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I am trying to study the function $f(x,y) = (x^2 + y^2)\sin(\frac{1}{x^2 + y^2})$ for $(x,y) \neq (0,0)$ and $f(x,y) = 0$ otherwise.

I've been asked to show that the function is differentiable in $R^2$ but the partial derivatives at $(0,0)$ do not exist.

What I know, what I tried:

For a function from $R^2 \rightarrow R$, if the partial derivatives exist in the neighbourhood of $(x_0,y_0)$, and are continuous at $(x_0,y_0)$, then the function is differentiable at $(x_0,y_0)$

However, the partial derivatives at $(0,0)$ do not even exist here, so what's going on?

What is the sufficient condition to ensure the differentiability of a multivariate function?

P.S. If partial derivatives at a point exist, and if they are also continuous at that point, can we say that the function is differentiable? Is the converse valid, that is, is it necessary for the partial derivatives to be continuous if a function is differentiable at a point? (From this example, clearly not so)

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    $\begingroup$ Partial derivatives exist at $(0,0)$ and both partial derivatives are $0$. $\endgroup$ – Kavi Rama Murthy Sep 29 '19 at 5:57
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You are right, if partial derivatives do not exist then $f$ can't be differentiable, infact existence of partial derivatives is a necessary condition since differentiability implies their existence.

But are you sure that in this case partial derivatives don't exist?

If partial derivatives are not continuos at the point you can't yet conclude anything about differentiability. You need to check directly differentiability by definition:

$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=0$$

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  • $\begingroup$ The problem asks us to show that partial derivatives are NOT continuous at (0,0), but the function is differentiable in $R^2$, seems a little weird to me. Also, I've updated the function definition in the problem, please check! $\endgroup$ – epsilon-emperor Sep 29 '19 at 5:52
  • $\begingroup$ Alright, got it! This was for (0,0) - how do I explain that the function is differentiable for $R^2$? Is it enough to say that the partial derivatives exist at all points in $R^2$ and that they are also continuous? Please also check the postscript of the question, thank you! $\endgroup$ – epsilon-emperor Sep 29 '19 at 6:10
  • $\begingroup$ @arya_stark In general if partial derivatives exist and are continuos you are done, indeed if all the partial derivatives exist and are continuous in a neighborhood of the point then the function is differentiable at that point. $\endgroup$ – user Sep 29 '19 at 6:18
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At $(0,0)$ both partial derivatives exist and are $0$. Also $f$ is differentiable at the origin. This follows from the fact $\frac {|f(x,y)|} {\sqrt {x^{2}+y^{2}}} \leq \sqrt {x^{2}+y^{2}}$. However the partial derivatives are not continuous at the origin. To see this take limit through $y=0$ of the partial derivative w.r.t. $x$.

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  • $\begingroup$ Okay, one question. If partial derivatives at a point exist, and if they are also continuous at that point, can we say that the function is differentiable? Is the converse valid, that is, is it necessary for the partial derivatives to be continuous if a function is differentiable at a point? (From this example, clearly not so) $\endgroup$ – epsilon-emperor Sep 29 '19 at 6:12
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    $\begingroup$ @arya_stark For the first question the answer is YES. For the second, NO by the above example. $\endgroup$ – Kavi Rama Murthy Sep 29 '19 at 6:19

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