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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?

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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 $$ f(x) = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ 2\arctan\left(\frac{1}{x}\right) &\!\!\!, |x|>1 \end{cases} = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ \operatorname{sgn}(x)\pi -2\arctan(x) &\!\!\!, |x|>1 \end{cases} \, .$$

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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 $$\arcsin\dfrac{2x}{1+x^2}=\begin{cases}2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2 \\ \pi-2\arctan x & \mbox{if } 2\arctan x>\dfrac\pi2\\-\pi- 2\arctan x & \mbox{if } 2\arctan x<-\dfrac\pi2\end{cases}$$

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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:

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