by David
Last Updated January 17, 2018 10:20 AM

In this exercise one must show that $$f(x,y) = \begin{cases} 0 \text{ if } (x,y) = (0,0)\\ (x^2 + y^2) \sin\left(\frac{1}{x^2 + y^2}\right) \text{ else} \end{cases}$$ is differentiable for all $(x,y) \in \mathbb{R^2}$

In the lecture, we have proven that if all partial derivatives of $f$ exist and are continuous, then the derivative of $f(x,y)$ exists everywhere.

I start by calculating the partial derivatives of $f$ at $(0,0)$: $$\partial_xf(0,0) = \lim_{x\to0} \frac{f(x,0)-f(0,0)}{x} = \lim_{x\to0} \frac{x^2\sin({1 \over x^2})}{x} = 0$$ Similar for $\partial_y$: $$\partial_yf(0,0) = \lim_{y\to0} \frac{f(0,y)-f(0,0)}{x} = \lim_{y\to0} \frac{y^2\sin({1 \over y^2})}{y} = 0$$

We know that for all other values of $x$, $\partial_x f = 2x\sin\left(\frac{1}{x^2+y^2}\right)-\frac{2x\cos \left(\frac{1}{x^2+y^2}\right)}{x^2+y^2}$

I have tried to show that $\lim _ {x,y \to (0,0)} \partial_x f = 0$. (to show f is continuous) this however is problematic, since this limit doesn't exist. This implies that the partial derivates are not continuous at all points. Which makes this approach of proving that $f$ is differentiable useless.

How do I prove that $f$ is differentiable on all of $\mathbb{R}^2$.

I am open for all kind of suggestions. Thanks for your time.

Note that$$\tag1\lim_{(x,y)\to0}\frac{\bigl|f(x,y)\bigr|}{\bigl\|(x,y)\bigr\|}=\lim_{(x,y)\to0}\sqrt{x^2+y^2}\left|\sin\left(\frac1{x^2+y^2}\right)\right|=0.$$Therefore, $f$ is differentiable at $(0,0)$ and $D_{(0,0)}f$ is the null function, since, if $L\colon\mathbb{R}^2\longrightarrow\mathbb{R}$ is the null function, then $(1)$ means that$$\lim_{(x,y)\to0}\frac{\bigl|f(x,y)-f(0,0)-L(x,y)\bigr|}{\bigl\|(x,y)\bigr\|}.$$So, by the definition of differentiable function, $f$ is differentiable at $(0,0)$ and $D_{(0,0)}f=L\equiv0$.

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