Answer
$r^{\prime}(x)=0.5(0.1+4.2x^{-2})(0.1x-4.2x^{-1})^{-0.5}$
Work Step by Step
Let $u(x)=x^{0.5},\qquad v(x)=0.1x-4.2x^{-1}$
$\displaystyle \frac{du}{dx}=0.5x^{-0.5}, \quad$
$\displaystyle \frac{dv}{dx}=0.1(1)-4.2(-x^{-2})=0.1+4.2x^{-2}$
Then,$\quad y=r(x)=u(v(x))$ and $r^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dv}=0.5v^{-0.5}=0.5(0.1x-4.2x^{-1})^{-0.5}$
$r^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}$
$=0.5(0.1x-4.2x^{-1})^{-0.5}\cdot(0.1+4.2x^{-2})$
$r^{\prime}(x)=0.5(0.1+4.2x^{-2})(0.1x-4.2x^{-1})^{-0.5}$
