Differentiability in $\mathbb{R}^2$

Question

I was given the equation:

$$f(x,y)=\begin{cases}x+y+\dfrac{xy^2\cos y}{x^2+y^2} & \text{for } x^2+y^2\neq 0\\ 0& \text{for }x^2+y^2=0\end{cases} $$

I need to show whether it is differentiable at $(0,0)$.

I managed to show that $0<|f(x,y)|\leq 2|x|+|y|$ which I then used to show continuity at $(0,0)$. Then I found the partial derivatives at $(0,0)$ which turned out to be both $1$ but I got these values by taking a limit to zero of terms which were like $0/x^4$ and $0/x^8$ which tend to $0$ (thus the PDs tend to $1$). Then I was hoping to use the fact that: Continuity & continuous PDs $\implies$ differentiable.

Now assuming I evaluated the PDs correctly at $(0,0)$ (which I am also not too sure about), since it's taking a limit rather than being a trivial evaluation, would that make it continuous at that point?

Answer 1

Your computation of the partial derivatives at $(0,0)$ is correct, but it doesn't tell you whether or not the partial derivaties are continuous there.

Actually, they aren't. If they were, $f$ would be differentiable at $(0,0)$. But it is not. If it was, $Df_{(0,0)}(x,y)=x+y$ and we would have$$\lim_{(x,y)\to(0,0)}\frac{\bigl\|f(x,y)-(x+y)\bigr\|}{\sqrt{x^2+y^2}}=0.$$But this is not true. See what happens when $y=x$.

Answer 2

HINT : Yes it's called a continuation.

You need to calculate $\displaystyle \frac{\partial f }{\partial x}(x,y)$ and $\displaystyle \frac{\partial f }{\partial y}(x,y)$ and to show that they tend to $1$ as $(x,y) \rightarrow (0,0)$