Answer
$$y'=\frac{1}{2y(1-x)^2}\sqrt{\frac{1-x}{1+x}}$$
Work Step by Step
$$y^2=\sqrt{\frac{1+x}{1-x}}$$
$$y^2=\Big(\frac{1+x}{1-x}\Big)^{1/2}$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$2y\times y'=\frac{1}{2}\Big(\frac{1+x}{1-x}\Big)^{-1/2}\times\Big(\frac{1+x}{1-x}\Big)'$$
$$2yy'=\frac{1}{2}\Big(\frac{1-x}{1+x}\Big)^{1/2}\times\frac{(1+x)'(1-x)-(1+x)(1-x)'}{(1-x)^2}$$
$$2yy'=\frac{1}{2}\sqrt{\frac{1-x}{1+x}}\times\frac{1\times(1-x)-(1+x)\times(-1)}{(1-x)^2}$$
$$2yy'=\frac{1}{2}\sqrt{\frac{1-x}{1+x}}\times\frac{(1-x)+(1+x)}{(1-x)^2}$$
$$2yy'=\frac{1}{2}\sqrt{\frac{1-x}{1+x}}\times\frac{2}{(1-x)^2}$$
$$2yy'=\frac{2}{2(1-x)^2}\sqrt{\frac{1-x}{1+x}}$$
$$2yy'=\frac{1}{(1-x)^2}\sqrt{\frac{1-x}{1+x}}$$
Finally, we calculate $y'$: $$y'=\frac{1}{2y(1-x)^2}\sqrt{\frac{1-x}{1+x}}$$
