Answer
$\dfrac{2x-2x\ln (x^2-1)}{(x^2-1)^2}$
Work Step by Step
We have: $f(x)=\dfrac{\ln (x^2-1)}{x^2-1}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [\dfrac{\ln (x^2-1)}{x^2-1}] \\=\dfrac{(x^2-1)\times \dfrac{2x}{x^2-1}-2x\ln (x^2-1)}{(x^2-1)^2}$
Simplify to obtain:
$f^{\prime}(x)=\dfrac{2x-2x\ln (x^2-1)}{(x^2-1)^2}$
