Answer
$\frac{dy}{dx}=(\frac{1-x}{1+x})^{1/2}\times(\frac{1}{2y(1-x)^2})$
Work Step by Step
Rewrite the equation:$y^2=(\frac{1+x}{1-x})^{1/2}$
Take the derivative of the equation on each side separately. Apply chain rule when differentiating the "y" variables since we are differentiating with respect to x:
$2y\frac{dy}{dx}=\frac{1}{2}(\frac{1+x}{1-x})^{-1/2}\times(\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2})$
$2y\frac{dy}{dx}=\frac{1}{2}(\frac{1+x}{1-x})^{-1/2}\times(\frac{1-x+1+x}{(1-x)^2})$
$2y\frac{dy}{dx}=\frac{1}{2}(\frac{1+x}{1-x})^{-1/2}\times(\frac{2}{(1-x)^2})$
$2y\frac{dy}{dx}=(\frac{1-x}{1+x})^{1/2}\times(\frac{1}{(1-x)^2})$
Move all terms with dy/dx to one side of the equation, and isolate dy/dx:
$\frac{dy}{dx}=(\frac{1-x}{1+x})^{1/2}\times(\frac{1}{2y(1-x)^2})$
