# Derivative Rule Issue

1. Jan 6, 2015

### Euler2718

1. The problem statement, all variables and given/known data

Finding the derivative of an inverse trigonometric function

2. Relevant equations

*This is the problem*

3. The attempt at a solution

In my text book, Single Variable Essential Calculus, Second Edition, by James Stewart, the derivative rules for the inverse trigonometric functions are causing me great pain, as it seems there are different variations depending on where you look. For instances, take the derivative rule for arc-secant...

$$\frac{d}{dx} [arcsec(x)] = \frac{1}{x\sqrt{x^{2}-1}}$$

This differs from a hand out that I obtained that claims the rule is...

$$\frac{d}{dx} [arcsec(u)] = \frac{1}{|u|\sqrt{u^{2}-1}}\frac{du}{dx} , |u|>1$$

My question is which one should I be using? Does the absolute sign make a difference? I was working on finding the tangent to

$$y=arcsec(4x), x=\frac{\sqrt{2}}{4}$$

and when I got the derivative using the hand out rule...

$$\frac{dy}{dx} = \frac{1}{|x|\sqrt{16x^{2}-1}}$$

The book yields the exact same thing, but in less steps, as you don't have to take ' du/dx '

So, is it more appropriate to write it in terms of a kind of u-substitution with the absolutes, or just in terms of 'x' with no absolutes?

2. Jan 6, 2015

### Staff: Mentor

The rule from the handout, I believe, is more general, in that it handles angles in the second quadrant (i.e., $\pi/2 < x < \pi$). The principal domain for the arcsec function is $[0, \pi/2) \cup (\pi/2, \pi]$.

3. Jan 6, 2015

### Euler2718

So it would be safer to stick with the general case I suppose. Thanks for your help.

4. Jan 7, 2015

### vela

Staff Emeritus
The handout's formula include an application of the chain rule. If you use the book's formula, you have to recognize the need to apply the chain rule. Either way, you're essentially doing the same thing. Personally, I prefer the book's approach since you need to know when to apply the chain rule in general anyway, and including it in the formula just clutters things up.

That said, the book's formula isn't really correct since it doesn't work for negative values of $x$. If you look at a plot of arcsec x, you'll see that the derivative is positive for every point in its domain. The book's formula, however, will give you a negative answer for x<-1.