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