Answer
$\displaystyle \frac{dy}{dx}=\frac{2x}{(x^{2}+1)^{3/2}(x^{2}-1)^{1/2}}$
Work Step by Step
$y=\sqrt{\frac{x^{2}-1}{x^{2}+1}}$
$...$apply ln( ) to both sides
... on the RHS, apply $\ln M^{n}=n\ln M,$ and $\displaystyle \ln(\frac{M}{N})=\ln M-\ln N$
$\displaystyle \ln y=\frac{1}{2}[\ln(x^{2}-1)-\ln(x^{2}+1)]\qquad $...$/\displaystyle \frac{d}{dx}$
... chain rule...
$\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}[\frac{2x}{x^{2}-1}-\frac{2x}{x^{2}+1}]$
$\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}[\frac{2x(x^{2}+1)-2x(x^{2}-1)}{(x^{2}-1)(x^{2}+1)}]$
$\displaystyle \frac{dy}{dx}=\sqrt{\frac{x^{2}-1}{x^{2}+1}}\cdot\frac{4x}{2(x^{2}-1)(x^{2}+1)}$
$\displaystyle \frac{dy}{dx}=\sqrt{\frac{x^{2}-1}{x^{2}+1}}\cdot\frac{2x}{(x^{2}-1)(x^{2}+1)}$
$\displaystyle \frac{dy}{dx}=\frac{2x(x^{2}-1)^{1/2}}{(x^{2}+1)^{1/2}(x^{2}-1)(x^{2}+1)}$
$\displaystyle \frac{dy}{dx}=\frac{2x}{(x^{2}+1)^{3/2}(x^{2}-1)^{1/2}}$
