$f(x) = \arccos {\frac{1-x^2}{1+x^2}}$; 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.

Note $f(x)=f(-x)$. By the Chain Rule, $f'(-x)=-f'(x)$. That gives your formula for $f'(x)$ for $x\lt0$. 
