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Thread: derevatitive

  1. #1
    Member
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    Apr 2010
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    135

    derevatitive

    hello how would i find dy/dx of $\displaystyle y = x^x$
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  2. #2
    Senior Member
    Joined
    May 2010
    Posts
    251
    Hi there. You must think about logaritms.

    This is the way you can solve it:

    $\displaystyle y=x^x$

    $\displaystyle \ln(y)=\ln(x^x)$

    $\displaystyle \ln(y)=x\ln(x)$

    Now the derivatives:

    $\displaystyle \frac{1}{y}y'=\ln(x)+x\frac{1}{x}$

    $\displaystyle y'= y(\ln(x)+1)$
    Last edited by Ulysses; May 26th 2010 at 04:49 AM. Reason: Fixed a typo.
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