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?

Implicit differentiation:
$$\sin(f(x)) = \frac{2x}{1 + x^2}$$
$$\cos(f(x)) f'(x) = \frac{2(1 - x^2)}{(1 + x^2)^2}$$
$$
f'(x) = \frac{2(1 - x^2)}{\cos(f(x))(1 + x^2)^2} =
\frac{2(1 - x^2)}{\sqrt{1 - \sin^2(f(x))}(1 + x^2)^2} = 
\frac{2(1 - x^2)}{\sqrt{(1 - x^2)^2/(1 + x^2)^2}(1 + x^2)^2}\cdots
$$
and:


*

*$1 - x^2 < 0\hbox{ for }|x| > 1$;

*$1 - x^2 > 0\hbox{ for }|x| < 1$ .

