Answer
$y'=\frac{-1}{x^2+1}$
Work Step by Step
Let $u=\frac{1-x}{1+x}$.
We have $y=\arctan(u)$.
Find $\frac{du}{dx}$ using the Quotient Rule:
$\frac{du}{dx}=\frac{\frac{d}{dx}(1-x)\cdot (1+x)-(1-x)\cdot \frac{d}{dx}(1+x)}{(1+x)^2}$
$=\frac{-1(1+x)-(1-x)1}{(1+x)^2}$
$=\frac{-1-x-1+x}{(1+x)^2}$
$=\frac{-2}{(1+x)^2}$
Find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
$=\frac{d}{du}(\arctan(u))\cdot \frac{-2}{(1+x)^2}$
$=\frac{1}{1+u^2}\cdot \frac{-2}{(1+x)^2}$
$=\frac{1}{1+(\frac{1-x}{1+x})^2}\cdot \frac{-2}{(1+x)^2}$
$=\frac{1}{1+\frac{(1-x)^2}{(1+x)^2}}\cdot \frac{-2}{(1+x)^2}$
$=\frac{-2}{(1+x)^2+(1-x)^2}$
$=\frac{-2}{1+2x+x^2+1-2x+x^2}$
$=\frac{-2}{2x^2+2}$
$=\frac{-1}{x^2+1}$
