Answer
$$\frac{{dy}}{{dx}} = 30{\sec ^2}x{\left( {1 + 2\tan x} \right)^{14}}$$
Work Step by Step
$$\eqalign{
& y = {\left( {1 + 2\tan x} \right)^{15}} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {1 + 2\tan x} \right)}^{15}}} \right] \cr
& {\text{Use the chain rule}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = 15{\left( {1 + 2\tan x} \right)^{15 - 1}}\frac{d}{{dx}}\left[ {1 + 2\tan x} \right] \cr
& \frac{{dy}}{{dx}} = 15{\left( {1 + 2\tan x} \right)^{14}}\left( {0 + 2{{\sec }^2}x} \right) \cr
& \frac{{dy}}{{dx}} = 30{\sec ^2}x{\left( {1 + 2\tan x} \right)^{14}} \cr} $$
