Question

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

Answer 1

We derive the derivative of the secant function as follows.

`y = 1/cosx`

By the quotient rule:

`y = (0 xx cosx - 1 xx -sinx)/(cosx)^2`

`y = sinx/cos^2x`

`y = tanxsecx`

We now apply this to our problem.

Let `y = secu` and `u = secx`.

By the chain rule:

`dy/dx = dy/(du) xx (du)/dx`

`dy/dx = tanusecu xx tanxsecx`

`dy/dx = tan(secx)sec(secx)tanxsecx`

Hopefully this helps!