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

##### 1 Answer

**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

#1/cos y = x#

And solve for

#1 = xcosy#

#1/x = cosy#

#y = arccos(1/x)#

Now this looks much easier to differentiate. We know that

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

#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.