Answer
$r^{\prime}(x)=1.5(0.2x-4.2)(0.1x^{2}-4.2x+9.5)^{0.5}$
Work Step by Step
Let $u(x)=x^{1.5},\qquad v(x)=0.1x^{2}-4.2x+9.5$
$\displaystyle \frac{du}{dx}=1.5x^{0.5}, \quad \frac{dv}{dx}=0.1(2x)-4.2=0.2x-4.2$
Then,$\quad y=r(x)=u(v(x))$ and $r^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=1.5v^{0.5}=1.5(0.1x^{2}-4.2x+9.5)^{0.5}$
$r^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=1.5(0.1x^{2}-4.2x+9.5)^{0.5}\cdot(0.2x-4.2)$
$r^{\prime}(x)=1.5(0.2x-4.2)(0.1x^{2}-4.2x+9.5)^{0.5}$
