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Math Help - 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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    ln(x)^x = x\times ln(x)

    now use the product rule for y=uv then 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
    y=(\ln(x))^{\ln(x)}

    This is going to need the chain rule and this identity

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

    So now we get

    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)

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