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