# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$(\tan x)^{1 / x}$

Solution:

Let $y=(\tan x)^{\frac{1}{x}}$

Taking log both the sides:

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

$\Rightarrow \log y=\frac{1}{x} \log \tan x\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

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

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

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

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

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

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

$\left\{\frac{d(\tan x)}{d x}=\sec ^{2} x\right\}$

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

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

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