I am trying to derive the taylor expansion for the potential $U(\vec r + \delta \vec r)$.

The general expression for the taylor expansion is: $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$.

I am trying to derive the same expression as in the wikipedia article : https://en.wikipedia.org/wiki/Fine_structure#:~:text=This%20yields%20a,can%20be%20estimated%3A

But my derivation is not proper and it's messy, with multiple punctuality mistakes etc. From my understanding the variable is $\delta\vec r$. So we need to derivate with respect to it.

$U(\vec r + \delta \vec r)=U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec r'-\delta \vec r)^0 + \sum_i \frac{\partial}{\partial x_i}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)^1|_{\delta x_i=0}+\frac 1 2\sum_{i,j}\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}U(\vec r + \delta \vec r)|_{\delta \vec r=0}(\delta \vec x'_i-\delta x_i)|_{\delta x_i=0}(\delta \vec x'_j-\delta x_j)|_{\delta x_j=0} +...$

I have the following 3 questions for what I wrote above:

1- If we want to be precise, shouldn't I write $\frac{\partial}{\partial \delta x_i}$ instead of $\frac{\partial}{\partial x_i}$ ? Because if am writing $\frac{\partial}{\partial x_i}$ than it looks as if I am derivating with respect to $x_i$.

2- In the wikipedia article the 2nd term is written as the product of the Nabla operator and the potential, so a gradient. How do I cleanly produce that expression?