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

Sep 6, 2014

You would use the [chain rule] for this
The derivative of a composite function F(x) is:
$F ' \left(x\right) = f ' \left(g \left(x\right)\right) \left(g ' \left(x\right)\right)$
(Where $f \left(u\right)$ is the outer function and $u = g \left(x\right)$ is the inner function)
Or, in words:
the derivative of the outer function (with the inside function left alone!) times the derivative of the inner function.

In this case $\tan \left(u\right)$ is the outer function and $u = {x}^{2}$ is the inner function. You don't have to worry about the "u" too much, it's just there to show you that the ${x}^{2}$ is "composed" in the $\tan$ and that, according to the chain rule, you should derive the outer function without doing anything to the inner function.

The derivative of the outer function (leaving the inner function alone) is:

$\frac{d}{\mathrm{dx}} \tan \left(u\right) = {\sec}^{2} \left(u\right)$

Then, the derivative of the inner function is:

$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$

Combining the two by multiplying them together, we get:

$\frac{d}{\mathrm{dx}} \tan \left({x}^{2}\right) = 2 x {\sec}^{2} \left({x}^{2}\right)$