Posted: a few days before June 8, 2004
In mathematics, an absolute value (always plus) is denoted by a quantity like x or f(x) flanked by two vertical lines: |x|. This function can be dealt with in any equation except that I have not yet seen its derivative used as a tool in the calculus. I propose to correct this lack.
I start by giving my (current) derivation of this derivative.
To the left, I state a conventional definition of the derivative of the function in question. When x is negative as it is on the second line of the derivation, the numerator (|x + dx| - |x|) is -dx because dx is always a positive number. Thus the derivative is -1. On the third line when x is positive, this simplifies to +1. I would like to end with a formula for the derivative that does not depend on the sign of x. Thus, I choose to multiply by x/x in each case since this is, in effect, multiplying by 1.
When x is negative, minus a negative number is a positive numerator that can be represented here by |x|. When x is positive, likewise you can represent it as |x|. Now we have one expression that represents the derivative of the absolute value function over the whole number range. The reader is left to show that |x|/x = x/|x| for him- or herself.
Notice that the derivative is not defined when x = 0 as is required.
In the references to this derivative I have seen, the formula has been obtained generally by differentiating y = (x^2)^1/2 (the square root of x squared). I am told now by competent authority (Ken Watson, Prof of Physics, emeritus) that this latter derivation is common knowledge to mathematicians today.
This derivative of the absolute value function can be used in any differentiation where an absolute value appears, using the chain rule as needed. Two examples follow:
y = |x|^3 (y = abs value of x quantity cubed)
y' = (3 |x|^2) |x|/x = (3 |x|^2) x/|x| = 3 x |x|
We will differentiate the function graphed below:
Applying the differentiation formula for the absolute-value function
using the chain rule, we get the result shown in the following graph:
It is clear that, although the first graph is continuous, it is not
differentiable at every point. Whenever x is evenly divisible by π,
the derivative doesn't exist because sin x is 0 at these values.
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especially corrections to the WEBMASTER
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