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File0041

# File0041 - 62 63 64 65 66 67 68 69 70 71 Section 3.5 The...

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Unformatted text preview: 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. Section 3.5 The Chain Rule and Parametric Equations 129 .y:4sin(W>:%¥=4cos( 1+\/E)'d%<\/T\/E)=4COS<W)- m-d— d(1+\/) y=<1+i =>y=3<1+1>2<—12)=—3—2<1+1>2=>y=1 - =<—;1>(2(1+1>Hum?)(1+1)2=§11+1>+%(1+1)2=511+)< +1+) ~%<1+1)(1+%> y = (1 - x/i)‘l :1 y’ 2 - (1 — \/§)‘2(—§x-1/2) 21(1~ ﬁ)“2x—1/2 . . . ﬁ)73 (‘1 2 x‘1/2)] = % [—71X—3/2 (1 _ ﬁ)—2 +x'1 (1 _ \$04] = %x‘1 (1_ ﬁ)—3 [_%X—1/2(1 _ ﬂ) +1] *< I 531... 1—1 A .— I E 1 [\J 1 m1...- >< I o. \ (x. V + >4 | 1: A l N V A '—d I y : 1mm — 1) :> y’: —-%csc2(3x—1)(3):— —12csc (3x — 1) :> y” - (— -) (csc(3x — 1) — csc(3x — 1)) = — % csc(3x —1)(-—csc(3x —1)c0t(3x — 1) - d—dx (3x —1))= 2051:2 (3x — 1) c0t(3x — 1) ) y = 9 me) : y' = 91sec? (1)) (1) = 3 (1) j y” = 3-2 we) (sec (1) (1)) (1) = 2m? (1) tan< ulx y = ex2 + 5x => 31’ = 2x1:x2 + 5 => y” = 2x - e":(2x) + 2ex2 = (4x2 + 2)ex2 y = sin(xze" )=> y’ — —cos(x2 ex)- (x 26" + 2xe") = (x2 + 2x)excos(xze") => (Use triple product rule: D(fgh) = f’gh + fg’h + fgh’) y” = (2x + 2)excos(x2e") + (x2 + 2x)excos(x2e") + (x2 + 2x)e" (—— sin(x2 e ) (x2 e" + er" )) : (x2 + 4x + 2)excos(x2ex) — xe2"(x3 + 4x2 + 4x)sin(xzex) ;f(u )=u5+1 => f’(u)=5u4 ;» f’(g(1))=f’(1)=5; Nip—1 g(x>= \ﬂ => g’<x)= ﬁx- => g(1)=1andg<1)= therefore, (f0 g)’(1)= f’(g(1)) -g’(1)= 5 . 5 = 5 ECO—(1 X)_1 gI(X) (1 X) 2( 1)_(1_1X) _> g( 1)—§andg’-( 1):}7; f(u):1—% —> f’(u)— ﬁ —> f’(g( 1))—f’(2) —4, therefore, (fo g)’( 1)=f’(g(— 1))g(— 1)=4 -% 1 g(x)=5\/§ => g’(x) = ﬁ- :> g(1):5andg(1)=g; f(u)— cot(71'—(1“-)) => f’(u)— —csc2 (7;) (ﬂ :j csc2 (%) :> f/(g(1)) = fl(5) : ~ m csc2 (g) 2 _ _ therefore (f0 g)’ (1)_ _ f (g(1))g (1)_ _ _ 7% . 10’ I: MIL/1 #1)! O g(x):7rx :> g’(x)=7r :> g( 1 Z ndg'Gi) =7r;f(u)=u+sec2u :> f’(u)=l+23ecu- secutanu :1+Zsec2 utanu:>f’(g( ) :1+2sec2—tan§=5;therefore,(fog)’(1)=f’(g())g() 1’u2 + 1)(2) V (2u)(2u) a“? (u2 + 1)" g(x)=10x2 +x+l :> g’(x) : 20x +1 => g(0):1andg’(0):1;f(u): “121—1 :> f’(u) : = '2,"—+% => f'(g(0)) = f'(1) = 0; therefore, (fo g)’(0) = f/(g(0))g’(0) : 0 . 1 z 0 (u2 +1) ...
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