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?
 A: 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
$$ f(x) = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ 2\arctan\left(\frac{1}{x}\right) &\!\!\!, |x|>1 \end{cases} = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ \operatorname{sgn}(x)\pi -2\arctan(x) &\!\!\!, |x|>1 \end{cases} \, .$$
A: 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 $$\arcsin\dfrac{2x}{1+x^2}=\begin{cases}2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2 \\
\pi-2\arctan x & \mbox{if } 2\arctan x>\dfrac\pi2\\-\pi- 2\arctan x  & \mbox{if } 2\arctan x<-\dfrac\pi2\end{cases}$$
A: 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:


*

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

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

