## Calculus 10th Edition

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