# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$x^{\tan ^{-1} x}$

Solution:

Lety $=x^{\tan ^{-1} x}$

Taking log both the sides:

$\Rightarrow \log y=\log x^{\tan ^{-1} x}$

$\Rightarrow \log y=\tan ^{-1} \times \log \times\left\{\log x^{2}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

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

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

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

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

$\left\{\frac{d(\log u)}{d x}=\frac{1}{u} \frac{d u}{d x} ; \frac{d\left(\tan ^{-1} u\right)}{d x}=\frac{1}{u^{2}+1} \frac{d u}{d x}\right\}$

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

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

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

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