 0, 0 ftsec., 7, 8 7 7 ftsec. since d. r ft sec 5 ft

Review Eercises for Chapter 77 d 9. (a) L, and 50, 600, When 50 and 00, L 50 and (b) t 50 750 hr 0 min dl d L dl 5050 00600 750 mph. 50 00 00 L 00 00. d s 90, 6, 0. V r h, h 0.08, V 0.08r, dv dr 0.6r s ds d ds d s When r 750 and dr, When 6, s 90 6 9.68 ds 6 0 8. ftsec. 9.68 dv 0.6750 60 88.5 ft min. nd rd s st 90 ft Home 5. P R C p C 6000 0. 00 500 0. 600 500 dp d d. 600 d dp 600. dp d When and units per week. 68, 600. 68 5 Review Eercises for Chapter. Slope. Slope 0 5. t 7: slope $5500 million per ear per ear 7. At t 0, slope 80. (sales are increasing) t 0: slope $7500 million per ear per ear (sales are increasing) At t, slope 70. At t 6, slope 900. t : slope $500 million per ear per ear (sales are increasing)

78 Chapter Differentiation 9. f f f lim 0. f f f lim 0 5 5 lim 0 lim 0 lim 0 lim 0 f lim 0 lim 0 f. f f f lim 0 9 9 9 9 lim 0 9 9 lim 0 lim 0 9 9 9 9 9 9 9 f5 5. f f f lim 0 5 5 lim 0 7. f 8 5 f 5 f 5 5 5 lim 0 5 5 lim 0 5 5 5 f6 9. f. f. f 5 5 f f 0 f f9 9 6 f 5 5 f 5 5 5. is not differentiable at. 7., 0 is not differentiable at 0., > 0

Review Eercises for Chapter 79 9. gt t. gt t t (, g t ( f,, 6 f f (, ) 0 5 t. 0 0 5. f 7 7 (, 7) 0 f (, 0) f 0 7. f f f 6 7 9. f, 0, Average rate of change f f0 f 5 f f 0 0 0 (, ) 8. (a) 998-00: (b) 998: 00: 58,60 0,57 8 S8 $77 million per ear per ear $6976 million per ear per ear S $876 million per ear per ear (c) Sales were increasing in 998 thru 00, and grew at an average rate of $6976 million over the period 998 00.

80 Chapter Differentiation. P 0.00059t 0.005t 0.850t.007t 0.50 (a) (b) 997: (c) Pt 0.006t 0.0905t 0.57t.007 000: 00: 6 P7 $0.00 per pound P0 $0.6 per pound P $0.8 per pound Price is increasing on 7,, or 997 to 00. Decreasing on 6, 7 and,, or 996-997 and 0-00. (d) Positive slope: 7 < t < Negative slope: 6 < t < 7 and < t < (e) When prices increase, slope is positive. When prices decrease, slope is negative. 5. (a) (b) (c) (d) (e) st 6t 76 Average velocit vt t v 6 ftsec v 96 ftsec s s0 0 ftsec st 6t 76 0 t 76 6 v.5.5.8 velocit Speed.8 ftsec at impact 6 t.5 sec 7.5 7. R 7.50 9. C 5 500 P R C 7.50 5 500.50 500 dc 0 d 5. C 70.55 5. C.55.75 R 00 5 R 00 5 55. R dr d 5 5 5 5 5 57. 5 dp d 0.0006 59. f 5 5 5 6. f 5 5 5 6 6

Review Eercises for Chapter 8 6. f 6 5 f 6 6 5 65. f 5 67. f 5 0 05 h h 6 0 6 5 69. g 7. g f 8 6 8 6 5 f 80 7. h 6 75. h 6 6 5 6 5 8 5 f 5 f 5 5 5 7 77. ht ht t t t t t t t t 6 t t t 9t 5 t t 79. (a) T 00 t t 5 00t t 5 Tt 00t t 5 t T 5 9 T T5 T0 8 6.6 6.5..6 600t t t 5 (b) 60 0 0 The rate of decrease is approaching zero.

8 Chapter Differentiation 8. f 7 8. f 6 7 f 6 f 6 f 5 f 5 0 6 0 6 85. f 7 5 87. f 5 f 6 f 89. (a) st 6t 5t 0 9. (b) st 0 6t 5t 0 Using the Quadratic Formula or a graphing utilit, t.5 seconds. (c) vt st t 5 v.5.09 ftsec (d) at vt ftsec 0 d d 0 d d 9. 8 9 0 95. 8 9 0 9 8 8 d 9 At,,. 97. 99. b 8h, 0 h 5 At,,. V bh0 0bh 08hh 80h dv dh 60h dh dv 60h 6h When h, dh 6 6 dv since 0 ft/min. 0 ft 0 ft min 9 ft 0 ft ft

Practice Test for Chapter 8 Practice Test for Chapter. Use the definition of the derivative to find the derivative of f 5.. Use the definition of the derivative to find the derivative of f.. Use the definition of the derivative to find the equation of the tangent line to the graph of f at the point 6,.. Find f for f 5 6 5 9. 5. Find for f 6 f. 6. Find f for f 5. 7. Find the average rate of change of f over the interval 0,. Compare this to the instantaneous rate of change at the endpoints of the interval. 8. Given the cost function C 600. 0.000, find the marginal cost of producing units. 9. Find f for f 7 9. 0. Find f for f 7 8.. Find f for f 5.. Find f for f.. Find f for f 6 5.. Find f for f 8. 5. Find f for f. 6. Find f for f 0. 7. Find f for f 9 7. 8. Find f for f.

8 Chapter Differentiation 9. Use implicit differentiation to find for 5 5 00. d 0. Use implicit differentiation to find for 0. d. Use implicit differentiation to find for 5. d. Use implicit differentiation to find for d. d. Let Find when and. 5.. The area A of a circle is increasing at a rate of 0 in. min. Find the rate of change of the radius r when r inches. 5. The volume of a cone is V dv Find the rate of change of the height when and h 0 inches. r 00, h r h., Graphing Calculator Required 6. Graph f and its derivative on the same set of coordinate aes. From the graph of f, determine an points at which the graph has horizontal tangent lines. What is the value of f at these points? 7. Use a graphing utilit to graph. Then find and sketch the tangent line at the point 8,.