Differentiation involving ln x

**puggie** · June 3rd 2009, 06:34 PM

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

y = ( ln x ) to the power ln x

Thanks

**pickslides** · June 3rd 2009, 06:39 PM

$ln(x)^x = x\times ln(x)$

now use the product rule for $y=uv$ then $y' = v'u+u'v$

**TheEmptySet** · June 3rd 2009, 06:44 PM

Originally Posted by puggie

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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