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:
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.
Best Answer
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 $(0,0)$.