How do we differentiate ? I know that the derivative of itself is , and the derivative with respect to x and y are respectively and . But I don't understand how to find the derivatives with the absolute value signs. Can anyone show me?

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- April 16th 2010, 01:43 PMdemodeDifferentiating Absolute Values
How do we differentiate ? I know that the derivative of itself is , and the derivative with respect to x and y are respectively and . But I don't understand how to find the derivatives with the absolute value signs. Can anyone show me?

- April 16th 2010, 01:54 PMDiegoConsider a separate function
To make myself clearer I am going to consider the function . This function can be rewritten as:

So the derivative is:

It can be shown that in x = 0 the derivative does not exist. Apply this idea to your function. - April 17th 2010, 12:32 AMdemode
- April 17th 2010, 07:01 AMDiegoApply the definition of derivative to different parts of the function
Suppose x > 0 then just apply the definition:

Suppose x < 0 then:

Now if x = 0

and:

thus the derivative on zero does not exist.

, provided x is not zero, is just a compact way of writing the derivative, but is still the same function as:

Maybe what you wanted to know is if you can obtain the derivative in the form x/|x| by consdering any x, I do not know if this is possible, but in any case I don't think it is that important, because you still obtain the same result. I hope this was the answer you were looking for, if not do tell me. - April 17th 2010, 07:37 AMdrumist
Well, you could apply the chain rule.

Let and , so

- April 17th 2010, 07:45 AMHallsofIvy
First let me point out that the statement of the "one thing you know" was wrong. The derivative of xy is NOT xy. If you are differentiating with respect to x and y is a function of x, it is x+ xy'. If you are differentiating with respect to y and x is a function of y, it is y+ yx'. If x and y are both functions of some parameter and you are differentiating with respect to that parameter it is x'y+ xy'. If x and y are independent variable, the partial derivatives are and .

For |xy|, assuming x and y are independent variables,

In the first quadrant, x> 0, y> 0 so |xy|= xy and the partial derivatives are and

In the second quadrant, x< 0 and y> 0 so |xy|= -xy and the partial derivatives are and .

In the third quadrant, x< 0 and y< 0 so xy> 0, |xy|= xy and the partial derivatives are and .

In the fourth quadrant, x> 0 and y< 0 so |xy|= -xy and the partial derivatives are and .

Of course, |xy| is not differentiable if either x or y is 0- that is, on the axes.