# How do you find the derivative of f(x) = sec(x^2 + 1)^2?

$f ' \left(x\right) = 4 x {\sec}^{2} \left({x}^{2} + 1\right) \tan \left({x}^{2} + 1\right)$
Let $\left({x}^{2} + 1\right) = u$
$\frac{d}{\mathrm{dx}} \left({\sec}^{2} \left(u\right)\right) = 2 {\sec}^{2} u \tan u \cdot \frac{d}{\mathrm{dx}} \left(u\right)$