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