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Thread: Differentiation involving ln x

  1. #1
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    Differentiation involving ln x

    I am stuck at this! Differentiate with respect to x:

    y = ( ln x ) to the power ln x

    Thanks
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  2. #2
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    $\displaystyle ln(x)^x = x\times ln(x)$

    now use the product rule for $\displaystyle y=uv$ then $\displaystyle y' = v'u+u'v$
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  3. #3
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    Quote Originally Posted by puggie View Post
    I am stuck at this! Differentiate with respect to x:

    y = ( ln x ) to the power ln x

    Thanks
    $\displaystyle y=(\ln(x))^{\ln(x)}$

    This is going to need the chain rule and this identity

    $\displaystyle \ln(x)^{\ln(x)}=e^{ln(\ln(x)^{\ln(x)})}=e^{\ln(x)\ cdot \ln(\ln(x))}$

    So now we get

    $\displaystyle y'=e^{\ln(x)\cdot \ln(\ln(x))}\left( \frac{1}{x}\ln(\ln(x)+\ln(x)\left(\frac{1}{\ln(x)} \frac{1}{x}\right)\right)$

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