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Concerning partial derivatives

  1. Feb 2, 2008 #1
    f(x,y) = (2xy)/(x2 + y2), 0 if (x,y) = (0,0)

    Now I'm supposed to evaluate this at (0,0). I take the first partial derivative and I get 0/0 but when I use the definition of derivatives I get a whole number. Why the hell is this?
  2. jcsd
  3. Feb 2, 2008 #2


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    What "definition" of the derivatives?

    That function cannot be given a continuous extension at the origion.
  4. Feb 2, 2008 #3
    I'm sorry. I'm talking about the definition of derivative, the newton quotient. Letting the limit go to zero.

    lim (f(x+h,y) - f(x,y))/h
    (x,y) -> (0,0)
  5. Feb 2, 2008 #4


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    Surely you do not mean "I'm supposed to evaluate this at (0,0)". That's easy: it's 0!

    As for the derivatives, I'm afraid I get exactly the opposite result to you.
    The derivative with respect to x is taken for y fixed so it is the limit of [f(0+h,0)- f(0,0)]/h= 0. The derivative with respect to y is the same. (If you just "plug in" h=0 you get 0/0 but you know from Calculus I that that happens for any derivative.) The derivative is 0 because f(h,0)= 0 for any h. 0/h= 0 which has limit 0 as h goes to 0. Of course, the same is true the partial derivative with respect to y.

    In order that the "derivative" (really the "gradient"), as opposed to the partial derivatives, the limit of [itex][f(x+h_1,y+h_2)- f(x,y)]/\sqrt{h_1^2+ h_2^2}[/itex] must exist as [itex]h_1[/itex] and [itex]h_2[/itex] go to 0 independently. Since we are talking about (x,y)= (0,0) this is just the same as taking the limit of
    [tex]\frac{2xy}{(x^2+ y^2)^{3/2}}[/tex]
    as (x, y) goes to (0, 0). (I've replaced [itex]h_1[/itex] and [itex]h_2[/itex] by x and y for convenience.)
    Best way to find a limit like that is to convert to polar coordinates:
    [tex]\frac{2(r cos(\theta))(r sin(\theta))}{r^3}= \frac{2 sin(\theta)cos(\theta)}{r}[/tex]
    which clearly does not exist as r goes to 0.

    I suspect that your text book is trying to convince you that "having partial derivatives" and "being differentiable" are not the same for functions of more than one variable. A function of several variables is "differentiable" at a point if and only if its partial derivatives are continuous at that point.
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