## Calculus (3rd Edition)

$$y'=3\tan^2x \sec^2 x+3x^2\sec^2 x^3.$$
Recall that $(\tan x)'=\sec^2 x$. Recall that $(x^n)'=nx^{n-1}$ Since $y=\tan^3 x+\tan x^3$, by using the chain rule, the derivative $y'$ is given by $$y'=3\tan^2x \sec^2 x+3x^2\sec^2 x^3.$$