<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] Note that


*

*$(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}$


then apply chain rule $f(g(x))'=f'(g(x))g'(x)=\frac{\frac{2(1+x^2)-4x^2}{(1+x^2)^2}}{\sqrt{1-{\left(\frac{2x}{1+x^2}\right)}^2}}=\frac{2(1-x^2)}{(1+x^2)|1-x^2|}$
</s>