Derivative of

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

Use the following identities and formulas.

As Derivative of $\tan x=se{c}^{2}x$

As we know that $1+{\tan }^{2}x=se{c}^{2}x$

Also $\tan \left({\tan }^{-1}x\right)=x$

Given: $f\left(x\right)={\tan }^{-1}x$

Let $y={\tan }^{-1}x$

$\Rightarrow \tan y=x$

Differentiate both sides$w.r.t.x$

$1=se{c}^{2}y\left(\frac{dy}{dx}\right)$

$\Rightarrow 1=(1+{\tan }^{2}y)\left(\frac{dy}{dx}\right)$

Put $\tan y=x$

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

$\Rightarrow \frac{dy}{dx}=\frac{1}{(1+x^2)}$

Hence proved.