<s>[INST] 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,
 [/INST] $$
f'(x)= \frac{-1}{1+{(\frac{1-x}{1+x}})^2}.\frac{d}{dx}(\frac{1-x}{1+x})
$$
$$f'(x)=\frac{-(1+x)^2}{2(1+x^2)}.\frac{-1(1+x)-(1-x)}{(1+x)^2}$$
$$f'(x)=\frac{-1(-1-x-1+x)}{2(1+x^2)}$$
$$f'(x)=\frac{1}{1+x^2}$$
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