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Help with partial derivative.

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data

    The question asks:

    f(x,y) = 3xy+5y^3/[x^2+y^2] when (x,y) =! (0,0)
    f(x,y) = 0 when (x,y) = (0,0)

    what is df/dy at (0,0)?

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure what the answer is. At 0,0 f(x,y) is 0, so it's simply a point and the function is not continuous at the point, therefore df/dy doesn't exist?
  2. jcsd
  3. Dec 19, 2011 #2

    Ray Vickson

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    There are functions for which all partial derivatives exist at a point of discontinuity, so lack of continuity cannot be relied upon in this question.

  4. Dec 19, 2011 #3
    So then it must be 0? It's the partial derivative of a point.
  5. Dec 19, 2011 #4


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    Can you show that f is not continuous at (0,0) ?
  6. Dec 19, 2011 #5


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    Don't get confused by how the function is defined. It doesn't make sense to talk about the "partial derivative of a point." You find the partial derivative of a function at a point.

    Go back to the basic definition. You want to find
    [tex]\left.\frac{\partial f}{\partial y}\right|_{(x_0,y_0)} = \lim_{h \to 0} \frac{f(x_0,y_0+h)-f(x_0,y_0)}{h}[/tex]when (x0,y0)=0. You need to find the limit or show it doesn't exist.
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