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Homework Help: Partial Derivatives of Discontinuous Fcn?

  1. Sep 27, 2010 #1
    f(x,y)= xy2/(x2+y2) if (x,y)[tex]\neq[/tex](0,0)
    =0 if (x,y)=(0,0)
    Show that the partial derivatives of x and y exist at (0,0).

    This may be a really stupid question, but would the partial derivatives of x and y at (0,0) just be 0? I tried taking that partial derivatives of xy2/(x2+y2) and got:
    df/dx=[(x2+y2)(y2)-xy2(2x)]/(x2+y2)2
    and
    df/dy=[(x2+y2)(2xy)-xy2(2y)]/(x2+y2)2
    which i don't believe cancel out.
    Please help?
     
  2. jcsd
  3. Sep 28, 2010 #2

    HallsofIvy

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    Science Advisor

    You are asked for the partial derivatives at (0, 0) so use the definition directly:

    [tex]\displaytype\frac{\partial f}{\partial x}(0, 0)= \lim_{h\to 0}\frac{f(h, 0)- f(0, 0)}{h}[/tex]
    [tex]\displaytype = \lim_{h\to 0}\frac{\frac{h(0)}{h^2+ 0}}{h}= \lim_{h\to 0} 0= 0[/tex]

    Why do you say this function is discontinuous?
     
  4. Sep 28, 2010 #3
    Thanks so much!
    I realized after I posted the question that it's not discontinuous. Oops. >.<
     
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