<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] Do you know that $(\cos^{-1}t)'=-\frac{1}{1+t^2}$?
So $$f'(x)=-\frac{1}{1+(\frac{1-x}{1+x} )^2} \cdot (\frac{1-x}{1+x})'
=-\frac{1}{1+(\frac{1-x}{1+x} )^2} \cdot \frac{-(1+x)-(1-x)}{(1+x)^2} $$
Is it okay?
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