Answer
$$y' = \sqrt x {e^{\sqrt x }} + 2{e^{\sqrt x }}$$
Work Step by Step
$$\eqalign{
& y = 2x{e^{\sqrt x }} \cr
& {\text{find the derivative}} \cr
& y' = \left( {2x{e^{\sqrt x }}} \right)' \cr
& {\text{product rule}} \cr
& y' = 2x\left( {{e^{\sqrt x }}} \right)' + {e^{\sqrt x }}\left( {2x} \right)' \cr
& {\text{compute derivatives}} \cr
& y' = 2x\left( {{e^{\sqrt x }}} \right)\left( {\sqrt x } \right)' + {e^{\sqrt x }}\left( 2 \right) \cr
& y' = 2x\left( {{e^{\sqrt x }}} \right)\left( {\frac{1}{{2\sqrt x }}} \right) + {e^{\sqrt x }}\left( 2 \right) \cr
& {\text{simplifying}} \cr
& y' = \frac{{x{e^{\sqrt x }}}}{{\sqrt x }} + 2{e^{\sqrt x }} \cr
& y' = \sqrt x {e^{\sqrt x }} + 2{e^{\sqrt x }} \cr} $$
