# What is the derivative of tan(x^2)?

Oct 3, 2016

$y = \tan \left({x}^{2}\right) = \tan \left(u\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{du}} = {\sec}^{2} \left(u\right) = {\sec}^{2} \left({x}^{2}\right)$

$u = {x}^{2} , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = 2 x$

Use the chain rule...

$\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = 2 x \cdot {\sec}^{2} \left({x}^{2}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$

Remember that:

If $y = \tan x$, $\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(x\right)$

It's a general rule.