Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,

Hint: Let $f(x) = \cot^{-1} x$ then $f'(x) = -\frac{1}{1+x^2}$ and if $g(x) = \frac{1-x}{1+x}$ then $g'(x) = -\frac{2}{(x+1)^2}$
Now use the chain rule $$[f(g( x ))]' = f'(g(x))\ g'(x) $$
Notice that $$-\frac{1}{1+\Big(\frac{1-x}{1+x}\Big)^2} = -\frac{(1+x)^2}{(1+x)^2+(1-x)^2} = -\frac{(1+x)^2}{1+2x + x^2 +1-2x+ x^2} = -\frac{1}{2}\frac{(1+x)^2}{(1 + x^2)} $$
