Let $f$ be $f: \mathbb R^2 \to \mathbb R$.
What I recall/understand:
- That $f$ is differentiable at a point $(a,b)$ is defined as this weird thing from James Stewart, Calculus (similar for $v$)
- Sufficient condition: That $f: \mathbb R^2 \to \mathbb R$ is differentiable a point $(a,b)$ has a sufficient condition from this theorem also from James Stewart, Calculus
which I read as
If $f_x$ and $f_y$ exist in an open disc containing $(a,b)$ and are continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$.
Questions:
Does $f$ differentiable at $(a,b)$ imply $f_x$ and $f_y$ exist at $(a,b)$?
For $f: \mathbb R^2 \to \mathbb R$ differentiable at $(a,b)$, what exactly its 'derivative' anyway?
Update: It's Jacobian. The thing is wiki doesn't say the jacobian of $f$ is the derivative of $f$. Anyway, I'll tag this question with jacobian.
- What I understand is for just $g: \mathbb R \to \mathbb R$, we have $g$ 'differentiable' at $a$ if $\lim_{x \to a} \frac{g(x)-g(a)}{x-a}$ exists and then define the 'derivative' $g'(x)$ as the limit. Similar for even complex derivative, for example.
So what about for $f$? What is the 'derivative' of $f$? You might argue that $f$ doesn't really have 'a'/'the' derivative but rather has infinite derivatives, eg using gradient based on $f_x$ and $f_y$.
But even for just $f_x$ and $f_y$ these derivatives exist based on the existence of a limit. Like you could define something like 'differentiable in the $x$-direction at $(a,b)$' if $\lim_{x \to a} \frac{f(x,b)-f(a,b)}{x-a}$ exists and then you define 'the derivative in the $x$-direction at $(a,b)$' as the limit.
So, what, differentiable for $f: \mathbb R^2 \to \mathbb R$ doesn't really have like 'a derivative' ?