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,
 A: 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.
A: 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 $$-\frac{1}{1+\Big(\frac{1-x}{1+x}\Big)^2} = -\frac{(1+x)^2}{(1+x)^2+(1-x)^2} = -\frac{(1+x)^2}{1+2x + x^2 +1-2x+ x^2} = -\frac{1}{2}\frac{(1+x)^2}{(1 + x^2)} $$
A: $$
f'(x)= \frac{-1}{1+{(\frac{1-x}{1+x}})^2}.\frac{d}{dx}(\frac{1-x}{1+x})
$$
$$f'(x)=\frac{-(1+x)^2}{2(1+x^2)}.\frac{-1(1+x)-(1-x)}{(1+x)^2}$$
$$f'(x)=\frac{-1(-1-x-1+x)}{2(1+x^2)}$$
$$f'(x)=\frac{1}{1+x^2}$$
A: 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?
