Chapter 3 96 - 246 CHAPTER 3 DIFFERENTIATION RULES “who...

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Unformatted text preview: 246 CHAPTER 3 DIFFERENTIATION RULES _ “who ”IWI‘nm+mm i m fl[email protected][email protected][email protected] Wm=um+mmt If (96) + 9%)]2 f If(é10)I2 flit) + f(w)g(w)f’(0c) + (109(909' : .Jc_'(~”0)Ig(flc)I2 + 9/(96)If(:v)I2 If (a?) + 9%)]2 mhmyz £2 : [email protected]%,Nmmmafommg:fomorwwmo) up 2 fovmmaon 2umWfl¢flm 79. Using the Chain Rule repeatedly, h(ac) 2 f (g(sin 430)) 2> / I ~ d ~ / - I . . h (w) 2 f (g(srn4m)) ~ Em— (g(sm4:r)) 2 f (g(sm4:r)) -g (811141!) - % (Sln4m) 2 f’(g(sin4x))g’(sin4a:)(cos4$)(4) 30. (a) 8 (b) The average rate of change is larger on [2, 3]. (c) The instantaneous rate of change (the slope of the tangent) is larger at m 2 2. (d) flat) 2 a: — 2:9in 2> f’(:1:) 2 1 — 2cosw, so 0 8 f’(2) = 1 — 2cos2 x 1.8323 and f’(5) 2 1 # 2c0s5 % 0.4327. So f'(2) > f’(5), as predicted in part (c) 1 1 4 1. 1—23I—xi—Iandy’:0 ¢2>1n(a:+4)20 1:) ar+4i $+4 81. y : [111(9; + 4)]2 :> y' = 2mg; + 4)] a: + 4 2 1 2) m 2 —3. so the tangent is horizontal at the point (—3, 0). 1 Z => (— 1114, i). Thus, an 331—4260 2> 82. (a) The line m - 4y 2 1 has slope %. A tangent to y 2 6”” has slope 711 when 3/ 2 ea” 2 m 2 lnél1 2 - 1114. Since y 2 er, the y—Coordinate is i and the point of tangency is equation of the tangent line is y 2 i 2 %($ + ln4) org 2%:1: + }1(ln4 + 1). 1 1 d , . 1 (b) The slope of the tangent at the pomt (6116“) [S d— 61 2 6“. Thus. an equation of the tangent line IS ac I : a. y 2 0 into this equation, since we want the line to pass through the y — e“ 2 e“(ac - a). We substitute m 2 O. 2) a 2 1. So an equation of the tangent line at the point origin: 0 2 e“ 2 e“(0 — a) 2) —e“ 2 ea(#a) (1113“)2(1.e)isyie2e($—1)ory2e$1 2> f’(;v) 2 2am + b_ We know that f’(#1) 2 6 and f’(5) 2 28 2 a 2 —%. Substituting 83.y2f(:c)2a$2+b$+c 22,5022a+b:6 and 10a + b 2 #2. Subtracting the first equation from the second gives 12a 2 2 1 —§forainthefirstequationgivesb2133.N0wf(1)24 2> 42a+b+c.s0024+§ #34 20and hence, f(:c) 2 1%} + 337431. 84. (a) 11m C(t) 2 tlim [K(e’“t — €th 2 K tlirn (e’at — 6—1”) 2 K(0 - 0) 2 Obecause —at ——> -00 and t—>oo ——->o<> —+oo #btA—ooast—aoo. ...
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