Answer
$$\frac{{dy}}{{dx}} = 8x{\sec ^2}\left( {4{x^2} + 3} \right)$$
Work Step by Step
$$\eqalign{
& y = \tan \left( {4{x^2} + 3} \right) \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\tan \left( {4{x^2} + 3} \right)} \right] \cr
& {\text{using the chain rule for }}{D_x}\left( {\tan u} \right) = {\sec ^2}u \cdot {D_x}\left( u \right).{\text{ for this exercise }}u = 4{x^2} + 3 \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = {\sec ^2}\left( {4{x^2} + 3} \right)\frac{d}{{dx}}\left[ {4{x^2} + 3} \right] \cr
& \frac{{dy}}{{dx}} = {\sec ^2}\left( {4{x^2} + 3} \right)\left( {8x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 8x{\sec ^2}\left( {4{x^2} + 3} \right) \cr} $$
