Answer
$\frac{{dy}}{{dx}} = {x^3}{e^x} + 3{x^2}{e^x}$
Work Step by Step
$$\eqalign{
& y = {x^3}{e^x} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^3}{e^x}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {fg} \right)' = fg' + gf' \cr
& {\text{Let }}f = {x^3}{\text{ and }}g = {e^x} \cr
& \frac{{dy}}{{dx}} = {x^3}\left( {{e^x}} \right)' + {e^x}\left( {{x^3}} \right)' \cr
& {\text{Computing derivatives}} \cr
& \frac{{dy}}{{dx}} = {x^3}\left( {{e^x}} \right) + {e^x}\left( {3{x^2}} \right) \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dx}} = {x^3}{e^x} + 3{x^2}{e^x} \cr} $$
