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