# Why is the derivative of absolute value of x (x*x')/abs(x)

I have no problem that I am trying to solve but simply a question about the derivative of an absolute value equation. I know that the derivative of and absolute value function is (x*x')/(abs(x)) and I understand the process of reaching this equation through the process shown here.

http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/Calculus/DerivativesInvolvingAbsoluteValue.pdf [Broken]

I would like to know why one cannot give the answer of (abs(x))/(x*x') instead of the equation I mentioned earlier. Their graphs are exactly the same and do not have any differences I can find other than the equation.

For example the derivative of abs(x) should be x/abs(x) but the graph of abs(x)/x is defined for all the same values and also returns all the same values and the proper answer. Please help me understand why the latter equation is considered incorrect and not the derivative of the abs(x).

## The Attempt at a Solution

Last edited by a moderator:

jbunniii
Homework Helper
Gold Member
You are correct that in general, if ##y## is real-valued and nonzero,
$$\frac{y}{|y|} = \frac{|y|}{y}$$
To see why, start with ##|y|^2 = y^2##, and divide both sides by ##y|y|##.

So why would abs(y)/y be considered incorrect for the derivative of abs(y) if you say they're equal?

jbunniii
Homework Helper
Gold Member
It's not incorrect. It is a perfectly valid alternative expression.

 What I mean to say is that
$$\frac{x \cdot x'}{|x|} = \frac{|x| \cdot x'}{x}$$
However, I don't think either of these is equal to
$$\frac{|x|}{x \cdot x'}$$
which is the expression you gave above.

Great. Now I just need to get that into the head of my Calculus teacher. Thanks!

jbunniii
Homework Helper
Gold Member
I edited my previous post with a clarification. I'm not sure about the expression (abs(x))/(x*x') you proposed above.

Yeah sorry. I meant the derivative of x being x'(x/abs(x)) also being equal to x'(abs(x)/x)

Still have to prove that my answers right to my teacher though. Thank you again. I know it might seem stupid but this question had been bothering me for a while and finding out I'm right feels good.

jbunniii
$$\frac{y}{|y|} = \begin{cases} 1 & \text{ if }y > 0\\ -1 & \text{ if }y < 0\\ \text{undefined} & \text{ if }y = 0 \end{cases}$$