Let $$f$$ be $$f: \mathbb R^2 \to \mathbb R$$.

What I recall/understand:

1. That $$f$$ is differentiable at a point $$(a,b)$$ is defined as this weird thing from James Stewart, Calculus (similar for $$v$$) 1. 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 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:

1. Does $$f$$ differentiable at $$(a,b)$$ imply $$f_x$$ and $$f_y$$ exist at $$(a,b)$$?

2. 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' ?

• 1. is true but not its reciprocal see -> math.stackexchange.com/questions/2569710/…
– zwim
Oct 27, 2021 at 19:30
• @zwim thanks. as for why the reciprocal (converse?) is false...well that's the whole point of sufficient condition right? i don't know any explicit counter-example, but i just find it weird if a book will call it sufficient condition instead of equivalent condition, if they're actually equivalent...?
– BCLC
Oct 27, 2021 at 19:57

The usual definition of derivative of a function $$f$$ from $$\mathbf{R}^m$$ to $$\mathbf{R}^n$$ at a point $$x \in \mathbf{R}^m$$ is that it is the linear map that approximates $$f$$ to order greater than one near $$x$$. Precisely, a linear map $$L:\mathbf{R}^m \to \mathbf{R}^n$$ is the derivative of $$f$$ at $$x$$ provided

$$\lim_{h \to 0} \frac{f(x+h)-f(x)-L(h)}{|h|}=0,$$ where $$h$$ is a vector in $$\mathbf{R}^m$$ tending towards $$0$$ and $$|h|$$ denotes its length. If you define the quantity in the limit as $$\epsilon(h)$$, then this is the same as saying that $$f(x+h)=f(x)+L(h)+|h| \epsilon(h),$$ where $$\epsilon(h)$$ goes to $$0$$ with $$h$$ (in this sense $$L$$ approximates $$f$$ to order greater than $$1$$). Moreover if a linear map with this property exists, it is unique.

This is what the first definition you mention is saying (in coordinates, and with $$m=2$$ and $$n=1$$). But you are right that you could also define directional derivatives: for a vector $$v$$ you could put $$\partial_v(f)(x)=\lim_{h \to 0} \frac{f(x+hv)-f(x)}{h}$$ where $$h$$ now runs over $$\mathbf{R}$$, when this limit exists and call it the derivative of $$f$$ in the direction $$v$$. The relationship between these is that if $$f$$ has a derivative in the first sense, it has derivatives in all directions $$v$$ and $$\partial_v(f)(x)=L(v).$$

Conversely, the last assertion is that if the derivatives in the directions $$v_1,\dots,v_m$$ exist in a neighborhood of $$x$$ and are continuous at $$x$$, where $$v_1,\dots,v_m$$ is a basis of $$\mathbf{R}^m$$, then $$f$$ has a derivative $$L$$ in the first sense. All this is completely standard and can be found, for instance, in Spivak's nice book Calculus on manifolds.

• wait so jacobian is the derivative? it seems your $L$ is the same as $J$ in wiki, but i didn't see anything in wiki like $J$ is the 'derivative' of $f$. thanks Stephen
– BCLC
Oct 27, 2021 at 19:34
• I usually prefer to just call $L$ the derivative, as distinct from the Jacobian matrix and especially the Jacobian determinant, which sometimes people shorten to Jacobian. But yes, the Jacobian matrix as defined at your link is the matrix of the derivative with respect to the canonical bases of $\mathbf{R}^m$ and $\mathbf{R}^n$. Oct 27, 2021 at 19:35
• ...and for reasonable functions, the columns of the Jacobian matrix are the directional derivatives corresponding to the canonical basis (the partial derivatives of $f$). Oct 27, 2021 at 19:36
• Right! And the best you can say for the converse is that if the derivatives in all directions are well-defined in a neighborhood of your point and continuous at it, then it has a derivative in the first sense whose matrix is given by them in the way I mentioned above. Oct 27, 2021 at 19:55
• @JohnSmithKyon Sure, I just posted an answer there! I hope it helps. Oct 28, 2021 at 0:19