Answer
$$\frac{{dy}}{{dx}} = 2{e^{2x}}{\sec ^2}2x + 2{e^{2x}}\tan 2x$$
Work Step by Step
$$\eqalign{
& y = {e^{2x}}\tan 2x \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{e^{2x}}\tan 2x} \right] \cr
& {\text{Use product rule}} \cr
& \frac{{dy}}{{dx}} = {e^{2x}}\frac{d}{{dx}}\left[ {\tan 2x} \right] + \tan 2x\frac{d}{{dx}}\left[ {{e^{2x}}} \right] \cr
& \frac{{dy}}{{dx}} = {e^{2x}}\left( {{{\sec }^2}2x} \right)\left( 2 \right) + \tan 2x\left( {2{e^{2x}}} \right) \cr
& \frac{{dy}}{{dx}} = 2{e^{2x}}{\sec ^2}2x + 2{e^{2x}}\tan 2x \cr} $$
