What is the partial derivative in this expression? The question is similar to this one Implicit derivative?
Let $x_1, x_2, x_3$ be three points in $\mathbb{R}^3$, $A=(a_{ij})$ is a $3\times 3$ matrix with $a_{ii}=0$ and $a_{ij}=\frac{1}{|x_i-x_j|}$ for $1\le i,j\le 3$. My question: what is $\frac{\partial A}{\partial x_1}$, $\frac{\partial A}{\partial x_2}$ and $\frac{\partial A}{\partial x_3}$? I came across this in a paper <'Note on an inequality' by Y.Xu 2006. Annales de l'Institut Henri Poincare>, but I was puzzled with this expressions.
 A: I didn't look at the paper in too much detail, but my best guess is that $\partial_{x_i}A$ is a rank-3 tensor, in the sense that we can write $$v\cdot\partial_{x_i}A = \lim_{h\to 0} \frac{1}{h}(A[x_1, \ldots,x_i + hv, \ldots,x_p] - A[x_1,\ldots, x_p])$$
Or, in index notation, writing $a_{ij} = \frac{1}{x_i - x_j}$, we have that $$(\partial_{x_l}A)_{ijk} = \partial_{(x_l)^k} (\sum_{m} [(x_i)^m - (x_j)^m]^2)^{-1/2}$$
where $(x_l)^k$ denotes the $k$-th component of the point $x_l$. So in conjecture 1 of the linked paper, the term inside the absolute value sign in the first supremum is in fact, a vector, with the $i$ and $j$-th component in the above expression contracted against a vector $u$. 
More intuitively, you can think of it as the gradient on the total space projected to the $k$-th copy of $\mathbb{R}^3$. 
It is, admittedly, somewhat questionable and sloppy notation. I may have written it as the variation $\delta A / \delta x_i$ instead to get over the cognitive disconnect of taking a partial relative to a vector, but still it implicitly depends on the Euclidean structure of $\mathbb{R}^p$. 
