Question

What is the derivative of `f(x)=sec^-1(x)` ?

Answer 1

`d/dx[sec^-1x] = 1/(sqrt(x^4 - x^2))`

Process:

First, we will make the equation a little easier to deal with. Take the secant of both sides:

`y = sec^-1 x`

`sec y = x`

Next, rewrite in terms of `cos`:

`1/cos y = x`

And solve for `y`:

`1 = xcosy`

`1/x = cosy`

`y = arccos(1/x)`

Now this looks much easier to differentiate. We know that
`d/dx[arccos(alpha)] = -1/(sqrt(1-alpha^2))`
so we can use this identity as well as the chain rule:

`dy/dx = -1/sqrt(1 - (1/x)^2) * d/dx[1/x]`

A bit of simplification:

`dy/dx = -1/sqrt(1 - 1/x^2) * (-1/x^2)`

A little more simplification:

`dy/dx = 1/(x^2sqrt(1 - 1/x^2))`

To make the equation a little prettier I will move the `x^2` inside the radical:

`dy/dx = 1/(sqrt(x^4(1 - 1/x^2)))`

Some final reduction:

`dy/dx = 1/(sqrt(x^4 - x^2))`

And there's our derivative.

When differentiating inverse trig functions, the key is getting them in a form that's easy to deal with. More than anything, they're an exercise in your knowledge of trig identities and algebraic manipulation.