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Homework Help: Check differentiability in function

  1. Jan 18, 2010 #1


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    1. The problem statement, all variables and given/known data

    Hello, I can't find any way to prove if this funtion is or isn't differentiable in if [tex](x,y)=(0,0)[/tex] :
    [tex]{f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}[/tex] if [tex](x,y) \neq(0,0)[/tex]

    [tex]f(x,y)=0[/tex] if [tex](x,y)=(0,0)[/tex]

    3. The attempt at a solution

    Partial derivatives don't exist in (0,0), so i have calculate definition of differentiability, and i get:

    [tex]\displaystyle\lim_{x,y \to{0,0}}{} \displaystyle\frac{x^{3}}{(x^{2}+y^{2})^{3/2}}[/tex]

    Which limit doesn't exit, so it is not differentiable in (0,0).

    Is this right?

  2. jcsd
  3. Jan 18, 2010 #2
    If we approach the origin along the x-axis, we find that the limit is 0. But if we approach the origin along the curve y = x^(3/2), the limit is 1. Hence the function is not continuous at the origin, which of course means it's not differentiable there.

    If you were trying to apply the definition of the partial derivative to show non-differentiability at (0,0), then the approach would fail since this is an example of a function which has a directional derivative in every direction at the origin (and of course partial differentiability is the specific case where the unit vector in the definition of the directional derivative is taken to be a standard basis vector). You should demonstrate this.
  4. Jan 18, 2010 #3


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    That isn't correct. You may be thinking of the limit as x goes to infinity, not zero. The function is continuous at (0,0).

    To the OP: Try looking at the partial fy along y = 0 and y = x.
  5. Jan 18, 2010 #4
    Sorry you're right, I was thinking about infinity; the epsilon and deltas check out. Incidentally, I found an error in my untested lecture notes.

    As for your hint, presumably you want to show that the partial derivative of y is not continuous at (0,0), but this doesn't guarantee non-differentiability right? The theorem that all the partial derivatives of a function exist and are continuous implies differentiability does not have a converse, as can be demonstrated via the following function:

    [tex]\[ f(x,y) =\begin{cases}x^2\sin(\frac{1}{x}) + y^2 &\text{if }x \neq 0 \\
    y^2 &\text{if }x = 0 \end{cases} \][/tex]
  6. Jan 18, 2010 #5


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    Yes, you're correct about the converse. On looking at it a bit more I think it is differentiable at (0,0). One can calculate fx(0,0) and fy(0,0) then check the condition for differentiability:

    [tex]f(0+h,0+k) - f(0,0) = f_x(0,0) h + f_y(0,0)k +\varepsilon_1h + \varepsilon_2k[/tex]

    where ε1 and ε2 → 0 as h and k → 0 directly by writing it out.
  7. Jan 19, 2010 #6


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    Sorry I said partial derivatives don't exist, I wanted to say they are not continuous in (0,0).

    I have calculated [tex]f_{y}[/tex] and it is not continuous in (0,0).

    How can you calculated the condition for differentiability that have written?

  8. Jan 19, 2010 #7
    This function is both continuous and differentiable at (0,0). o But let's check these two conditions:

    1- Continuity: since the numerator x3 in [tex]{f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}[/tex] tends to zero faster than the denominator, so [tex]\displaystyle\lim_{x,y \to{(0,0)}}{} \displaystyle\frac{x^{3}}{x^{2}+y^{2}} = 0 [/tex] and by f(0,0)=0, this condition is met by our function.

    2- Differentiablity: first start with the derivative of [tex]{f(x,y)=\displaystyle\frac{x^{3}}{x^{2}+y^{2}}}[/tex] wrt x and after calculating it, find the value for [tex](x=\epsilon,y=0)[/tex]. Repeat this calculation for y and then substitute [tex](x=0,y=\epsilon)[/tex] in the result. This value must match the preceding one and of course f'(x=0,y=0) which is clearly zero from the second part of the definition of f(x,y), BUT IT DOESN'T.

    [Edit]: fx([tex]\epsilon[/tex],0) = 1 so it does not match fy(0,[tex]\epsilon[/tex])=0 and therefore the function is NOT differentiable at (0,0). For a better understanding of why this condition does not hold for f, see its graph:

    http://img149.imageshack.us/img149/627/plotg.jpg [Broken]

    Sounds like we have a cusp there.

    Last edited by a moderator: May 4, 2017
  9. Jan 19, 2010 #8


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    The first step is to calculate fx and fy at (0,0). I get that fx(0,0)=1 and fy(0,0)=0; you can check this. Then substituting everything into the equation:

    f(0+h,0+k) - f(0,0) = f_x(0,0) h + f_y(0,0)k +\varepsilon_1h + \varepsilon_2k


    [tex]\frac {h^3}{h^2+k^2} = 1h + 0k + \varepsilon_1h + \varepsilon_2k[/tex]

    Solve this for the epsilon terms:

    [tex]\varepsilon_1h + \varepsilon_2k = \frac {h^3}{h^2+k^2} -h= \frac{hk^2}{h^2+k^2}= 0h+\left(\frac{hk}{h^2+k^2}\right)k[/tex]

    That will do it if you can show

    [tex]lim_{(h,k)\rightarrow(0,0)}\frac{hk}{h^2+k^2} = 0[/tex]

    Can you take it from there?

    [Edit] And then again, maybe that doesn't go to zero :frown:
    Last edited: Jan 19, 2010
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