Find the second order derivative of $y={\mathrm{Tan}}^{-1}\left(\frac{2x}{1-x^2}\right)$

Given $y={\mathrm{Tan}}^{-1}\left(\frac{2x}{1-x^2}\right)$

Put $x=\tan \theta \Rightarrow \theta ={\mathrm{Tan}}^{-1}x$

$\therefore y={\mathrm{Tan}}^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)\Rightarrow y={\mathrm{Tan}}^{-1}(\tan 2\theta )$

$\Rightarrow y=2\theta \Rightarrow y=2{\mathrm{Tan}}^{-1}x$

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

$\therefore \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left[2\left(\frac{1}{1+x^2}\right)\right]$

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

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

$=-2{\left(1+x^2\right)}^{-2}(0+2x)$

$=\frac{-4x}{{\left(1+x^2\right)}^2}$