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Sqrt(|xy|) derivatives at (0,0)

  1. Apr 23, 2012 #1
    [tex] f(x,y)=\sqrt{|xy|} [/tex]
    Do the partial derivatives of f exist at x=0, y=0?
     
  2. jcsd
  3. Apr 23, 2012 #2


    Let's see: [itex]\displaystyle{f_x(0,0):=\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x}=\lim_{x\to 0}\frac{0}{x}}=0[/itex] , and of course the same is true for [itex]f_y(0,0)[/itex]

    DonAntonio
     
  4. Apr 23, 2012 #3
    Mathematica could not do it.

    And if I write the function as ((xy)^2)^(1/4) then the partial derivative is
    [tex]\frac{x y^2}{2 \left(x^2 y^2\right)^{3/4}}[/tex]
    which is undefined at (0,0). It means it is not differentiable at 0,0 correct?
     
  5. Apr 23, 2012 #4

    micromass

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    Mathematica is not always correct.

    Not necessarily. It just means that the usual way of computing the derivative fails at 0.
     
  6. Apr 23, 2012 #5


    What's not clear in what I wrote to you? I think it is pretty straightforward, though I can be wrong, of course. Check this.

    DonAntonio
     
  7. Apr 24, 2012 #6
    that's essentially the point i'm trying to get at. If the "usual way" of computing derivatives fails at a point, doesn't that mean that the function is not differentiable at that point?

    Even though partials exist at (0,0) as DonAntonio showed, the function is still not differentiable. So I'm trying to find out to what extent this is true:

    "If the "usual way" of taking partials does not yield the correct result at a point, the function is not differentiable at that point".
     
  8. Apr 24, 2012 #7


    There is only one way to calculate the (partial or not) derivatives wrt the definition. Now, if these derivatives exist

    AND they're continuous at some point then the function is differentiable at that point. This is precisely what does

    NOT happen in this case as the partial derivatives aren't cont. at (0,0), and to call something "the usual way" invites confusion and

    problems when dealing with mathematics, as what is usual for you might not be so for me and the other way around.

    DonAntonio
     
  9. Apr 24, 2012 #8
    Just to clarify for the OP: This does not mean that if the partial derivatives are not continuous at a point, then the function is not differentiable there, which is the converse statement. There are many functions whose derivatives exist everywhere, but the derivative function is not continuous. Likewise, there are nondifferentiable functions whose partial derivatives exist everywhere. They are rarely more than pathological curiosities, however, as the derivative in these cases fail to meet the ideal of being a linear approximation to the function.
    In summary, the existence or non-existence of the partial derivatives at a point will not tell you whether the function is differentiable there. If they exist and are continuous, however, then you have a derivative.
     
  10. Apr 24, 2012 #9
    The simple question I was asking is this: if the partial derivatives at a point DISAGREES with what is obtained by using the partial derivative formulas (the way we learned in highschool with all the differentiation rules etc etc) then does that mean that the function is not differentiable at that point?

    In this example the partial derivatives are both 0 at (0,0), but using the usual "formula" I get
    [tex]\frac{x y^2}{2 \left(x^2 y^2\right)^{3/4}}[/tex]
    which is undefined at (0,0). So "undefined" and "0" are not the same thing, so there is disagreement, and "coincidentally" the function happens to be non-differentiable at (0,0). Is this a general result? Will the partial derivative "formulas" always fail at non-differentiable points?

    Its really a simple question but I dont know if I'm explaining it very well.
     
  11. Apr 25, 2012 #10
    Your expression disagrees with the values of the partial derivatives at the point because the operation of the chain rule is not valid there.
    When you take the partial derivative of f(x, y) = ((xy)^2)^(1/4) at the point x = 0, y = 0, with respect to x, you applied the chain rule to the single-variable function f(x) = h(g(x)) where h(g) = g^(1/4) and g(x) = (xy)^2, treating y as a constant. The chain rule states that if f(x) = h(g(x)) and h is differentiable at the point g(0) and g is differentiable at the point x = 0, then f'(0) = h'(g(0))*g'(0). However, h(g) is not differentiable at g(0), so the chain rule is not applicable.
    You have to use another method to find the partial derivative of f there, such as the definition of the partial derivative, as applied by DonAntonio.
     
    Last edited: Apr 25, 2012
  12. Apr 25, 2012 #11

    HallsofIvy

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    That is the "formula" for x and y not equal to 0. If you are asking if the limit, as (x, y) goes to 0 for derivative must give the derivative at (0, 0), the answer is no- a derivative of a function is not necessarily continuous. However, even so a derivative must satisfy the "intermediate value property". If the limit exists, it must be equal to the derivative at that point.

     
  13. Apr 25, 2012 #12

    micromass

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