If $\mathrm{y}={\cos }^{-1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right),-1<\mathrm{x}<1$,then $\frac{\mathrm{dy}}{\mathrm{dx}}$is equal to 

 Substitute $\mathrm{x}=\tan \mathrm{\theta }\Rightarrow \mathrm{\theta }={\tan }^{-1}\mathrm{x}$ $\mathrm{y}={\cos }^{-1}\left(\frac{2\tan \mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}\right)$

$\Rightarrow \mathrm{y}={\cos }^{-1}(\sin 2\mathrm{\theta })\left[\because \sin 2\mathrm{\theta }=\frac{2\tan \mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}\right]$

$\Rightarrow \mathrm{y}={\cos }^{-1}\left[\cos \left(\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\right)\right]$ $\left[\because \sin 2\mathrm{\theta }=\cos \left(\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\right)\right]$

$\Rightarrow \mathrm{y}=\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\Rightarrow \mathrm{y}=\frac{\mathrm{\pi }}{2}-2{\tan }^{-1}\mathrm{x}\left[\because \mathrm{\theta }={\tan }^{-1}\mathrm{x}\right]$

On differentiating  both sides  vv.r.t.x, we get

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=0-\frac{2}{1+{\mathrm{x}}^2}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{1+{\mathrm{x}}^2}\left[\because \frac{\mathrm{d}}{\mathrm{dx}}\left({\tan }^{-1}\mathrm{x}\right)=\frac{1}{1+{\mathrm{x}}^2}\right]\end{array}$