$f(x) = \arccos {\frac{1-x^2}{1+x^2}}$; f'(0+), f'(0-)? $f(x) = \arccos {\frac{1-x^2}{1+x^2}}$
$f'(x) = 2/(1+x^2)$, 
but I see graphic, and it is true only for x>=0.
For x<=0 => $f'(x) = -2/(1+x^2)$
How can I deduce the second formula or proof that it is.

Method $\#1:$
Let $\displaystyle\arccos\frac{1-x^2}{1+x^2}=y$
$\displaystyle\implies(1) \cos y=\frac{1-x^2}{1+x^2}\ \ \ \ (i)$
Using the definition of Principal values,
$\displaystyle \implies(2)0\le y\le\pi \implies 0\le\frac y2\le\frac\pi2\implies \tan\frac y2\ge0$
Applying Componendo and dividendo on $(i),$
$\displaystyle x^2=\frac{1-\cos y}{1+\cos y}=\tan^2\frac y2$ (using $\displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$)
As $\displaystyle\tan\frac y2\ge0,\implies \tan\frac y2=|x|$ 

Method $\#2:$
Let $\displaystyle z=\arctan x$
$\displaystyle\implies -\frac\pi2\le z\le\frac\pi2\iff -\pi\le 2z\le\pi$
and $\displaystyle\tan z=x,\frac{1-x^2}{1+x^2}=\cos2z$
$\displaystyle\arccos(2z)=\begin{cases} 2z=2\arctan x &\mbox{if } 0\le 2z\le\pi\iff 0\le z\le\frac\pi2\implies x=\tan z\ge0\ \\-2z=-2\arctan x & \mbox{if } -\pi\le 2z<0\iff -\frac\pi2\le z<0\implies x<0 \end{cases} $
