CHAPTER 3: DERIVATIVES

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Candace Copeland
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1 (Answers to Exercises for Capter 3: Derivatives) A.3.1 CHAPTER 3: DERIVATIVES SECTION 3.1: DERIVATIVES, TANGENT LINES, and RATES OF CHANGE 1) a) f ( 3) f f 3.01 f ( 3) f f ( 3) f ( 3) f f.99 f ( 3).99 3 f.999 f ( 3) ; tis is f 3+ f ( 3) ; tis is f 3+ f ( 3) ; tis is f 3+ ; tis is f 3+ f ( 3) f ( 3) ; tis is f 3+ f ( 3) ; tis is f 3+ f ( 3) wit = wit = wit = wit = wit = wit = b) No c) 30 ) a) f ( a) = 3. Hint: Rationalize te numerator of te difference quotient. 3a b) Point-Slope Form: y 5 = 3 ( 10 x 9 ), Slope-Intercept Form: y = 3 10 x c) Point-Slope Form: y 5 = 10 ( 3 x 9 ), Slope-Intercept Form: y = 10 3 x ) a) i. 4.3 cm cm cm, ii ; b) 4 sec sec sec

2 (Answers to Exercises for Capter 3: Derivatives) A.3. SECTION 3.: DERIVATIVE FUNCTIONS and DIFFERENTIABILITY 1) Hint: f x + ) Hint: r x + 3) f 4) 0 ( f 4 ) 5) a) v t f ( x) r ( x) ( x) = 6 x 6x 1/3 3 b) v 1 c) a t ( x) = 56 = 1 ( x + ) 1 x ; f ( x) = x 3 x 4 = x4 + 4x 3 + 6x + 4x ; r ( x) = 4x 3, f ( x) = x, f ( x) = 8 4/3 3 x 4 3x 8 7/3 3 3 x 7 9x 56 10/3 9 3 x 10, = 0t 4 = 0 mp, v( ) = 30 mp, v( 4.7) = mp ; mp = miles per our = 80t 3 d) a( 1) = 80 mi r, a ( ) = 640 mi r, a ( 4.7 ) = mi r 6) a) Yes; b) No; c) No; d) No (observe tat p is discontinuous at 1); e) No; f) Yes 7) a) Yes, tere is a vertical tangent line; a cusp b) Yes, tere is a vertical tangent line; neiter a corner nor a cusp c) No, tere is not a vertical tangent line; a corner 8)

3 (Answers to Exercises for Capter 3: Derivatives) A.3.3 SECTION 3.3: TECHNIQUES OF DIFFERENTIATION g ( w) 1) Hint: g w + g ( w) = 6w 5 = 5( w + ) w + 3w 5w + 4 ; ) a) 15x + 6 x 3 1 x 3/ x /3 15x + 6 x 3 1 x 3 + b) 13 ( 5 t) 13 t 5 c) 4z 3 16z ; tis can be factored as 4z( z + ) ( z ). d) ( w 3) ( w 3 ) + ( w 3w + 1) ( 3w ) e) i. 1 x 5 (Hint: First, reexpress using algebra.), 3 x ii. 5x, wic is equivalent to 1 x 3 x 5 3 x x f), wic could be simplified to ( 3x + x ) ask your instructor if s/e as a preference. g) 18x x x ( 3+ x) ; x ) 6 ( 3x + 1) 3 3) a) v( t) = 1t + 30t 18 b) v 1 c) a t = 4 ft min, v = 4t + 30 = 90 ft min, v 4.7 = ft min d) a 1 = 54 ft min, a = 78 ft min, a 4.7 = 8.8 ft min

4 (Answers to Exercises for Capter 3: Derivatives) A.3.4 4) a) Point-Slope Form: y 1 = 1 ( x ), Slope-Intercept Form: y = 1 x + b) Point-Slope Form: y 1 = ( x ), Slope-Intercept Form: y = x 3 5) Hint: D x f x g x x = D x ( g ( x) ). f x x 6) 3 f x f ( x). Hint: Assume tat f, g, and are equivalent functions. 7) a) 1 3, and (, 5) b) Point-Slope Form: y 5 = 4 x 1, Slope-Intercept Form: y = 4x + 3 c) Point-Slope Form: y 5 = 1 ( 4 x 1 ), Slope-Intercept Form: y = 1 4 x 11 4 d) 3, 1 and 4 3, ) a) ( 10, 00) and 10, 00 ( ) on te fligt pat. Hint: Find te point(s) a, f a were te slope of te tangent line tere equals te slope of te line connecting te point and te target. b) (,104) and ( 50, 600)

5 (Answers to Exercises for Capter 3: Derivatives) A.3.5 SECTION 3.4: DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 1) a) 1 4 b) 5 3 c) 1000 sin( 5x). Hint: lim = 1. x 0 5x d) 0. Hint: Factor te numerator. ) a) 5x 4 cos x x 5 sin x x 4 ( 5cos x x sin x) b) 1 1+ cos w c) csc r ( cscr) ( cot r) ( csc r) ( csc r cot r) d) 7( secα )( tanα ) + 8α e) θ tanθ + θ sec θ θ tanθ + θ sec θ f) 0, Dom k = β β π + π n ( n ) 3) x x = π 4 + π n x = 3π 4 4) a) { x x = π n ( n ) } + π n n b) x x = 4π 3 + π n x = 5π 3 c) Tangent line: y = 3, Normal line: x = 0 + π n n

6 5) Tangent line: Point-Slope Form: y = 3 x 3π 4, Slope-Intercept Form: y = 3x + 9π + 8 ; 4 Normal line: Point-Slope Form: y = 1 3 x 3π 4, Slope-Intercept Form: y = 1 3 x + 8 π 4 (Answers to Exercises for Capter 3: Derivatives) A.3.6 6) Most efficiently: x x = π + π 3 n x = π + π n ( n ). Equivalently, x x = π + π n x = π 6 + π n x = 7π + π n ( n ) 6. Hint 1: Use a Double-Angle ID. Hint : sin x = ( sin x) ( sin x). 7) Hint: D x cot x 8) Hint: D x csc x sin x = D x cos x sin x = D x 1 1) ) SECTION 3.5: DIFFERENTIALS and LINEARIZATION OF FUNCTIONS 3) π 180 = π

7 (Answers to Exercises for Capter 3: Derivatives) A.3.7 SECTION 3.6: CHAIN RULE 1) a) 3( x 3x + 8) ( x 3) 3( x 3) ( x 3x + 8) 10m b) ( m + 4) 6 c) 6 x 1 x 5 x + x 3 1 x + 1 x 3 x 1 x 1( x 4 + 1) ( x 4 1) 5 x 13 d) 18 tan 6t sec e) 3x sin x f) ( 6t) + 4x 3 sin( x) cos( x) 3sin( x) + 4xcos( x) + xsin( 4x) x sin x x 3sin x 8t ( 8t 3 + 7) 8t /3 3 8t x sin 8t 3 ( 8t ) 8t g) 7 π 6θ 5 + csc( 5θ ) 6 ( 30θ 4 5 csc ( 5θ ) cot( 5θ ) ) ( ) 35 π 6θ 5 + csc 5θ 6 6θ 4 + csc 5θ cot 5θ ) sec( w) tan( w) + ( sec w) sec w i) Same as ). sin( φ ) j) sinφ φ cosφ 1 sin( φ ) + sinφ φ φ sin( φ ) + ( sinφ) φ secφ φ secφ 1 x4 + 1 x 4 1 x 3 x ( x) + x 3 sin( 4x) tan( w) + sec( w) 5

8 8x + 9 k) 18 6x 7 l) (Answers to Exercises for Capter 3: Derivatives) A x( 6x 7) 3 ( 8x + 9) 3, wic factors and simplifies as ( 6x 7) ( 8x + 9) 3. By te Test for Factorability from 64x 4x + 81 Section 0.7 in te Precalculus notes, te discriminant of 64x 4x + 81 is not a perfect square, so it cannot be factored furter over te integers. ( 3x + 3) x( x 3) ( 3x + 3) ( x + 5) 3/ ( x + 5) 3 x( x 3) ( 3x + 3) x + 5 ( x + 5) x x 3 6 ) ( 3x + 1) 3 3) 3 f x f ( x), just like in Section 3.3, Exercise 4. ( 3x + 3) ( x + 5) x + 5 x x 3 4) Most efficiently: x x = π + π 3 n x = π + π n ( n ). Equivalently, x x = π + π n x = π 6 + π n x = 7π + π n ( n ) 6, just like in Section 3.4, Exercise 6. 5) 4 5 6) a) D x x 3 5 b) D x x 3 c) dy dx = dy du = D x 15 x = 15x14 = 5 4 x3 3x du dx = 5u4 3x = 15x 14 = 15x u 4 = 15x ( x ) 3 4 = 15x 14

1 7) a) D x x + 1 = D x + 1 x x + 1 1 b) D x x + 1 = D x + 1 x c) dy dx = dy du du dx = 1 u (Answers to Exercises for Capter 3:

9 1 7) a) D x x + 1 = D x + 1 x x b) D x x + 1 = D x + 1 x c) dy dx = dy du du dx = 1 u (Answers to Exercises for Capter 3: Derivatives) A.3.9 x = ( x + 1) 1 x = x x = x + 1 u = x x + 1 = x ( x + 1) 8) Hints: How are slopes of perpendicular lines (or line segments) related? = x D x a x special cases. a x. Horizontal and vertical tangent lines correspond to 9) Hint: You will need te Cofunction Identities again: sec π x = csc x and tan π x = cot x. 10) a) cos( x) b) D x sin( x) = D x sin x ( cos x) = cos x =... = cos x sin x c) Te range of D x sin( x) is [, ]. Te range of D x ( sin x) is [ 1, 1]. Tis tells us, among oter tings, tat te steepest tangent lines to te grap of y = sin( x) are twice as steep as te steepest tangent lines to te grap of y = sin x. More incisively, te slope of te tangent line to te grap of y = sin( x) at x = a is twice te slope of te tangent line to te grap of y = sin x at x = a, were a is any real value.

10 (Answers to Exercises for Capter 3: Derivatives) A.3.10 SECTION 3.7: IMPLICIT DIFFERENTIATION 1) a) dy dx = y xy + 1 b) 1 + ( ) = 4 c) 4 3 d) Point-Slope Form: y = 4 3 x 1, Slope-Intercept Form: y = 4 3 x ) a) dy dx = y 10x 9x y 4 6x 3 y 3 x 4y b) 5( ) ( ) ( 1) + 3( ) 3 ( 1) 4 4( 1) = 44 c) 9 46 d) Point-Slope Form: y 1 3) a) dy dx = y cos y x y sin y cos y b) sin( 0) + 3cos0 = 3 c) 0 = 9 ( 46 x ), Slope-Intercept Form: y = 9 46 x 5 3 4) Hints: Consider te equation x + y = a, were a > 0. How are slopes of perpendicular lines (or line segments) related? Horizontal and vertical tangent lines correspond to special cases.

11 (Answers to Exercises for Capter 3: Derivatives) A SECTION 3.8: RELATED RATES 1) 4 17 ) Te radius is increasing at about cm sec. Note: π ) Te top of te ladder is sliding down at about 5.78 ft 5 1. Exact: min 4 4) Te distance is increasing at about 3.59 in Exact: min 745 5) Te volume is decreasing at 4 in3 min. 6) Te base radius is srinking at 1 cm 4000π r. in min. 7) Te total resistance is increasing at about om sec. Exact: ft min. om sec. Hint: From te given equation, R = 1 7 oms at tat moment. 8) Te plane s speed is about ft about mp. sec Exact: 00π ft 500π mp. 3 sec

CHAPTER 3: DERIVATIVES

(Exercises for Section 3.1: Derivatives, Tangent Lines, and Rates of Change) E.3.1 CHAPTER 3: DERIVATIVES SECTION 3.1: DERIVATIVES, TANGENT LINES, and RATES OF CHANGE In these Exercises, use a version