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Continuity of partial derivatives in a ball implies differentiability

  1. Dec 19, 2009 #1
    Hi all, I'm looking at the following problem:
    Suppose that f:[itex]\mathbb{R}^2\to\mathbb{R}[/itex] is such that [itex]\frac{\partial{f}}{\partial{x}}[/itex] is continuous in some open ball around (a,b) and [itex]\frac{\partial{f}}{\partial{y}}[/itex] exists at (a,b): show f is differentiable at (a,b).

    Now I know that if both partial derivatives are continuous in a ball around (a,b) it is differentiable, but I don't know how to deal with the case where we only have continuity of one partial derivative and existence of the other - could anyone help me with this?

    I've thought about it for a good hour or so now and don't seem to be getting anywhere (I know I'm meant to post what I've got so far but I literally have nothing!) so the more help you can give me the better!

    Many thanks :)
    Last edited: Dec 19, 2009
  2. jcsd
  3. Dec 20, 2009 #2


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    Use the definition of differentiable.

    A function of two variables, x and y, is "differentiable" at (a, b) if and only if there exist a function [itex]\epsilon(x,y)[/itex] such that [itex]f(a+ x, b+ y)= f(a, b)+ x (\partial f/\partial x(a,b))[/itex][itex]+ y(\partial f/\partial y(a,b))+ \epsilon(x,y)[/itex] and
    [tex]\lim_{(x,y)\to (a,b)}\frac{\epsilon(x,y)}{\sqrt{x^2+ y^2}}= 0[/tex]

    Since both [itex]\partial f/\partial x(a,b)[/itex] and [itex]\partial f/\partial y(a,b)[/itex] exist, we can define [itex]\epsilon(x,y)[/itex] as [itex]f(a+ x, b+ y)- f(a, b)[/itex][itex]+ x (\partial f/\partial x(a,b))+ y(\partial f/\partial y(a,b))[/itex].

    Try to use the fact that [itex]\partial f/\partial x[/itex] is continous to show that goes to 0 as (x,y) goes to (0, 0).
  4. Dec 20, 2009 #3
    Got it, thanks very much! :)
  5. Dec 20, 2009 #4


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    I'm impressed! Even with my hint, that was not a simple problem.
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