Answer
$4(s^{2}-s^{0.5})^{3}\cdot(2s-0.5s^{-0.5})$
Work Step by Step
$r(s)$ is a composite function.
Let $u(x)=x^{4},\qquad v(s)=s^{2}-s^{0.5}$
$\displaystyle \frac{du}{dx}=4x^{3}, $
Then,$\quad r(s)=u(v(s))$ and $\displaystyle \frac{dr}{ds}=\frac{du}{dv}\frac{dv}{ds}$
$\displaystyle \frac{du}{dv}=4v^{3}=4(s^{2}-s^{0.5})^{3}$
$\displaystyle \frac{dv}{ds}=2s-0.5s^{-0.5}$
$\displaystyle \frac{dr}{ds}=\frac{du}{dv}\frac{dv}{ds}=4(s^{2}-s^{0.5})^{3}\cdot(2s-0.5s^{-0.5})$
