Thanks for reading this post. I'm studying Vector Calculus by myself(for a while) and have a question about the differentiability of the function. The study material is Vector Calculus by Susan Jane Colley and here's the theorem.
Theorem 3.5
Suppose X is open in $\mathbb{R^2}$. If $f: X \rightarrow \mathbb{R}$ has continuous partial derivatives in a neighborhood of (a, b) in X, then $f$ is differentiable at (a, b).
My question is very simple but I am studying by myself (because of winter vacation), so I have nowhere to ask my idea is right or not. I want to ask why the neighborhood of (a, b) is used instead of the point (a, b).
I looked up the proof of the theorem and I found the mean-value theorem is used in $\epsilon - \delta$ method at the end. In my guess, this is the reason the neighborhood of (a, b) is used. And I searched google, I found no clear answer for my question but the Theorem was stated like I wrote.
I think this can be answered easily by someone who already knew. I guess it is already well-known problem so I saved my time for doing Tex job.
Edit: Here's the link
https://cpb-us-w2.wpmucdn.com/u.osu.edu/dist/6/3494/files/2014/04/13-4-to-13-6-handout-x02n2s.pdf
The same problem that Theorem 1.5 & Warning 1.6 says.
Continuity of partial derivatives in a neighbourhood is not needed for the proof of differentiability at the point. $f$ will be differentiable at $(a,b)$ as soon as the partial derivatives exist in a neighbourhood of $(a,b)$ and they are continuous at $(a,b)$. The usual estimate, which I'm stating for $f:\Bbb R^n\to\Bbb R$ and which I presume the proof you know uses, $$\frac{f(x_1,\cdots, x_n)-f(a_1,\cdots,a_n)-\sum_{j=1}^n (x_j-a_j)\frac{\partial f}{\partial x_j}(a_1,\cdots,a_n)}{\sqrt{\sum_{j=1}^n (x_j-a_j)^2}}=\\=\frac{\sum_{j=1}^n (x_j-a_j)\left(\frac{\partial f}{\partial x_j}(a_1,\cdots, a_{j-1},\xi_j,x_{j+1},\cdots, x_n)-\frac{\partial f}{\partial x_j}(a_1,\cdots,a_n)\right)}{\sqrt{\sum_{j=1}^n (x_j-a_j)^2}}$$
works with just continuity of partial derivatives at $(a_1,\cdots,a_n)$.