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Differentiable function of 2 variables

  1. Nov 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that function has directional derivative in every direction, but is not differentiable in (0,0):

    [tex]f(x,y)=\begin{cases}\frac{x^3}{x^2+y^2},&(x,y)\neq(0,0)\\ \\0,&(x,y)=(0,0)\end{cases}[/tex]

    3. The attempt at a solution

    I have already proved that it has directional derivative in every direction. But in my opinion, if it has directional derivative in direction of OX and OY, it has partial derivatives, so that it is differentiable. Lets derive it:

    [tex]\frac{\partial f}{\partial x}=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{\frac{h^3}{h^2+0}}{h}=\lim_{h\to0}\frac{h^3}{h^3}=1[/tex]
    [tex]\frac{\partial f}{\partial y}=\lim_{h\to0}\frac{f(0,h)-f(0,0)}{h}=\lim_{h\to0}\frac{\frac{0}{0+h^2}}{h}=\lim_{h\to0}\frac{0}{h^3}=0[/tex]

    Function is also continuous in (0,0) (although it does not affect differentiability). Is there a mistake in this exercise?
  2. jcsd
  3. Nov 23, 2013 #2


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    hold on a second, are you sure that
    [tex]\lim_{h\rightarrow 0} \frac{0}{h^3} = 0[/tex]
    I think you might have been going through this too quickly, and put an incorrect answer.

    edit: blech, oh no, I am the one that has gone through it too quickly and come to the wrong conclusion. ah, yeah you're right, it is zero, since the numerator is zero.

    Hmm. so yeah, it does look like the function is differentiable. wikipedia says: "If the partial derivatives of a function all exist and are continuous in a neighbourhood of a point, then the function must be differentiable at that point". So I guess this is the condition to check.
  4. Nov 23, 2013 #3


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    yeah, my hint is to check if the partial derivatives behave nicely in a neighbourhood of the point (0,0)
  5. Nov 23, 2013 #4

    Ray Vickson

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    In the multivariable case there is a difference between "having partial derivatives" and "being differentiable". Look in your textbook, or Google 'differentiable function + multivariable'. I am not saying your function is, or is not differentiable; I am just saying that your justification is faulty.
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