Answer
$y'=-\dfrac{1}{s^2}+\dfrac{3}{2\sqrt{x^5}}$
Work Step by Step
Reduce the expression
$y=\dfrac{s-\sqrt{s}}{s^2}=\dfrac{s}{s^2}+\dfrac{\sqrt{s}}{s^2}$
$y=s^{-1}-s^{-3/2}$
Apply power rule
$y'=(-1)s^{-1-1}-(-\dfrac{3}{2})s^{-3/2-1}$
$y'=-s^{-2}+(\dfrac{3}{2})s^{-5/2}$
$y'=-\dfrac{1}{s^2}+\dfrac{3}{2\sqrt{x^5}}$