## Calculus (3rd Edition)

$$y''= 2\sec x^2+4x^2 \sec x^2 \tan x^2.$$
Recall that $(\tan x)'=\sec^2 x$. Recall that $(x^n)'=nx^{n-1}$ Given $y=\tan x^2$, then, by using the chain rule, we have $$y'=(2x)\sec x^2=2x\sec x^2,$$ and, by using the product rule, the second derivative $y''$ is given by $$y''=2\sec x^2+2x (2x) \sec x^2 \tan x^2 \\=2\sec x^2+4x^2 \sec x^2 \tan x^2.$$