Answer
$$\frac{{dy}}{{dx}} = \left( {{x^3} + {x^2} - 4x + 1} \right){e^x}$$
Work Step by Step
$$\eqalign{
& y = \left( {{x^3} - 2{x^2} + 1} \right){e^x} \cr
& {\text{Find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy}}{{dx}}\left[ {\left( {{x^3} - 2{x^2} + 1} \right){e^x}} \right] \cr
& {\text{By the product rule}} \cr
& \frac{{dy}}{{dx}} = \left( {{x^3} - 2{x^2} + 1} \right)\frac{d}{{dx}}\left[ {{e^x}} \right] + {e^x}\frac{d}{{dx}}\left[ {\left( {{x^3} - 2{x^2} + 1} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \left( {{x^3} - 2{x^2} + 1} \right){e^x} + {e^x}\left( {3{x^2} - 4x} \right) \cr
& {\text{Factor out }}{e^x} \cr
& \frac{{dy}}{{dx}} = \left( {{x^3} - 2{x^2} + 1 + 3{x^2} - 4x} \right){e^x} \cr
& \frac{{dy}}{{dx}} = \left( {{x^3} + {x^2} - 4x + 1} \right){e^x} \cr} $$
