$\begin{align}  & y=\sqrt{\frac{x-1}{x+1}} \\ & y={{\left( \frac{x-1}{x+1} \right)}^{\tfrac{1}{2}}} \\ & \text{Take the logarithm of both sides} \\ & \log y=\log {{\left( \frac{x-1}{x+1} \right)}^{\tfrac{1}{2}}} \\ & \log y=\frac{1}{2}\log \left( \frac{x-1}{x+1} \right) \\ & \log y=\frac{1}{2}\left[ \log (x-1)-\log (x+1) \right] \\ & \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left( \frac{1}{x-1}-\frac{1}{x+1} \right) \\ & \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left( \frac{x+1-x+1}{(x-1)(x+1)} \right) \\ & \frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[ \frac{2}{(x-1)(x+1)} \right]=\frac{1}{(x-1)(x+1)} \\ & \frac{dy}{dx}=\frac{y}{(x-1)(x+1)}={{\left( \frac{x-1}{x+1} \right)}^{\tfrac{1}{2}}}\frac{1}{(x-1)(x+1)} \\ & \frac{dy}{dx}=\frac{1}{{{(x-1)}^{\tfrac{1}{2}}}{{(x+1)}^{\tfrac{3}{2}}}} \\\end{align}$