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