Answer
${dy}=(\frac{x-3}{2(x^2-1)}){dx}$
Work Step by Step
We evaluate the function: $y=\ln\frac{(x+1)}{\sqrt{(x-1)}}$
$y=\ln{(x+1)}-ln{(\sqrt{(x-1)})}$
$y=\ln{(x+1)}-\frac{1}{2}ln{{(x-1)}}$
on differentiating the above:
$\frac{dy}{dx}=\frac{d({\ln{(x+1)}-\frac{1}{2}ln{{(x-1)}}})}{dx}$
$\frac{dy}{dx}=\frac{1}{(1+x)}\frac{d(1+x)}{dx}-\frac{1}{2}\frac{1}{x-1}\frac{d(x-1)}{dx}$
or $\frac{dy}{dx}=\frac{1}{1+x}-\frac{1}{2(x-1)}$
or $\frac{dy}{dx}=\frac{2x-2-1-x}{2(x^2-1)}$
or ${dy}=(\frac{x-3}{2(x^2-1)}){dx}$
