Problem Set 08s

F x log 5 xe x solution b 1 d 1 f x

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Unformatted text preview: ྎ − sin x ln x ⎥ྏ y ⎣ྏ x ⎦ྏ cos x − x sin x ln x ⎤ྏ ⎡ྎ y ' = x cos x ⎢ྎ ⎥ྏ x ⎣ྏ ⎦ྏ d. f ( x ) = log 5 ( xe x ) Solution: b. 1 d 1 f (' x ) = x ( xe x ) = xe x ln 5 (e x + xe x ) xe ln 5 dx e x + xe x =x xe ln 5 1+ x f ' ( x) = x ln 5 ln x y = (sin x ) Solution: 1 ) Hung Nguyen c. Problem Set 08 Solution GE 207C 3.6 Derivatives of Logarithmic Functions ln x a. f ( x ) = 2 + ln x y = (sin x ) Solution: ln y = ln (sin x ln x ) = ( ln x ) ln (sin x ) f ( x) = 2 + ln x !1$ 1 1 y ' = ln (sin x ) + ln x # & cos x 1 1 ⎞ྏ 1 −1 / 2 ⎛ྎ " sin x % y x f ' ( x) = (2 + ln x ) ⎜ྎ 0 + ⎟ྏ = 2 x ⎠ྏ 2 x 2 + ln x ⎝ྎ ! cos x $ ln(sin x ) = + ( ln x ) # & 1 " sin x % x f ' ( x) = 2 x 2 + ln x 1 ln(sin x ) y' = + ( ln x ) c...
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This document was uploaded on 02/27/2014 for the course M 408c at University of Texas.

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