# How do you find the derivative of y=sec(sec x) using the chain rule?

Dec 17, 2016

We derive the derivative of the secant function as follows.

$y = \frac{1}{\cos} x$

By the quotient rule:

$y = \frac{0 \times \cos x - 1 \times - \sin x}{\cos x} ^ 2$

$y = \sin \frac{x}{\cos} ^ 2 x$

$y = \tan x \sec x$

We now apply this to our problem.

Let $y = \sec u$ and $u = \sec x$.

By the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \tan u \sec u \times \tan x \sec x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \tan \left(\sec x\right) \sec \left(\sec x\right) \tan x \sec x$

Hopefully this helps!