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,

The derivative is incorrect.
The usual way is to use the chain rule on $\cot^{-1}$ then the quotient rule on the fraction. You would then get
$$f'(x)=(\cot^{-1})'\left(  \frac{1-x}{1+x} \right) \cdot \left(  \frac{1-x}{1+x} \right)'$$
$$=-\frac{1}{1+\left(\frac{1-x}{1+x} \right)^2} \cdot \frac{(1-x)'(1+x)-(1-x)(1+x)'}{(1+x)^2}$$
$$=-\frac{1}{1+\left(\frac{1-x}{1+x} \right)^2} \cdot \frac{-1(1+x)-(1-x)1}{(1+x)^2}$$
You can finish it from here.
