Unformatted text preview: ( x )) = ln ³ x ( x +1) ´ , and then simplifying the righthand side, ln( y ( x )) = ( x + 1) ln x. Now diﬀerentiate this last equation implicitly: 1 y ( x ) y ( x ) = ( x + 1) · 1 x + 1 · ln x, so y ( x ) = y ( x ) ± x + 1 x + ln x ² . Now substituting back in the value of y ( x ) = x ( x +1) , we have y ( x ) = x ( x +1) ± ln x + 1 + 1 x ² ....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.
 Fall '10
 KIHYUNHYUN
 Calculus, Chain Rule

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