Differentiate the following functions with respect to x :

Let $y=x^{\frac{1}{x}}$

Taking log both the sides:

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

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

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

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

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

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

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

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{x} \times \frac{1}{x} \frac{d x}{d x}+\log x\left(\frac{-1}{x^{2}}\right)$

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} ; \frac{d\left(u^{n}\right)}{d x}=n u^{n-1} \frac{d u}{d x}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1-\log x}{x^{2}}$

$\Rightarrow \frac{d y}{d x}=y\left(\frac{1-\log x}{x^{2}}\right)$

Put the value of $y=x^{\frac{1}{x}}$ :

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\frac{1}{\mathrm{x}}}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\right)$