Unformatted text preview: x = Ï€ 2 , we get f ( Ï€ 2 ) = e Ï€ 2 1 Ï€ 2 Â· (1 + ln(1) + Ï€ 2 Â· 1 ) = e Ï€ 2 . Solution 2: Take logarithms of both sides of y = e x (sin x ) x , to get ln y = ln( e x (sin x ) x ) = ln( e x ) + ln((sin x ) x ) = x + x ln(sin x ) . Now take derivatives of both sides to get 1 y y = 1 + ln(sin x ) + x cos x sin x . Multiplying both sides by y = e x (sin x ) x gives exactly the same derivative formula as in solution 1, and the rest follows indentically. 1...
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This note was uploaded on 09/17/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at Berkeley.
 Spring '08
 WILKENING
 Math, Calculus

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