The derivative of ${\tan }^{-1}\frac{2x}{1-x^2}$ with respect to ${\sin }^{-1}\frac{2x}{1+x^2}$, is

Let $y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ and  $z={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ Then,

$y=\left\{\begin{array}{l}2{\tan }^{-1}x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }-1<x<1\\ 2{\tan }^{-1}x-\pi ,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }x>1\\ 2{\tan }^{-1}x+\pi ,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }x<-1\end{array}\right.$

and  $z=\left\{\begin{array}{l}2{\tan }^{-1}x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }-1\le x\le 1\\ \pi -2{\tan }^{-1}x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }x>1\\ -\pi -2{\tan }^{-1}x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if }x<-1\end{array}\right.$

$\therefore \frac{dy}{dz}=\left\{\begin{array}{r}1,\text{ if }-1<x<1\\ -1,\text{ if }\left|x\right|>1\end{array}\right.$