Differentiate the following functions with respect to x :

Let $y=e^{x \log x}$

Taking log both the sides:

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

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

$\Rightarrow \log y=x \log x\{\log e=1\}$

Differentiating with respect to $\mathrm{x}$ :

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

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

$\left\{\right.$ Using product rule, $\left.\frac{\mathrm{d}(\mathrm{uv})}{\mathrm{dx}}=\mathrm{u} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{v} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

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

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

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

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

Put the value of $y=e^{x \log x}:$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x} \log \mathrm{x}}\{1+\log \mathrm{x}\}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}^{\mathrm{x}}}\{1+\log \mathrm{x}\}\left\{\mathrm{e}^{\log \mathrm{a}}=\mathrm{a} ; \mathrm{alog} \mathrm{x}=\mathrm{x}^{\mathrm{a}}\right\}$

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