Answer
$$\frac{{dy}}{{dx}} = \sec 4x\left( {8x\tan 4x + 2} \right)$$
Work Step by Step
$$\eqalign{
& y = 2x\sec 4x \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {2x\sec 4x} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dx}} = 2x{D_x}\left( {\sec 4x} \right) + \sec 4x{D_x}\left( {2x} \right) \cr
& {\text{use the chain rule for }}{D_x}\left( {\sec u} \right) = \sec u\tan u \cdot {D_x}\left( u \right) \cr
& \frac{{dy}}{{dx}} = 2x\left( {\sec 4x} \right)\left( {\tan 4x} \right){D_x}\left( {4x} \right) + \sec 4x{D_x}\left( {2x} \right) \cr
& {\text{solve the derivatives}} \cr
& \frac{{dy}}{{dx}} = 2x\left( {\sec 4x} \right)\left( {\tan 4x} \right)\left( 4 \right) + \sec 4x\left( 2 \right) \cr
& {\text{multiply}} \cr
& \frac{{dy}}{{dx}} = 8x\sec 4x\tan 4x + 2\sec 4x \cr
& {\text{factor out sec}}4x \cr
& \frac{{dy}}{{dx}} = \sec 4x\left( {8x\tan 4x + 2} \right) \cr} $$
