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Check whether the function is differentiable: $$f:\mathbb{R}^2\rightarrow \mathbb{R}$$ $$f= \begin{cases} \frac{x^3-y^3}{x^2+y^2} & (x,y)\neq (0,0) \\ 0 & (x,y) = (0,0) \\ \end{cases} $$

So what I did is I calculated the partial derivatives of the function in point $(0,0)$. I got: $$\frac{∂f}{∂x}\left(0,0\right)=lim_{t\rightarrow 0}\left(\frac{f\left(t,0\right)-f\left(0,0\right)}{t}\right)=lim_{t\rightarrow 0}\left(\frac{t^3}{t^3}\right)=1$$and $$\frac{∂f}{∂y}\left(0,0\right)=lim_{t\rightarrow 0}\left(\frac{f\left(0,t\right)-f\left(0,0\right)}{t}\right)=lim_{t\rightarrow 0}\left(\frac{-t^3}{t^3}\right)=-1$$

And since the answers I got are not equal, that means the function isn't partially derivable in point $(0,0)$ so it isn't differentiable either?

I'm not sure whether what I did was right, differentiability is still a little unclear to me, for multivariable functions. I also asked about it here Differentiability of function definition but have yet to get an answer. Can someone tell me if I'm on the right track at least?

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The partial derivatives don't need to be equal.

take for example: $f(x) = x - y$

$\frac {\partial f}{\partial x} = 1, \frac {\partial f}{\partial y} = -1$

However in the example at hand:

$\frac{\partial f}{\partial x} = \frac {x^2(x^2 + 3y^2)}{(x^2 + y^2)^2}$

Which is not continuous at (0,0).

So, your conclusion is correct, but your reasoning is not.

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  • $\begingroup$ How do I know whether the partial derivatives are continuous or not? $\endgroup$ – MikhaelM Jun 22 '16 at 20:48
  • $\begingroup$ To show that they are not continuous is a little easier then to prove that they are. Find two paths through $(0,0)$ such that $\frac {\partial f}{\partial x}$ is different on each path. e.g. plug $y = x$ and plug $y = 0$. To prove continuity you would need to show that $|\frac {\partial f}{\partial x} - L| < \epsilon$ at every point in a neighborhood of $(0,0)$ $\endgroup$ – Doug M Jun 22 '16 at 20:52
  • $\begingroup$ What do you mean by $\frac{∂f}{∂x}$ different on each path? Don't I get the same answer(0) if I plug in $y=x$ or $y=0$? First would give me $y$ and the second gives me $0$, which are both 0 at $(0,0)$, no? $\endgroup$ – MikhaelM Jun 22 '16 at 20:57
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    $\begingroup$ My bad, $y=x$ is a poor choice... if you plug $y = 2x$ , into $\frac{\partial f}{\partial x} = \frac{x^2(x^2+3y^2)}{(x^2+y^2)^2} \implies \frac {13x^4}{25x^4} = \frac{13}{25}$, if you plug $y = 0, \frac{\partial f}{\partial x} = \frac {x^4}{x^4} = 1.$ $\endgroup$ – Doug M Jun 22 '16 at 21:02
  • $\begingroup$ Oh, I understand now! Thank you. $\endgroup$ – MikhaelM Jun 22 '16 at 21:04
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Yes, I believe your conclusion is correct.

To check that the function is differentiable at $(0,0)$ we have to show that the derivative is continuous at that point.

We know that to check continuity at a point, say $(0,0)$, we need

$$\lim_{(x,y)\rightarrow (0,0)} f(x,y)=f(0,0)$$

However, since the derivative is not continuous, we know that the function is not differentiable.

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  • $\begingroup$ But isn't $\lim_{x\rightarrow 0} f(x,0)=\lim_{y\rightarrow 0}f(0,y)=f(0,0) = 0$, which would mean continuity? How do I know that the partial derivative isn't continuous? $\endgroup$ – MikhaelM Jun 22 '16 at 20:53
  • $\begingroup$ f(x,y) is continuous $\implies \lim_{x\rightarrow 0} f(x,0)=\lim_{y\rightarrow 0}f(0,y)=f(0,0)$ but it is not sufficient to go the other way. $\endgroup$ – Doug M Jun 22 '16 at 20:55
  • $\begingroup$ Yes, my mistake, but DougM explained it well in the comment to his answer. $\endgroup$ – Hannah Jun 22 '16 at 20:58
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This is wrong. Being partially differentiable means that the partial derivatives exist, and you have shown this by showing the limits to exist. The partial derivatives need not coincide! To show that $f$ is differentiable a sufficient conditon is that the partial derivatives exist and are continous. To show that $f$ is not differentiable, it suffices to show that the partial derivatives not not exist.

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