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

1 Answer
Jul 31, 2014

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.