## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = - 4x{e^{ - 2x}}$$
\eqalign{ & y = \left( {1 + 2x} \right){e^{ - 2x}} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {1 + 2x} \right){e^{ - 2x}}} \right] \cr & {\text{use the product rule }} \cr & \frac{{dy}}{{dx}} = \left( {1 + 2x} \right)\frac{d}{{dx}}\left[ {{e^{ - 2x}}} \right] + {e^{ - 2x}}\frac{d}{{dx}}\left[ {\left( {1 + 2x} \right)} \right] \cr & {\text{use the formula }}\frac{d}{{dx}}{e^u} = {e^u}\frac{{du}}{{dx}}{\text{ for }}\frac{d}{{dx}}\left[ {{e^{ - 2x}}} \right] \cr & \frac{{dy}}{{dx}} = \left( {1 + 2x} \right)\left( {{e^{ - 2x}}} \right)\frac{d}{{dx}}\left[ { - 2x} \right] + {e^{ - 2x}}\frac{d}{{dx}}\left[ {\left( {1 + 2x} \right)} \right] \cr & {\text{solve the derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = \left( {1 + 2x} \right)\left( {{e^{ - 2x}}} \right)\left( { - 2} \right) + {e^{ - 2x}}\left( 2 \right) \cr & \frac{{dy}}{{dx}} = - 2{e^{ - 2x}}\left( {1 + 2x} \right) + 2{e^{ - 2x}} \cr & {\text{factor}} \cr & \frac{{dy}}{{dx}} = 2{e^{ - 2x}}\left[ { - \left( {1 + 2x} \right) + 1} \right] \cr & \frac{{dy}}{{dx}} = 2{e^{ - 2x}}\left( { - 2x} \right) \cr & \frac{{dy}}{{dx}} = - 4x{e^{ - 2x}} \cr}