Answer
$f^{\prime}(x)=-1.5(-x+2)^{0.5}$
Work Step by Step
Let $u(x)=x^{1.5},\qquad v(x)=-x+2$
$\displaystyle \frac{du}{dx}=1.5x^{0.5}, \quad \frac{dv}{dx}=-1$
Then,$\quad y=f(x)=u(v(x)$ and $f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=1.5v^{ 0.5}=1.5(-x+2)^{0.5}$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}$
$=1.5(-x+2)^{0.5}\cdot(-1)=-1.5(-x+2)^{0.5}$
