Solve this


Differentiate $(\log x)^{x}$ with respect to $\log x$


Let $u=(\log x)^{x}$

Taking log on both sides,

$\log u=\log (\log x)^{x}$

$\Rightarrow \log u=x \log (\log x)$

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=x \frac{d}{d x}\{\log (\log x)\}+\log (\log x) \frac{d}{d x}(x)$

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=x\left(\frac{1}{\log x}\right) \frac{d}{d x}(\log x)+\log \log x(1)$

$\Rightarrow \frac{d u}{d x}=u\left[\frac{x}{\log x}\left(\frac{1}{x}\right)+\log \log x\right]$

$\Rightarrow \frac{d u}{d x}=(\log x)^{x}\left[\frac{1}{\log x}+\log \log x\right]$                         ......(1)

Again, let $v=\log x$

$\Rightarrow \frac{d v}{d x}=\frac{1}{x}$           .....(2)

Dividing equation (i) by (ii), we get

$\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{(\log x)^{x}\left[\frac{1}{\log x}+\log \log x\right]}{\frac{1}{x}}$

$\Rightarrow \frac{d u}{d v}=\frac{(\log x)^{x}\left[\frac{1+\log x(\log \log x)}{\log x}\right]}{\frac{1}{x}}$

$\Rightarrow \frac{d u}{d v}=x(\log x)^{x^{-1}}(1+\log x \times \log \log x)$

Leave a comment