Is there a smarter way to differentiate the function $f(x) = \sin^{-1} \frac{2x}{1+x^2}$? 
Given $f(x) = \sin^{-1} \frac{2x}{1+x^2}$,
Prove that $$f'(x) = \begin{cases}\phantom{-}\frac{2}{1+x^2},\,|x|<1 \\\\ -\frac{2}{1+x^2},\,|x|>1 \end{cases}$$

Obviously the standard approach would be to use the chain rule and simplify from there.
But I noticed that some of these expressions are familiar, specifically, from the tangent half-angle formulae:
If $x = \tan \frac \theta 2$, then $\sin \theta = \frac{2x}{1+x^2}$ and $\frac{d\theta}{dx} = \frac{2}{1+x^2}$.
So my question is: can this observation be used to construct a more elegant proof?

Method$\#1:$
By Chain rule,
$$\dfrac{d\arcsin\dfrac{2x}{1+x^2}}{dx}=\dfrac1{\sqrt{1-\left(\dfrac{2x}{1+x^2}\right)^2}}\left(\dfrac2{1+x^2}-\dfrac{4x^2}{(1+x^2)^2}\right)$$
$$=\dfrac{2(1+x^2)(1-x^2)}{|1-x^2|(1+x^2)^2}=?$$
Method$\#2:$
Let $\arctan x=y\implies x=\tan y$
Using principal values $$\arcsin\dfrac{2x}{1+x^2}=\begin{cases}2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2 \\
\pi-2\arctan x & \mbox{if } 2\arctan x>\dfrac\pi2\\-\pi- 2\arctan x  & \mbox{if } 2\arctan x<-\dfrac\pi2\end{cases}$$
