Answer
$$\frac{{dy}}{{dx}} = 108{\sec ^2}\left( {9x + 1} \right)$$
Work Step by Step
$$\eqalign{
& y = 12\tan \left( {9x + 1} \right) \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {12\tan \left( {9x + 1} \right)} \right] \cr
& {\text{use multiple constant rule}} \cr
& \frac{{dy}}{{dx}} = 12\frac{d}{{dx}}\left[ {\tan \left( {9x + 1} \right)} \right] \cr
& {\text{using the chain rule for }}{D_x}\left( {\tan u} \right) = {\sec ^2}u \cdot {D_x}\left( u \right).{\text{ consider }}u = 9x + 1 \cr
& \frac{{dy}}{{dx}} = 12{\sec ^2}\left( {9x + 1} \right)\frac{d}{{dx}}\left[ {9x + 1} \right] \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = 12{\sec ^2}\left( {9x + 1} \right)\left( 9 \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = 108{\sec ^2}\left( {9x + 1} \right) \cr} $$
