# Show that f is differentiable at $(x, y) = (0, 0)$ for the given function

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).

• You asked a different quesiton initally Feb 28 '21 at 13:17
• Yes I forgot to edit the given function since I copied the latex notation elsewhere, apologies for that. Feb 28 '21 at 13:18

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.
• 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. Feb 28 '21 at 14:29