1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial derivative

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Well, I'm not sure about this one. Its actually a differentiation problem, it asks me to determine if the function is differentiable at the given point.

    [tex]\sqrt[ ]{|xy|}[/tex] at [tex]P(0,0)[/tex]
    I think its not, but I must demonstrate, off course.

    So I try to solve the partial derivatives:

    [tex]f_x=\begin{Bmatrix} \displaystyle\frac{|y|}{2\sqrt[ ]{|xy|}} & \mbox{ if }& x>0\\\displaystyle\frac{-|y|}{2\sqrt[ ]{|xy|}} & \mbox{if}& x<0\end{matrix}[/tex]
    And here is the deal. Is this right? if it is, its easy to see that the partial derivative is not continuous at (0,0), its actually not defined at that point. So its not differentiable.

    Bye there, and thanks.
  2. jcsd
  3. Sep 21, 2010 #2


    Staff: Mentor

    This is similar to f(x) = sqrt(|x|), which is defined and continuous for all real x. The graph has a cusp at (0, 0), so the derivative is not defined there.
  4. Sep 21, 2010 #3
    Thanks Mark. So what I did is ok, right?
  5. Sep 21, 2010 #4


    Staff: Mentor

    I think you have a mistake because of the absolute value.
    For example, if
    [tex]f(x) = \sqrt{|x|} = |x|^{1/2}[/tex]

    [tex]f'(x) = (1/2) |x|^{-1/2} \cdot d/dx \left(|x|\right)[/tex]

    In the above, d/dx|x| = |x|/x.

    This derivative works out to sqrt(|x|)/(2x).

    In your partial, it looks like you're missing the step where you take the partial with respect to x of |xy|.
  6. Sep 21, 2010 #5
    So this:
    [tex]f'(x) = (1/2) |x|^{-1/2} \cdot d/dx \left(|x|\right)[/tex]

    Its equal to this:

    [tex]f'(x)=\begin{Bmatrix} \displaystyle\frac{1}{2\sqrt[ ]{|x|}} & \mbox{ if }& x>0\\\displaystyle\frac{-1}{2\sqrt[ ]{|x|}} & \mbox{if}& x<0\end{matrix}[/tex]

    Isn't it?
  7. Sep 21, 2010 #6


    Staff: Mentor

  8. Sep 21, 2010 #7
    So, if we take [tex]\sqrt[ ]{|xy|}[/tex]

    [tex]f_x=\begin{Bmatrix} \displaystyle\frac{|y|}{2\sqrt[ ]{|xy|}} & \mbox{ if }& x>0\\\displaystyle\frac{-|y|}{2\sqrt[ ]{|xy|}} & \mbox{if}& x<0\end{matrix}[/tex]
    Cause in the numerator I've got the derivative of |x| multiplied by |y|, which is a constant.

    I see the mistake now. Im considering just when x>0 and x<0, and not xy>0 and xy<0.
    Last edited: Sep 21, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Partial derivative
  1. Partial derivative (Replies: 1)

  2. Partial derivative (Replies: 2)

  3. Partial derivatives (Replies: 1)

  4. Partial derivative (Replies: 21)

  5. The partial derivative (Replies: 2)