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

  1. Sep 29, 2013 #1
    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).

    Thanks in advance for your help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 29, 2013 #2

    jbunniii

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    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|##.
     
  4. Sep 29, 2013 #3
    So why would abs(y)/y be considered incorrect for the derivative of abs(y) if you say they're equal?
     
  5. Sep 29, 2013 #4

    jbunniii

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    It's not incorrect. It is a perfectly valid alternative expression.

    [edit] 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.
     
  6. Sep 29, 2013 #5
    Great. Now I just need to get that into the head of my Calculus teacher. Thanks!
     
  7. Sep 29, 2013 #6

    jbunniii

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    I edited my previous post with a clarification. I'm not sure about the expression (abs(x))/(x*x') you proposed above.
     
  8. Sep 29, 2013 #7
    Yeah sorry. I meant the derivative of x being x'(x/abs(x)) also being equal to x'(abs(x)/x)
     
  9. Sep 29, 2013 #8
    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.
     
  10. Sep 29, 2013 #9

    jbunniii

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    Another thing to notice which might make it clearer is the following:
    $$\frac{y}{|y|} = \begin{cases}
    1 & \text{ if }y > 0\\
    -1 & \text{ if }y < 0\\
    \text{undefined} & \text{ if }y = 0
    \end{cases}$$
    and ##|y|/y## is exactly the same.
     
  11. Sep 29, 2013 #10
    Yeah, I tried explaining that to my calc teacher and he just said that it was incorrect because they were not the same. At the time I did not realize I could prove the abs(x)/x to be equal to x/abs(x) so I just let it go.
     
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