## Calculus: Early Transcendentals 8th Edition

$f'(x)=-\dfrac{2x}{3\sqrt[3] {(x^{2}-1)^{4}}}$
$f(x)=\dfrac{1}{\sqrt[3] {x^{2}-1}}$ Let's do some algebra to change the appearance of the function: $f(x)=\dfrac{1}{\sqrt[3] {x^{2}-1}}=\dfrac{1}{(x^{2}-1)^{1/3}}=(x^{2}-1)^{-1/3}$ Now, differentiate using the chain rule: $f'(x)=-\dfrac{1}{3}(x^{2}-1)^{-4/3}(x^{2}-1)'=...$ $...=-\dfrac{1}{3}(x^{2}-1)^{-4/3}(2x)=-\dfrac{2}{3}x(x^{2}-1)^{-4/3}=...$ $...=-\dfrac{2x}{3(x^{2}-1)^{4/3}}=-\dfrac{2x}{3\sqrt[3] {(x^{2}-1)^{4}}}$