Differentiate the following functions with respect to x :

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

Taking log both the sides:

$\Rightarrow \log y=\log (\log x)^{x}$

$\Rightarrow \log y=x \log (\log x)\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log y)}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{x} \log \log \mathrm{x})}{\mathrm{dx}}$

$\Rightarrow \frac{d(\log y)}{d x}=x \times \frac{d(\log \log x)}{d x}+\log \log x \times \frac{d x}{d x}$

$\left\{\right.$ Using product rule, $\left.\frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right\}$

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

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{x}{\log x} \times \frac{1}{x}+\log \log x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}\left\{\frac{1}{\log \mathrm{x}}+\log \log \mathrm{x}\right\}$

Put the value of $y=(\log x)^{x}$

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