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

Jul 31, 2014

$\frac{d}{\mathrm{dx}} \left[{\sec}^{-} 1 x\right] = \frac{1}{\sqrt{{x}^{4} - {x}^{2}}}$

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 $\cos$:

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

And solve for $y$:

$1 = x \cos y$

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

$y = \arccos \left(\frac{1}{x}\right)$

Now this looks much easier to differentiate. We know that
$\frac{d}{\mathrm{dx}} \left[\arccos \left(\alpha\right)\right] = - \frac{1}{\sqrt{1 - {\alpha}^{2}}}$
so we can use this identity as well as the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {\left(\frac{1}{x}\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left[\frac{1}{x}\right]$

A bit of simplification:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - \frac{1}{x} ^ 2}} \cdot \left(- \frac{1}{x} ^ 2\right)$

A little more simplification:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} \sqrt{1 - \frac{1}{x} ^ 2}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{{x}^{4} \left(1 - \frac{1}{x} ^ 2\right)}}$

Some final reduction:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{{x}^{4} - {x}^{2}}}$

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.