Chapter 3 99

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern