,= 2x.!(x2 + 1)-1/2(2x) + vx2 + 1 (2) = 2x + 2 VX2 + 1 = 2x + 2(x + 1) = 2(2x + 1) EXERCISES. vx ~ 3W. 4y- ::::} (1+ X 2)2

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Download ",= 2x.!(x2 + 1)-1/2(2x) + vx2 + 1 (2) = 2x + 2 VX2 + 1 = 2x + 2(x + 1) = 2(2x + 1) EXERCISES. vx ~ 3W. 4y- ::::} (1+ X 2)2"

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1 CHAPTER 3 REVIEW False. d~ io = io- In False. In 10 is a constant, so its derivative is O. 9. True. d~ (tan x) = tan x sec x, and d~ (sec'' x) = sec x (sec x tanx) = tan x sec x. d d d Or: dx (sec x) = dx (1 + tan x) = dx (tan x). 10. False. f(x) = Ix + xl = x + x for x :: 0 or x ::; -1 and Ix + xl = _(x + x) for -1 < x < o. So f' (x) = x + 1 for x > 0 or x < -1 and f' (x) = - (x + 1) for -1 < x < O. But 1x + 11 = x + 1 for x :: - ~ and 1x+ 1\ = - x - 1 for x < - ~ True. g(x) = x ::::} g' (x) = 5x 4 ::::} g' () = 5()4 = 80, and by the definition of the derivative, lim g(x) - g() = g' () = 80. x---+ X - 1. False. A tangent line to the parabola y = x has slope dy / dx = x, so at (-, 4) the slope ofthe tangent is (- ) = -4 and an equation of the tangent line is y - 4 = -4(x + ). [The given equation, y - 4 = x(x + ), is not even linear!] EXERCISES 1. y = (x 4-3x + 5)3 ::::} d y' = 3(x 4-3x + 5) dx (x 4-3x + 5) = 3(x 4-3x + 5)(4x 3-6x) = 6x(x 4-3x + 5)(x - 3). y = cos(tanx) =} y' = - sin(tanx) d~ (tan x) = - sin(tanx) (sec x) 1 1/ 4/3 3. y = ~ + + ::::} y' =!X-1/ _ ~x-7 /3 = _1 4_ 3r-:-:i = X X- vx 4 3 ~ 3W 3x - 4y- ::::}. - Vx + 1 I VX+1 (3) - (3x - )~(x + 1)-1/() (x + 1)1/ 3(x + 1) - (3x - ) 3x + 5 y = (Vx+l) (x+l)1/ (x + 1)3/ (x + 1)3/ 5. y = x Vx + 1 ::::},= x.!(x + 1)-1/(x) + vx + 1 () = x + VX + 1 = x + (x + 1) = (x + 1) y vx + 1 vx + 1 vx + 1 ex ex(x - x + 1) 6. y = -1- +x (1+ X ) 7. y = esin8 ::::} y' = e sin 8 51:- (sin 8) = e sin 8(cos 8) () = cos 8 e sin 8 d8

2 5 D CHAPTER 3 DIFFERENTIATION RULES 9. y==- I- t t t )(1) I (1 - - t( -t) :=:;> Y == (1 - t) rn x 10. y == e cos nx :=:;> y' == em:x ( cos nx)' + cos nx (ernx)' == ern.x ( - sin nx. n) + cos nx (ern:x. m) == erna: (m cos nx - n sin nx) 11. y == V;;cos V;; :=:;> y' = V;; (cos V;;)' + cos V;; (V;;)' = V;; [- sin V;; (~X-l/)] + cos V;; (~X-l/) 1-1/ ( i r: i) cos V;;- V;; sin V;; == " x - V x SIn V x + cos V x == r: vx I ( ) ( ) I 1 4 arcsin x 1. y == (arcsin X) :=:;> y == arcsin x. arcsin Zz == arcsin x.. == ~ J1-(x) vl-4x e 1 / x (e 1 X)(_1/x ) 1 x X(x) / - e / 13. y ==- X x 4.. ( ) 1 ) f' (x) USIng the Reciprocal Rule, 9 x == f(x) :=:;> 9 I ( x ==---, we have y ==---- [f (x )] sin (x - sin x ) y' == cos(x - sinx)(l - cos x) sin (x - sin x) d d y) 15. dx (x y4 + x == dx (x + 3y) :=:;> x 4y 3 y' + y x. y' + y. x == 1 + 3y' :=:;> I 1 - y4 - xy y'(4x y3 + x - 3) == 1 - y4 - xy :=:;> Y ==---- 4x y 3 + x y == In(csc5x) :=:;> y' == ~5 (-csc5x cot5x)(5) == -5 cot 5x esc x 17. == sec B :=:;> y 1 + tanb I (1 + tan B)(sec B tan B. ) - (sec B)(sec B. ) sec B [(1 + tan B) tan B - sec B] y == (1 + tan B) == (1 + tan B) sec B (tan B + tan B - sec B) sec B (tan B - 1) [1 + tan x == sec xj (1 + tan B) (1 + tan B) 18." d~ (x cos Y + sin y) = d~ (xy) =? x ( - sin y. y') + (cos y) (x) + cos y. y' = x. y' + Y. 1 =? Y - xcosy cos y - x sin y - x I y' (_x sin y + cos y - x) == y - x cos Y :=:;> Y == y == ecx(csinx - cos x) :=:;> y' == ecx(ccosx + sinx) + cecx(csinx - cos x) == e cx(c sinx - ccosx + ccosx + sinx)

3 1. y = 3 x 1n x =? y' = 3 x lnx(ln 3) d~ (x lnx) = 3 x lnx(ln3) (X. ~ + lnx 1) = 3 xlnx(ln3)(1 + lnx) CHAPTER 3 REVIEW D y = (1 - X -1) -1 =? y' = -1(1 X 1 )-[_(-lx- )] = -(1 l/x)-x- = -((x 1)/x)- X- = -(x - 1)- 4. y = ( ) -1/3 x + v;; =? y' ()-4/3 ( 1) = -~ x + v;; 1 + v;; 5. sin(xy) = x - Y =? cos(xy)(xy' + y 1) = x - y' =? X cos(xy)y' + y',, x - y cos(xy) y [xcos(xy) + 1] = x ycos(xy) =? Y = () 1 x cos xy + = x - y cos(xy) =? 7 1 (1 ), 1 ~ (1 x) _. y= ogs + x =? Y (1+x) In5 dx + - (1+x) In5 1-.(-sinx)+lncosx.l 8. y = (cos x) x =? In y = In(cos x) x = x In cos x =? y' =x._ =? y cosx y' = (cosx)x(lncosx - xtanx) 9. y = lnsinx - ~ sin x =? y' = Si~X. cos x - ~. sinx cosx = cot x - sinx cos x (x + 1)4 30. y = (x + 1)3(3x - I)S =? (x + 1)4 In y = In (x + 1)3(3x _ 1)5 = In(x + 1)4 - In[(x + 1)3(3x - 1)5] = 41n(x + 1) - [In(x + 1)3 + In(3x - 1)5] = 41n(x + 1) - 31n(x + 1) - 51n(3x -1) =? y' 1 1 1,(x + 1)4 (8X 6 15) Y =4 x+1 x-3 x+l -5 3x-l 3 =? y = (x+l)3(3x-l)s x+1 - x+l - 3x-l. (x + 56x + 9)(x + 1)3.. [The answer could be simplified to y' = ( )4( )6' but this IS unnecessary.] x + 1 3x - 1 I(4x) 1 1 4x y = xtan- =? y' = x 1 + (4X) 4+ tan- (4x) 1 = x + tan- (4x) 33. y = In [sec 5x + tan 5xl =? y, = 1 (5 sec x tan 5 x 5 + sec 5 x 5) = 5 sec 5x (tan 5x + sec 5x) = 5 sec 5 x sec 5x + tan 5x sec 5x + tan 5x (3x (3x + 5) 35. y = cot(3x + 5) =? y' = - csc + 5)(6x) = -6x csc

4 CHAPTER 3 REVIEW 0 55 y' = [sin ( cos vsin 7rX)] [sin ( cos vsin 7rX)]' = sin (cos vsin 7rX) cos (cos vsin 7rX) (cos vsin 7rX)' = sin(cos vsin 7rX) cos ( cos vsin 7rX) (-sin vsin 7rX) ( vsin 7rX)' = - sin ( cos vsin 7rX) cos (cos vsin 7rX) sin vsin 7rX. ~ (sin 7rX)-1/(sin 7rx)' - sin ( cos Vsin 7rX) cos (cos Vsin 7rX) sin Vsin 7rX --~--_""":""-_--' ' vsin 1rX, cos 1rX ' 1r -7rsin(cos~) cos(cos~)sin~ COS7rX Vsin 1rX 51. f(t) = V4t + 1 ~ f'(t) = ~(4t + 1)-1/,4 = (4t + 1)-1/ ~ f"(t) = ( -~)(4t + 1)-3/,4 = -4/(4t + 1)3/, so f"() = -4/9 3/ = --:1r. 5. g(8) = 8 sin 8 ~ g' (8) = 8 cos 8 + sin 8,1 ~ gil(8) = 8(- sin 8) + cos 8 ' 1 + cos 8 = cos 8-8 sin 8, so g" (1r /6) = cos(1r /6) - (1r/6) sin (1r / 6) = (V3 /) - (1r/6) (1/) = V3-1r /1, 5X 4 y4 [y - x( -X 5/y5)J y10 5x 4 [(y6 + X6)/y5J y6 54. f(x) = ( - X)-l ~ f'(x) = ( - X)- ~ f"(x) = ( - x)-3 ~ f'il(x) =, 3( - X)-4 ~ n! f (4) (x) =,3, 4( - X)-5 In general fcn)(x) =,3,4, '" 'n( - x)-cn+1) =---,, ( - x)cn+1) ' 55. We first show it is true for n = 1: f(x) = xe" ~ f' (x) = xe" + ex = (x + l)e X, We now assume it is true for n = k: fck)(x) = (x + k)e X, With this assumption, we must show it is true for n = k + 1: f Ck+1)(X) =!!..- [fck)(x)] =!!..- [(x + k)e X] = (x + k)e X + ex = [(x + k) + l]e X = [x + (k + l)]e X, dx dx Therefore, fc n) (x) = (x + n) e" by mathematical induction, u t3 u t 3 cos" t u 3 1 u cos" t m-- = 1m = 1m cos t ' = 1m = -- = t~o tan" t t~o sin" t t~o sin 3 t t~o (' sin t) 3 8 ' 1 3 _ 8 8 (t)3 8 1~ ~ 57. y = 4sin x ~ y' = 4, sinxcosx, At (~, 1), y' = 8, ~, v:} = V3, so an equation of the tangent line is y - 1 = V3 (x - ~), or y = V3x + 1-1r V3/3, x - 1 I (x + 1)(x) - (x - 1)(x) 4x 58. y = x + 1 '* Y = (x + 1) = (x + 1)' At (0, -1), y' = 0, so an equation of the tangent line is y + 1 = O(x - 0), or y = -1.

5 56 0 CHAPTER 3 DIFFERENTIATION RULES 59. y = y' sin x :::;- y ' =~(1 + 4 si n x ) - l /.4 c o s x = cos~ y'1 + 4 Sill X Jr = At (0, 1), y' =, so an equation of the tangent line is y - 1 = (x - 0), or y = x x + 4x y + y = 13 :::;- x + 4(x y' + y. 1) + yy' = 0 :::;- x + x y' + y + yy ' = 0 :::; I I '() I -x - y xy + yy = - x - y :::;- y x + y = - x - y :::;- Y =. x +y At (, 1), y' = -4 - = _i, so an equation of the tangent line is y - 1 = - t (x - ), or y = - tx The slope ofthe normal line is ~, so an equation ofthe normal line is y - 1 = ~ (x - ), or Y = ~x - ~. 61. y = ( + x) e- X :::;- y' = ( + x)(_e- X ) + e- x. 1 = e- X[ - ( + x) + 1] = e- X ( -x - 1). At (0, ), y' = 1(-1) = - 1, so an equation of the tangent line is y - = - l (x - 0), or y = -x +. The slope of the normal line is 1, so an equation ofthe normal line is y - = l (x - 0), or y = x +. s in 6. f( x) = xe x :::;- l'(x ) = x [e Sin x (C OS x) ] + e s in x (1) = e x (x cos x + 1). As a check on our work, we notice from the graphs that l'(x) > 0 when f is increas ing. Also, we see in the larger viewing rectangle a certain similarity in the graphs of f and 1': the sizes of the oscillations of f and l' are linked. s in -3 f (a) f (x ) = x y' 5 - x :::; 1' (x) =X [ ~( 5 -X) -l /( - 1) ] + y'5- x= -x + V 5-x. V5="X = -x + (5 - x ) V5 - x V5 - x V5="X V5 - x -x x 10-3x V5="X V5- x (b) At (1,): 1' (1) = ~. (c) 10 So an equation ofthe tangent line is y - = ~ ( x - 1) or y = ~ x + i. At (4,4): 1' (4) = - ~ = - 1. So an equation of the tangent line is y - 4 = - l (x - 4) or y = - x f ,.;----'--"""""-l 1O - 10 (d) 4.5 -""f The graphs look reasonable, since l' is positive where f has tangents with positive slope, and l' is negative where f has tangents with negative slope. - 1 f-- : ~--+--l 4.5 /

6 CHAPTER 3 REVIEW D (a) f (x ) = 4x - tan x => 1' (x ) = 4 - sec" x => f" (x ) = - sec x (sec x tan x ) = - sec x tan x.,,_--r--_ (b) 5, f',,......",,/ ">,, i " '\, - 5 \ f We can see that our answers are reasonable, since the graph of l' is 0 where f has a horizontal tangent, and the graph of l' is positive where f has tangents with positive slope and negative where f has tangents with negative slope. The same correspondence holds between the graphs of l' and i". 65. Y = sin x + cos x => y' = cos x - sin x = 0 } cos x = sin x and 0 ::::: x ::::: 7f } x = %or 54", so the points are (% ' J ) and (54", - J ). 66. x + y = 1 => x + 4yy' = 0 => y' = - x / (y) = 1 } x = - y. Since the points lie on the ellipse, we have (-y) +y =1 => 6y=1 => y = ± ~. T h e p o i n t s a re (-~,~ ) an d (~, -~ ). 67. f (x ) = (x - a)(x - b)(x - c) => 1' (x ) = (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b). So l'(x) = (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b) = _ 1_ + _ 1_ + _1_ f (x) (x - a)(x- b)(x - c) x - a x- b x -c' Or: f (x ) = (x - a)(x - b)(x - c) => l'(x) = f(x) x - a x - b x - c In If(x) 1= In Ix - al + In Ix - bl + In Ix - cl => 68. (a) cos x = cos X - sirr' x => - sin x = - cos x sin x - sin x cos x } sin x = sin x cos x (b) sin(x + a) = sinx cos a + cos x sin a => cos(x + a) = cos x cos a - sin x sin a. 69. (a)h(x) =f(x)g(x) => h'( x )=f(x)g'(x ) +g(x)1'(x) => h' () = f ()g'() + g() l' () = (3)(4) + (5)( - ) = 1-10 = (b) F (x ) = f (g(x )) => F' (x ) = l'(g(x )) g'(x) => F' () = l'(g()) g'() = l' (5)(4) = 11 4 = (a) P (x ) = f (x )g (x) => P' (x ) = f (x)g' (x ) + g(x ) 1' (x ) => P' () = f ()g' () + g() I ' () = (1) ( ~ =~) + (4) O= ~) = (1)( ) + (4)( -1) = - 4 = - (b) Q(x ) = f (x ) => Q' (x ) = g(x ) 1' (x ) - f (x ) g'(x ) => g(x ) [g(x) J Q'() = g() 1'() - f() g'() = (4)(-1) - (1)( ) = -6 = _~ [g()] (c) C(x) = f (g(x )) => C'(x ) = 1' (g(x ))g' (x ) => C'( ) = 1' (g())g' () = 1' (4)g' () = O= ~) ( ) = (3)() = f (x ) = x g(x ) => 1' (x ) = x g'(x ) + g(x )(x ) = x[xg' (x) + g(x )] 7. f (x ) = g(x ) => 1' (x ) = g' (x )(x ) = xg' (x ) 73. f (x) = [g(x)] => 1'(x) = [g(x) ]. g'( x ) = g(x) g'(x )

7 58 0 CHAPTER 3 DIFFERENTIATION RULES 74. f(x) = g(g(x)) =? fl(x) = gl(g(x)) gl(x) 75. f (x) = 9 (ex) =? fl (x) = gl(ex)ex 76. f(x) = e9 (x ) =? fl (x) = e9 (x ) gl(x) 77. f( x ) = In I9 ( x ) I f l() 1 I () gl(x) =? x = 9 (x) 9 x = 9 (x ) 78. f (x) = 9 (1n x) =? fl (x) = gl(1n x).! = gl(1n x) x x 79. h(x) = f (x ) 9 (x) =? f(x) + g(x) hl(x) - - [f(x) + g(x)] [f(x) + g(x)] [f(x) gl(x) + g(x) fl(x)] - f(x) g(x) [fl(x) + gl(x)] [f(x)] gl(x) + f(x) g(x) fl(x) + f(x) g(x) gl(x) + [g(x)] fl(x) - f(x) g(x) fl(x) - f(x) g(x) gl(x) [f(x) + g(x )] fl(x) [g(x)] + gl(x) [f(x)] [f(x) + g(x )] 80. h(x) = f(x) =? h'(x) = f'(x)g(x) - f(x)g'(~) = f'(x)g(x) - f(x)g'(x) Jg(x) J f(x) / g(x) [g(x)] [g(x)]3/ J f(x) 81. Using the Chain Rule repeatedly, h(x) = f(g(sin4x)) =? h' (x) = f' (g(sin 4x)). d~ (g(sin 4x)) = f' (g(sin 4x)). g' (sin 4x). d~ (sin 4x) = f' (g(sin 4x) )g'(sin 4x) (cos 4x) (4). 8. (a) _ (b) The average rate of change is larger on [,3]. (c) The instantaneous rate of change (the slope of the tangent) is larger at x =. (d) f (x) = x - sin x =? fl (x) = 1 - cos x, o ~~~ so fl () = 1 - cos ~ and fl (5) = 1 - cos 5 ~ So fl () > fl (5), as predicted in part (c). 83. y = [In(x+ 4)] =? yl = [ln(x + 4)]1. _1_.1 = In(x + 4) and yl = 0 {:? In(x + 4) = 0 {:? x+4 x+4 x + 4 = eo =? x + 4 = 1 {:? x = -3, so the tangent is horizontal at the point (-3,0). 84. (a) The line x - 4y = 1 has slope i. A tangent to y = e" has slope i when yl = e" = i =? x = In ~ = -In 4. Since y = e"; the y-coordinate is ~ and the point of tangency is (- In 4, ~). Thus, an equation of the tangent line is y - ~ = ~ (x + In 4) or y = ~ x + ~ (1n4 + 1). (b) The slope of the tangent at the point (a, e a) is d~ exix = a = ea. Thus, an equation of the tangent line is a y - e = e a (x - a). We substitute x = 0, y = 0 into this equation, since we want the line to pass through the origin: o- e a = ea(o - a) {:? _e a = ea(-a) {:? a = 1. So an equation ofthe tangent line at the point (a, e'") = (1, e) is y - e = e(x - 1) or y = ex.

8 CHAPTER 3 REVI EW D y = f(x) = ax + bx + c ::::}- f'(x) = ax + b. We know that I'!-1) = 6 and f'(5) = -, so -a + b = 6 and loa + b = -. Subtracting the first equation from the second gives 1a = -8 ::::}- a = - ~. Substituting - ~ for a in the first equation gives b =. Now f(l) = 4 ::::}- 4 = a + b + c, so c = 4 + ~ - = 0 and hence, f(x) = _~X + x. bt 86. (a) lim C(t) = lim [K(e- at - e- ) ] = K lim (e- at - e- ) = K(O - 0) = 0 because -at and -bt t---*cxl t---*cxl t-i--cx: as t bt (c) C I (t) = 0 {:} be- bt = ae- at {:} ~ = e( -a+b)t {:} In ḇ = (b - a)t t = In(b/ a) {:} b-a 87. s(t) = Ae- ct cos(wt + 6) ::::} a v(t) = S'(t) = A{e- ct [-wsin(wt+6)] + cos(wt + 6)(-ce- ct ) } = _Ae- ct [wsin(wt+6) + ccos(wt+ 6)] ::::} ct a(t) = v'(t) = _A{e- ct[w cos(wt + 6) - cwsin(wt + 6)] + [wsin(wt + 6) + ccos(wt + 6)](_ce- ) } = _Ae- ct [w cos(wt + 6) - cw sin(wt + 6) - cw sin(wt + 6) - c cos(wt + 6)] = _Ae- ct[(w - c ) cos(wt + 6) - cwsin(wt + 6)] = Ae- ct[(c - w ) cos(wt + 6) + cwsin(wt + 6)] t/jb + c t 88. (a) x = Vb + c t ::::}- v(t) = x' = [1/( Jb + c t ) ] c t = c ::::}- I c Jb + c t - c t (c t/ Vb + c t ) b c a(t) - v (t) b+ct - (b+ct )3/ (b) v (t) > 0 for t > 0, so the particle always moves in the positive direction. 89. (a) y = t 3-1t + 3 ::::}- v(t) = y' = 3t - 1 ::::}- a(t) = v' (t) = 6t (b) v(t) = 3(t - 4) > 0 when t >, so it moves upward when t > and downward when 0 ::; t <. (c) Distance upward = y(3) - y() = -6 - (-13) = 7, Distance downward = y(o) - y() = 3 - (-13) = 16. Total distance = = 3. a (d) 0_-----_ Y ~ *"----I (e) The particle is speeding up when v and a have the same sign, that is, when t >. The particle is slowing down when v and a have opposite signs; that is, when 0 < t < (a) V = ijrrh ::::}- dv/dh = ijrr [r constant] (b) V = ijrrh ::::}- dv/dr = ~Jrrh [h constant] 91. The linear density p is the rate of change of mass m with respect to length x. m. = x (1 +..j;;) = x + X 3/ =}- P = dm/dx = 1 + ~..j;;, so the linear density when x = 4 is 1 + ~V4 = 4 kg/rn,

9 60 D CHAPTER 3 DIFFERENTIATION RULES 9. (a) C(x) == 90 + x - 0.0x x 3 =* C' (x) == x x (b) C'(100) == == $0.10junit. This value represents the rate at which costs are increasing as the hundredth unit is produced, and is the approximate cost of producing the 101st unit. (c) The cost of producing the 101st item is C(101) - C(100) == == $ , slightly larger than C' (100). 93. (a) y(t) == y(o)e kt == 00e kt =* y(0.5) == 00eO. 5k == 360 =* eo. 5k == 1.8 =* 0.5k == In 1.8 =* k == ln 1.8 == In(1.8) == In 3.4 =* y( t) == 00e(ln 3.4)t == 00(3.4)t (b) y(4) == 00(3.4)4 ~,040 bacteria (c) y'(t) == 00(3.4)t ln3.4, so y'(4) == 00(3.4)4 ln3.4 ~ 5,910 bacteria per hour (d) 00(3.4)t == 10,000 =* (3.4)t == 50 =* tln3.4 == In50 =* t == In50jln3.4 ~ 3.33 hours 94. (a) Ify(t) is the mass remaining aftertyears, theny(t) == y(o)e kt == 100e kt. y(5.4) == 100e 5.4k == ~ 100 =* 5.4k == 1 e - =* 5.4k == -In =* k == - _1_ In =* y(t) == 100e-(ln )t/5.4 == t/ 5.4 Thus 5.4., y(0) == /5.4 ~ 7.1 mg. t/5.4 t/ t 1 In 100 (b) == 1 =* - == 100 =* In == In 100 =* t == 5.4 ~ ~ 34.8 years 95. (a) C' (t) == -kc(t) =* C(t) == C(O)e- kt by Theorem But C(O) == Co, so C(t) == c;«:", (b) C(30) == ~Co since the concentration is reduced by half. Thus, ~Co == Coe- 30k =* In ~ == -30k =* k == -fa In ~ == fa In. Since 10% of the original concentration remains if90% is eliminated, we want the value oft such that C(t) == foco. Therefore, foco == Coe-t(ln )/30 =* In 0.1 == -t(ln ) j30 =* t == -1~0 In 0.1 ~ 100 h. 96. (a) Ify == u - 0, u(o) == 80 =* y(o) == 80-0 == 60, and the initial-value problem is dyjdt == ky with y(o) == 60. So the solution is y(t) == 60e kt. Now y(0.5) == 60e k(0.5) == 60-0 =* eo. 5k == t == ~ =* k == 1n ~ == In~, so y(t) == 60e(ln4/9)t == 60(~)t. Thus, y(l) == 60(~)1 == == 6~ "C and u(l) == 46~ "C. (b) u(t) = 40 =?- y(t) = 0. y(t) = 60 GY= 0 =? or 81.3 min. In 1. t == ---t ~ 1.35 h In "9 97. Ifx == edge length, then V == x 3 =* dvjdt == 3x dxjdt == 10 =* dxjdt == 10j(3x ) and S == 6x =* ~ dsjdt == (1x) dxjdt == 1x[10j(3x )] == 40jx. When x == 30, dsjdt == ~ == ~ crrr' jmin. 98. Given dvj dt ==, find dh j dt when h == 5. V == ~ 7rr h and, from similar. I r 3 V == ~ (3h) h == ~h3 tnang es, h == 10 =* ' so when h == 5. dh dt = 91rh = 91r(5) = 91r cm/s

10 CHAPTER 3 REVIEW Given dh /dt = 5 and dx /dt = 15, find dz /dt. z = x + h => dz dx dh dz 1 z dt = x dt + h dt => dt = ~(1 5x + 5h). When t = 3, h = (5) = 60 and x = 15(3) = 45 => z = V = 75, x dz 1 so dt = 75 [15(45) + 5(60)] = 13 ft/s We are given dz / dt = 30 ft/s. By similar triangles, J!.. = ~ => z y41 4 dy 4 dz 10 y = V41 z, so dt = V41 dt = V41 ~ 7.7 ft/ s We are given db/dt = rad/h. t anb = 400 / x => dx db x = 400 cot B => dt = esc B dt When B = ~, ~~ = - 400(?(- 0.5) = 400 It/h. I~ x-i 10. (a) f(x) = V5 - x => j'(x ) = ~ = - x(5 _ X )-1 /. (b) 5.""5, x So the linear approximation to f (x ) near 3 is f (x ) ~ f (3) + j'(3 )(x - 3) = 4 - %(x - 3). (c) For the required accuracy, we want V5 - x < 4 - %(x - 3) and 4"".8, %(x - 3) < V5 - x From the graph, it appears that these both hold for.4 < x < I '--~~~--~-../ 4.5 '----~---~-.../ (a) f(x) = f/1 + 3x = (1 + 3X)1/3 => j'(x ) = (1 + 3X) - / 3, so the linearization of f at a = 0 is L (x ) = f (O) + j'(o)(x - 0) = 1 1 / / 3 x = 1 + x. Thus, f/1 + 3x ~ 1 + x => f/1.03 = VI + 3(0.01) ~ 1 + (0.01) = (b) The linear approximation is f/1 + 3x ~ 1 + x, so for the required accuracy 1.5 we want f/1 + 3x < 1 + x < f/1 + 3x From the graph, it appears that this is true when < x < f +OI ~ L I,~ Y = x - x + 1 => dy = (3x - 4x ) dx. When x = and dx = 0., dy = [3() - 4()] (0.) = 0.8.

11 6 D CHAPTER 3 DIFFERENTIATIOI\J RULES 105. A==X+~7f(~X) == (1+i)x =? da== (+~)xdx. When x ==60 and dx == 0.1, da == ( + ~)60(0.1) == 1 + 3;, so the maximum error is approximately 1 + 3; ~ 16.7 em". x 106. lim x 17-1 == [~X17J == 17(1)16 == 17 x X - 1 dx x = 1 x 107. lim ~- == [~ if;;j ==.!.X- 3 / 1 h----+o h dx x = 16 4 x = lim cos e- 0.5 == [~cos ej == _ sin ~ == _ V f/3 e -7f/3 de 8=1f/3 3. V1 + tan x - V1 + sin x. (v 1 + tan x - V1 + sin x ) (V1 + tan x + v1 + sin x ) 109. lim == lim --' ::-:-;---;:=======::::::--'---;=======~ :...- x----+o x 3 x----+o x 3 (V1 + tanx + V1 + sinx) == lim (1 + tanx) - (1 + sinx) == lim sinx (1/ cos x - 1) cos x x----+o x 3 (V1 + tan x + V1 + sin x ) x----+o x 3 (V1 + tan x + V1 + sin x) cos x. sin x (1 - cos x) 1 + cos x 1 + cosx == 1 un -:::-/r:::==========------r===:::::::::===-~x----+o x 3 (V1 + tan x + V1 + sin x ) cos x. sinx. sin' x == 1Iffl ----=--;--;.==========------r=========-~-----;--- :_ x----+o x 3 (V1 + tan x + V1 + sin x ) cos x (1 + cos x).)3 1 r SIn x r == ( x~ ~ x~ (V1 + tanx + V1 + sinx) cos x (1 + cosx) Differentiating the first given equation implicitly with respect to x and using the Chain Rule, we obtain f(g(x)) == x =? 1'(g(x)) g'(x) = 1 =? g'(x) =!'(;(x))' Using the second given equation to expand the denominator ofthis expression gives g'(x) = 1 + [f(~(x))j' But the first given equation states that f(g(x)) = x, so g'(x) = 1:x ' 111. d~ [f(x)] = x =? l'(x) = x =? l'(x) = ~X. Let t = x. Then l'(t) = H~t) = ~t, so l'(x) = ~X. / Let (b,c) be on the curve, that is, b / 3 + C = a /. Now X /3 + y/3 = a / =? ~X-l/3 + ~y-l/3 ~~ = 0, so d 1/3 ()1/3 d~ = - ~1/3 = - ~, so at (b, c) the slope ofthe tangent line is - (c/ b)1/ 3 and an equation ofthe tangent line is y - c == _(C/b)1/3(X - b) or y == _(C/b)1/3 x + (c + b /3C1/3). Setting y == 0, we find that the x-intercept is b 1/3 C /3 + b == b 1/3 (C /3 + b /3) b 1/3 == a /3 and setting x == 0 we find that the y-intercept is c + b /3C1/3 == C 1/3(C/3 + b /3) == c 1/3a/3. So the length ofthe tangent line between these two points is V(b 1/3a/3) + (c 1/3a/3) == vb /3a4 / 3 + c /3a4 / 3 == V(b /3 + c /3)a 4 / 3 == Va / 3a4 / 3 == R == a == constant

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