<s>[INST] Find derivative of $y=\sin^{-1}\frac{2x}{1+x^2}$ 
Find $\frac{dy}{dx}$ if $y=\sin^{-1}\frac{2x}{1+x^2}$

The solution is given as $\frac{2}{1+x^2}$. But is it a complete solution ?
My Attempt
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2},\quad |x|\leq 1\\
\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\; x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\;x>0\\
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x,\quad |x|\leq 1\\
\pi-2\tan^{-1}x,\quad |x|>1 \;\&\; x>0\\
-\pi-2\tan^{-1}x,\quad |x|>1 \;\&\;x>0\\
\end{cases}\\
$$
Thus,
$$
\frac{dy}{dx}=\frac{d}{dx}\bigg[\sin^{-1}\frac{2x}{1+x^2}\bigg]=\begin{cases}
\frac{d}{dx}[2\tan^{-1}x]=\frac{2}{1+x^2},\quad |x|\leq 1\\
\frac{d}{dx}[\pm\pi-2\tan^{-1}x]=\frac{-2}{1+x^2},|x|>1 
\end{cases}
$$
Is it correct ?
 [/INST] Hint
Your answer seems correct to me even I don't really understand the need of tan function
$$y=\sin^{-1}\frac{2x}{1+x^2}  \implies \sin y=\frac{2x}{1+x^2}$$
Differentiate 
$$y'\cos(y)=\frac {-2(x^2-1)}{(x^2+1)^2}$$
use the fact that $$\sin^2(y)+\cos^2(y)=1$$
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