# Derivative Rule Issue

## Homework Statement

Finding the derivative of an inverse trigonometric function

## Homework Equations

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*This is the problem*

## The Attempt at a Solution

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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?

Mark44
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]##.

Euler2718
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]##.

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

vela
Staff Emeritus