$f(x)=\sin^{-1}{\frac{2x}{1+x^2}} , x \in \mathbb{R}$ What is the derivative of $f(x)$?

$\begin{cases} \frac{2}{1+x^2}& \text{if}{\hspace.2cm} x^2<1\\ \frac{-2}{1+x^2}& \text{if}{\hspace.2cm} x^2>1 \end{cases}$

Let. $f(x)=sin^{-1}{\frac{2x}{1+x^2}} , x \in \mathbb{R}$ What is the derivative of $f(x)$?\\ Solution: $f$ is composition of two functions $h(x)=sin^{-1}x$ and $g(x)=\frac{2x}{1+x^2}$. Since $g(x)\leq 1$, $h(g(x))$ is well defined. Now $f'(x)=\frac{1}{\sqrt {1-(2x/1+x^2)^2}}\frac{d}{dx}(2x/1+x^2)=\frac{1+x^2}{\sqrt {(1-x^2)^2}}*\frac{2-2x^2}{1+x^2}$ provided $\frac{2x}{1+x^2} \neq 1$ i.e $x \neq 1,-1$. Now $\sqrt{(1-x^2)}=1-x^2 $ if $x^2<1$ and $x^2-1$ if $x^2>1$. So $f'(x)=\begin{cases} \frac{2}{1+x^2}& \text{if}{\hspace.2cm} x^2<1\\ \frac{-2}{1+x^2}& \text{if}{\hspace.2cm} x^2>1 \end{cases}$