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Suppose $f$ is a function defined on $\mathbb{R}^{2}$ as follows:

$$ f(x,y) = \frac{x^{3}-y^{3}}{x^{2}+y^{2}} \text{ if } (x,y) \neq (0,0)$$

and $0$ if $(x,y) = (0,0)$.

I am asked to prove that the function is not differentiable at $(0,0)$.

I calculated the partial derivatives and they are $1$ and $-1$ respectively, in the $x$ and $y$ directions. I am unable to proceed from here.

Could any shed some light?

EDIT:

Based on the comments below, this is what I did:

We must show that

$$ \frac{\frac{x^{3} - y^{3}}{x^{2} + y^{2}} -x + y}{\sqrt{x^{2}+y^{2}}}$$

does not tend to zero as $(x,y) \to (0,0)$. Now, if we take $y = \alpha x$, we get that this expression is equal to some non-zero constant involving $\alpha$ and hence does not tend to zero.

Vishal Gupta
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  • so your potential gradient in $(0,0)$ is $(-1,1)^T$. Have you tried showing that this gradient indeed is a linear approximation? If you fail, where and why? – Quickbeam2k1 Jul 19 '13 at 15:30
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    Try differentiating along radial lines, $y = \alpha x$, which generalizes what you've already done. – hardmath Jul 19 '13 at 15:30
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    http://math.stackexchange.com/questions/357433/let-fx-y-fracx3-y3x2y2-is-f-differentiable-in-0-0?rq=1 – Amzoti Jul 19 '13 at 15:32
  • to my point. If the function is differentiable any directional derivative has to be "gradient $\cdot$ direction" – Quickbeam2k1 Jul 19 '13 at 15:36
  • @Quickbeam2k1 I tried something. You might want to look. – Vishal Gupta Jul 19 '13 at 16:20
  • It appears this is homework, so I will merely suggest that you review the rules of differentiability, particularly paying attention to the limit values of the function as it approaches (0, 0) from different directions. – abiessu Jul 19 '13 at 15:32
  • if you choose $\alpha=0$ you find the limit 0. Furthermore, I don't see (directly) why the expression is equal to an non zero constant – Quickbeam2k1 Jul 19 '13 at 19:37
  • @Quickbeam2k1 The point is that $\alpha$ can be arbitrary and the constant is zero if and only if $\alpha \in {0,1}$. – Vishal Gupta Jul 20 '13 at 04:44
  • You didn't say that above. Is there really directly a constant for $\alpha\neq {0,1,}$ or do you need a limit first? And my above statement that you find the limit "0" is not really correct. You don't need a limit. But if you really get such a result. It is of course not differentiable. – Quickbeam2k1 Jul 20 '13 at 05:40

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