Answer
$\dfrac{-20x}{(x^2-1)^{11}} $
Work Step by Step
We have: $f(x)=\dfrac{1}{(x^2-1)^{10}}$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [(x^2-1)^{-10}] \\=(-10) (x^2-1)^{-11}\dfrac{d}{dx}(x^2-1)\\=-10(x^2-1)^{-11}\times 2x $
Simplify to obtain:
$f^{\prime}(x)=\dfrac{-20x}{(x^2-1)^{11}} $
