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I know that this question has been asked before for the given function, but I still have specific questions regarding this problem that hasn't been addressed and I would like to explain my specific reasoning as well.

We are given the following function: $$f(x,y) = \begin{cases}(x^2+y^2)\sin(\frac{1}{\sqrt{x^2+y^2}})&\text{ if }(x,y)\not =(0,0)\\0 &\text{ if }(x,y)=(0,0)\end{cases}$$

Show that $f$ is differentiable at $(x, y) = (0, 0)$ and determine the derivative.

In general, we know the following:

  • If the partial derivatives $f_x$ and $f_y$ exist near $(a, b)$ and are continuous at $(a, b)$, then $f$ is differentiable at $(a, b)$.
  • If $\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b)$, then $f$ is continuous at $(a, b)$.
  • If $f$ is differentiable at $(a, b)$, then $f$ is continuous at $(a, b)$.

I know the following about the function:

  • The partial derivatives $f_x$ and $f_y$ exist at every $(x,y)$, where $f_x(0, 0) = 0$ and $f_y(0, 0) = 0$.
  • The partial derivatives are not continuous at $(0, 0)$, since $\lim_{(x, y) \to (0, 0)} f_x(x, y)$ does not exist.

We can conclude the following:

  • We know that the partial derivatives are not continuous at $(0, 0)$, but this doesn't mean that $f$ is not differentiable at $(0, 0)$, since the reverse of the first general statement is not true.

So how can we show that $f$ is differentiable at $(0, 0)$, specifically using the following hint: $-(x^2+y^2) \leq f(x, y) \leq x^2 + y^2$ for all $(x, y$)? Checking whether $f$ is continuous at $(a, b)$ isn't enough right? Because if $f$ is continuous at $(a, b)$, then this doesn't mean that $f$ is differentiable at $(a, b)$ (since the reverse of the third general statement isn't true).

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  • $\begingroup$ You asked a different quesiton initally $\endgroup$ Feb 28 '21 at 13:17
  • $\begingroup$ Yes I forgot to edit the given function since I copied the latex notation elsewhere, apologies for that. $\endgroup$
    – Stallmp
    Feb 28 '21 at 13:18
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Since $f_x(0,0)=f_y(0,0)=0$, if $f$ is differentiable at $(0,0)$, then $f'(0,0)$ can only be the null function. And it is the null function if and only if$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$And this is true because you have$$-(x^2+y^2)\leqslant f(x,y)\leqslant x^2+y^2$$and therefore$$-\sqrt{x^2+y^2}\leqslant\frac{f(x,y)}{\sqrt{x^2+y^2}}\leqslant\sqrt{x^2+y^2}$$and so all you have to do is to apply the squeeze theorem.

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  • $\begingroup$ Thank you for your answer! Does your answer apply to the given function in the latest edit? (See comment above: Yes I forgot to edit the given function since I copied the latex notation elsewhere, apologies for that.) $\endgroup$
    – Stallmp
    Feb 28 '21 at 13:19
  • $\begingroup$ Nothing has to be changed from my answer; it is still correct after that edition. $\endgroup$ Feb 28 '21 at 14:02
  • $\begingroup$ Thank you! I don't quite understand the answer completely however. We know that the partial derivatives evaluated at the origin are both 0. Then you state that f'(0,0) can only be the null function. What exactly is f'(0,0) here? Because there are only partial derivatives right? And the derivative evaluated at a point (0, 0) is a number (slope) and not a function? And what exactly is a null function and how did you get the first equation from the given function? $\endgroup$
    – Stallmp
    Feb 28 '21 at 14:25
  • $\begingroup$ If $f$ is a differentiable function from $\Bbb R^n$ to $\Bbb R^m$ and if $x_0\in\Bbb R_n$, then $f'(x_0)$ is a linear map from $\Bbb R^n$ into $\Bbb R^m$. Actually it's the only linear map $L\colon\Bbb R^n\longrightarrow\Bbb R^m$ such that$$\lim_{x\to x_0}\frac{\|f(x)-f(x_0)-L(x-x_0)\|}{\|x-x_0\|}=0.$$It is not a number and it has nothing to do with slopes. $\endgroup$ Feb 28 '21 at 14:29
  • $\begingroup$ Thank you for your responses. How would you exactly determine the derivative in this case? For a function of two variables, you can only determine the partial derivatives and the gradient right? What would be the derivative of this function? $\endgroup$
    – Stallmp
    Feb 28 '21 at 18:22

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