Chapter 3 99 - CHAPTER 3 REVIEW 249 98 = m3 — 2102 1>...

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Unformatted text preview: CHAPTER 3 REVIEW 249 98. = m3 — 2102 +1 :> dy = (3x2 — 4:17) dac. When a: = 2 and d2: = 0.2. dy = [3(2)2 — 4(2)] (0.2) = 0.8. 99. A 2172+ énGxV 2 (1+%)x2 => dA= (2+ glrdfifi ' When a: = 60 and dm 2 0.1. dA = (2 +g)60(0.1) : 12 + 37”, so A . , . 2 the max1mum error is approxrmately 12 + 37" % 16.7 cm . 17_ 100.11mx 1 [d zfil x—l . 6/16+h—2¥ d 4 _1 _3/4 _ 1 :i 101. AET— dxfix—lfi 43: 1:16 4(416)3 32 cosl970.5 d - 77 fl . - ‘2 — 6 :7 -=-— 102 6221/3 (9—7r/3 [d0 cos ie:7r/3 Sln3 2 1031. \/1+tanx—\/1+sinw_l, (x/l-l-tanflfi—\/1—+Sinx)(\/1+tan$+\/I+sinx) III‘IR) $3 ”01310 w3(\/1+tanx+\/1+sinm) l' (1 +tanm) — (1 +sinm) lim sinm(1/cosa7 — 1) cosa: — 1m . ~ m—>0m3(\/1+tanw+\/l——sinac) w-‘O m3(\/1+tana:+\/1+sma:) cosm sin$(1—cosm) 1+cosm — lim 90—“) x3(\/1+tan$+\/1—— sinm)cosm . 1+COS$ . sinx-sin23: 1 11m z->0 9:3(«1 +tanm—l— V1 —— sinm) cosw(1 +Cosm) 3 sinx 1 : lim lim (1.4) x > xao(M1+tanw+v1+sinm)cosm(1+cosx) ‘ 13 1 l (fi+fi)-1-(1+1)‘4 104. Differentiating the first given equation implicitly with respect to m and using the Chain Rule, we obtain f(g(m)) : a? => f’(g(x))g’(a¢) = 1 => g’(a:) = m Using the second given equation to expand the 1 denominator ofthis expression ives g/ a: : \. But the first iven e uation states that f g as : 3:. g ( ) ”mgMHQ 5; q ( ( )) , 1 sog(m):1+$2. d l I 105. E [f(2:v)] : x2 :> f’(2:c) ~ 2 = x2 : f (2m) : émQ. Lett 2 23:. Then f (t) : %(%t)2 2 $152. so f'($) : §$2. ...
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