Answer
$f^{\prime}(x)=-5(2x-3)(x^{2}-3x-1)^{-6}$
Work Step by Step
Let $u(x)=x^{-5},\qquad v(x)=x^{2}-3x-1$
$\displaystyle \frac{du}{dx}=-5x^{-4}, \quad \frac{dv}{dx}=2x-3$
Then,$\quad y=f(x)=u(v(x))$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=-5v^{-6}=-5(x^{2}-3x-1)^{-6}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=-5(x^{2}-3x-1)^{-6}\cdot(2x-3)$
