Chapter 3 92 - _—’_—————————— 242...

Unformatted text preview: ______—’___________—————————— 242 CHAPTER 3 DIFFERENTlATION RULES 2 4 (2x +1)3(3ac — 1)5 2 4 lny = In W + 1) = ln(:r.2 +1)4 — lIl[(2$ +1)3(3x — 1)5] (23: +1)3(3a: — 1)5 = 4ln(m2 +1) — [111m +1)3 +1n(3x - 1)5) : 4111032 + 1) — 3ln(2$ + 1) — 5ln(3;v — 1) : 1 .2 -3. . . :c?+1 m 2m+1 333—1 , (x2 +1)4 81' 6 15 y _ (2x+1)3(3$—1)5 x2+1 2x+1 3a:#1 ‘ (1:2 + 569: + 9) ($2 + 1)3 (23: + 1)4(3m — 1)6 / y—=4- 2—5 -3 : y [The answer could be simplified to y' : — . but this is unnecessary) __1__ 1 + (4:10)2 43: ,1 _ -4 +tan (43:) ~ 1 — ————1 + 163:2 31. y = :ctan’1(4zc) => y' : :17- + tan—101:6) 32. y : em” +cos(e$) => y’ = ec°sx(— $111.23) + [— sin(ex) ~ez] : —Sina:ec°” — ex sin(ew) 33. y : lnlsec 5m +tan 526) :> 1 I _ 5 sec 526 (tan 5:1: + sec 530) y _ sec 53: + tan 535 (sec 53v tan 5m ~ 5 + 5e02 52: ~ 5) : : 5 sec 533 sec 53: + tan 51v 34. y 2 10mm "9 :> y' : 10tan ”0 ~ln 10 - sec2 71'6-71‘ : «(In 10)10ta“”0 sec2 7H9 35_ y : cot(3m2 + 5) => 34' : — csc2 (3232 + 5) (63:) = —61:csc2(3ar2 + 5) 36. y : «/tln(t4) => 31' : é [tln(t4)]_1/2 % [t1n(t4)] : m - [1 -1n(t4) +t- 111 -4t3 _ 1 . n 4 : ln(t4)+4 — 2 t1n(t4) [I (t )+4] 2 tln(t4) 0r: Since _v is only defined fort > 0. we can write y 2 wt - 4lnt : 2 x/tlnt. Then y' : 2 - 1 - <1 - lnt +t - l) — lnt + 1. This agrees with our first answer since 2x/tlnt t _ x/tlnt ln(t4)+4 4lnt+4 4(lnt-l—1) _lnt+1 2 tln(t4) _ 2x/t-4lnt I 2.2x/t1nt _ x/tlnt. 37.y:sin(tan\/1+CB3) :> y':cos(tan\/1+m3)(sec2v1+m3)[3w2/(2x/1+:c3)] 1 1 1 # 1+(a.rcsin\/a_:)2 x/l—x Zfi 39. y : tan2 (sin 6) : [tan(sir16)]2 => 7/ : 2[ta.n(sin 6)] - sec2 (sin 9) - cosO / 38. y : arctan(arcsin fl) y 40.$ey:y71 2 meyyl+ey:y' :> eyzy'—meyy' => y’:ey/(1-xey) \/m+1(2—;z:)5 1 _ l #—lnm+1 +5ln24$ —7ln($+3) => 41y (H3). my 2 r > < > y' 1 —5 7 , \/;z: + 1(2 — ac)5 1 5 7 or "' + ’ y 7 1 2— 33—1—3 y 2(w+1) 2fim 90—1—3 (:z;+3) 2(a:+) a: / _ (2 — $)4(3:62 — 55m — 52) y T 2x/w+1(1¢+3)8 ' ...
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