Answer
$\frac{{dy}}{{dx}} = \left( {3{x^2} + x - 5} \right){e^x}$
Work Step by Step
$$\eqalign{
& y = \left( {3{x^2} - 5x} \right){e^x} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {3{x^2} - 5x} \right){e^x}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {fg} \right)' = fg' + gf' \cr
& {\text{Let }}f = 3{x^2} - 5x{\text{ and }}g = {e^x} \cr
& \frac{{dy}}{{dx}} = \left( {3{x^2} - 5x} \right)\left( {{e^x}} \right)' + {e^x}\left( {3{x^2} - 5x} \right)' \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{dx}} = \left( {3{x^2} - 5x} \right){e^x} + {e^x}\left( {6x - 5} \right) \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dx}} = 3{x^2}{e^x} - 5x{e^x} + 6x{e^x} - 5{e^x} \cr
& \frac{{dy}}{{dx}} = 3{x^2}{e^x} + x{e^x} - 5{e^x} \cr
& or \cr
& \frac{{dy}}{{dx}} = \left( {3{x^2} + x - 5} \right){e^x} \cr} $$
