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Let $$f(x,y)=\begin{cases} xy\sin(x/y) & y\neq 0 \\ 0 & y=0\end{cases},$$ show whether $f(x,y)$ is differentiable at $(0,0)$.

It seems that there are multiple ways to do this but there is no clear example online or in texbooks..

The provided solution begins with showing that the two partial derivatives equal zero when at $(0, 0)$, to me they look like they're supposed to be undefined..

using the definition of partial derivatives, isn't this the case: $$ f_x=xcos(x/y)+ysin(x/y) \\ f_y=ysin(x/y)-\frac{x}{y}sin(x/y) $$

??

and even if they were true, the solution uses the definition of differential df to prove the case. Solution is this:

enter image description here

I vaguely understand what it's doing because the existence of a tagent plane approximation implies differentiability.

But I can't seem to understand where the $\epsilon_1$ and $\epsilon_2$ came from.

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  • $\begingroup$ The expressions you found for $f_x$ and $f_y$ are for $\{(x,y)\in \mathbb R^2\colon y\neq 0\}$. Write the definition of $f_x(0,0)$ and $f_y(0,0)$. Regarding the rest, what exactly is the definition of differentiability you're using? $\endgroup$ – Git Gud Mar 12 '15 at 11:44
  • $\begingroup$ just edited the stuff again.. thank you for trying to walk this lost soul through this $\endgroup$ – user223022 Mar 12 '15 at 11:50
  • $\begingroup$ You still haven't written the definition of $f_x(0,0)$ and $f_y(0,0)$. The definition of differentiability is too still lacking. I can't really comment on the solution given because I don't understand the language used and I don't know what definition of differentiability you're using. $\endgroup$ – Git Gud Mar 12 '15 at 11:52
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One certainly has $$f_x(0,0)=\lim_{x\to0}{f(x,0)-f(0,0)\over x}=\lim_{x\to0}{0-0\over x}=0$$ and $$f_y(0,0)=\lim_{y\to0}{f(0,y)-f(0,0)\over y}=\lim_{y\to0}{0-0\over y}=0\ .$$ From this we draw the following conclusion: If $f$ is in fact differentiable at $(0,0)$ then necessarily $$df(0,0)=0\ .$$ By definition of differentiability we have to check whether $$f(x,y)-f(0,0)- df(0,0).(x,y)=o\bigl(\sqrt{x^2+y^2}\bigr)\qquad\bigl((x,y)\to(0,0)\bigr)\ .\tag{1}$$ Here the left hand side is $=f(x,y)$. Now from $|xy|\leq{1\over2}(x^2+y^2)$ it follows that $$\left|{f(x,y)\over\sqrt{x^2+y^2}}\right|\leq {1\over2}\sqrt{x^2+y^2}\qquad\bigl((x,y)\ne(0,0)\bigr)\ ,$$ and this corroborates $(1)$. Therefore $f$ is indeed differentiable at $(0,0)$.

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