Answer
$$\frac{{dy}}{{dx}} = {e^{\left( {4\sqrt x + {x^2}} \right)}}\left( {\frac{2}{{\sqrt x }} + 2x} \right)$$
Work Step by Step
$$\eqalign{
& y = {e^{\left( {4\sqrt x + {x^2}} \right)}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{e^{\left( {4\sqrt x + {x^2}} \right)}}} \right] \cr
& {\text{we can use the formula }}\frac{d}{{dx}}{e^u} = {e^u}\frac{{du}}{{dx}};{\text{ where }}u{\text{ is any differentiable function of }}x. \cr
& {\text{For this exercise you can note that }}u = 4\sqrt x + {x^2},{\text{ then}} \cr
& \frac{{dy}}{{dx}} = {e^{\left( {4\sqrt x + {x^2}} \right)}}\frac{d}{{dx}}\left[ {4\sqrt x + {x^2}} \right] \cr
& \cr
& {\text{Solving }}\frac{d}{{dx}}\left[ {4\sqrt x + {x^2}} \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\left( {\frac{1}{{2\sqrt x }}} \right) + 2x \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{\sqrt x }} + 2x \cr
& \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = {e^{\left( {4\sqrt x + {x^2}} \right)}}\frac{d}{{dx}}\left[ {4\sqrt x + {x^2}} \right] = {e^{\left( {4\sqrt x + {x^2}} \right)}}\left( {\frac{2}{{\sqrt x }} + 2x} \right) \cr
& \frac{{dy}}{{dx}} = {e^{\left( {4\sqrt x + {x^2}} \right)}}\left( {\frac{2}{{\sqrt x }} + 2x} \right) \cr} $$
