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Is the function differentiable everywhere?

  1. Dec 3, 2011 #1
    I am hoping someone can help me with the following problem:

    Define f by:

    [tex]f(x, y) = 0 \ if \ (x, y) = (0,0) \ and \ f(x, y) = \frac{xy^{2}}{(x^{2}+y^{4})^{1/2}} \ otherwise[/tex]

    The problem is to determine (and prove) whether the function is differentiable everywhere.

    First of all, the partials exist everywhere (they are 0 at the origin in particular) but are not continuous at the origin. Since they are continuous everywhere else, the function is certainly differentiable everywhere, except possibly at the origin.

    If the function has a derivative at the origin then it is the zero vector since I have computed the partials to be 0 at the origin. Hence to determine differentiability at the origin I must check whether the following holds:

    [tex] \lim_{(x, y)\rightarrow(0,0)} \frac{f(x, y) - f(0, 0) - <0,0> \cdot <x, y>}{||(x, y)||} = \lim_{(x, y)\rightarrow(0,0)} \frac{xy^{2}}{(x^{2}+y^{4})^{1/2}(x^{2}+y^{2})^{1/2}} = 0[/tex]

    I'm asked to show whether this holds using the definition, i.e.: [tex] \forall \epsilon > 0 \ \exists\ \delta > 0 \ s.t. \ if \ ||(x, y)|| < \delta \ then \ \frac{|x|y^{2}}{(x^{2}+y^{4})^{1/2}} < \epsilon ||(x, y)||[/tex]

    This is the part where I am having some issues. In particular I tried choosing paths like x = y or x = y^2 to try and show that the inequality cannot hold for certain values of epsilon but this was not very helpful.

    Any tips/hints are greatly appreciated.
     
    Last edited: Dec 4, 2011
  2. jcsd
  3. Dec 4, 2011 #2
    I think you are supposing here that in the origin the tangent plane is the horizontal plane.
    But:
    [tex]\lim_{(x, y) \to (0,0)} \frac{xy^{2}}{(x^{2}+y^{4})^{1/2}(x^{2}+y^{2})^{1/2}} = \lim_{(x, y) \to (0,0)} \frac{xy^{2}}{(x^{4}+y^{6}+x^2y^2+x^2y^4)^{1/2}} [/tex]

    If you approach the origin by [itex]x=y^2[/itex]
    you have
    [tex]\lim_{(x) \to (0)} \frac{x^{2}}{(x^{2}(2+2x^2)^{1/2}} = \frac{1}{\sqrt2} [/tex]
    so there cannot be any horizontal plane there.
    I'm not sure 100%, I hope this helps.
     
    Last edited: Dec 4, 2011
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