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?

We have
$$ \frac{t}{1+\sqrt{1-t^2}} \Bigg \rvert_{t = \frac{2x}{1+x^2}} = \frac{2x}{1+x^2 + \lvert 1-x^2\rvert} = \begin{cases} x &\!\!\!, |x|<1 \\ \frac{1}{x} &\!\!\!, |x|>1 \end{cases} \, .$$
Therefore we can use
$$ \arcsin(t) = 2 \arctan\left(\frac{t}{1+\sqrt{1-t^2}}\right) \, ,$$
which follows from the half-angle formula, to obtain
$$ f(x) = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ 2\arctan\left(\frac{1}{x}\right) &\!\!\!, |x|>1 \end{cases} = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ \operatorname{sgn}(x)\pi -2\arctan(x) &\!\!\!, |x|>1 \end{cases} \, .$$
