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Unformatted text preview: u = e 4 x , u ± = 4 e 4 x and Power Chain Rule for the second term as cos 3 ( x ) = (cos x ) 3 y ± = 4e 4 x ƒ sec( e 4 x ) tan( e 4 x ) + 3 cos 2 x ƒ (sin x ) (d) h ( θ ) = θ 2 tan( √ θ ) Solution: Use Product Rule ( u ƒ v ) ± = u ƒ v ± + v ƒ u ± , with u = θ 2 , v = tan( √ θ ) , u ± = 2 θ, and v ± = sec 2 ( √ θ ) ƒ 1 2 √ θ h ± ( θ ) = θ 2 ƒ sec 2 ( √ θ ) ƒ 1 2 √ θ + 2 θ ƒ tan( √ θ ) (e) f ( u ) = eu 1 + u 2 Solution: Use Quotient Rule ‡ w v · ± = v ƒ w ±w ƒ v ± v 2 , with w = eu , v = 1 + u 2 , w ± =eu , and v ± = 2 u f ± ( u ) = (1 + u 2 ) ƒ (eu )eu ƒ (2 u ) (1 + u 2 ) 2 =eu (1 + u 2 + 2 u ) (1 + u 2 ) 2...
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 Spring '08
 Dr.Z
 Calculus, Chain Rule, Derivative, Integrals

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