Answer
$f'\left( x \right) = - \frac{3}{{2{x^{5/2}}}} - \frac{1}{{{x^2}}}$
Work Step by Step
$$\eqalign{
& y = \frac{{\sqrt x + x}}{{{x^2}}} \cr
& {\text{Rewrite the function}} \cr
& y = \frac{{\sqrt x }}{{{x^2}}} + \frac{x}{{{x^2}}} \cr
& y = \frac{1}{{{x^{3/2}}}} + \frac{1}{x} \cr
& y = {x^{ - 3/2}} + {x^{ - 1}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{ - 3/2}} + {x^{ - 1}}} \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{ - 3/2}} + {x^{ - 1}}} \right] \cr
& {\text{Use the constant multiple rule}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{ - 3/2}}} \right] + \frac{d}{{dx}}\left[ {{x^{ - 1}}} \right] \cr
& {\text{Apply the power rule: }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}{\text{ }} \cr
& f'\left( x \right) = - \frac{3}{2}{x^{ - 3/2 - 1}} - {x^{ - 1 - 1}} \cr
& f'\left( x \right) = - \frac{3}{2}{x^{ - 5/2}} - {x^{ - 2}} \cr
& {\text{Rewrite}} \cr
& f'\left( x \right) = - \frac{3}{{2{x^{5/2}}}} - \frac{1}{{{x^2}}} \cr} $$
