Answer
$f(x)=-\dfrac{15x^2}{(x^3-2)^2},f'(-2)=-\dfrac{3}{5}.$
Work Step by Step
$f(x)=5(x^3-2)^{-1}$
$u=x^3-2$; $\dfrac{du}{dx}=3x^2$
$f(u)=5u^{-1};\dfrac{d}{du}f(u)=-\dfrac{5}{u^2}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=-\dfrac{15x^2}{(x^3-2)^2}.$
$f'(-2)=-\dfrac{15(-2)^2}{((-2)^3-2)^2}=-\dfrac{3}{5}.$
A graphing utility was used to verify this result.
