Q:

Differentiating a product symbol

Can someone explain how to differentiate something like
$$\prod\limits_{i<j}^N {(x_i-x_j)}$$
with respect to $x_i$
The product starts from 1 and goes to N. I started off by ignoring the $x_j$ as it doesn't depend on $i$ but I have a feeling that's not correct. What confuses me is the $i<j$
I have tried this on Maple but the answer doesn't make sense to me. Could someone help me out? Is there an explicit formula I should use?

A:

Since $i$ and $j$ are both bound variables of the expression, you should at most differentiate with respect to something "new", $x_k$ say. Then use the product rule: Since $$\frac\partial{\partial x_k}(x_i-x_j)=\begin{cases}1&\text{if }k=i\\-1&\text{if }k=j\\0&\text{otherwise}\end{cases}$$
we obtain
$$\begin{align} \frac\partial{\partial x_k}\prod_{1\le i<j\le n}(x_i-x_j)&=\sum_{1\le i<j\le n} \frac{\partial(x_i-x_j)}{\partial x_k}\cdot\prod_{1\le \iota<\nu\le n\atop(\iota,\nu)\ne(k,j)}(x_\iota-x_\nu)\\
&=\sum_{k<j\le n} \prod_{1\le \iota<\nu\le n\atop(\iota,\nu)\ne(i,j)}(x_\iota-x_\nu)-\sum_{1\le i<k} \prod_{1\le \iota<\nu\le n\atop(\iota,\nu)\ne(i,k)}(x_\iota-x_\nu)\\
&=\sum_{j=k+1}^n \prod_{1\le \iota<\nu\le n\atop(\iota,\nu)\ne(i,j)}(x_\iota-x_\nu)-\sum_{i=1}^{k-1} \prod_{1\le \iota<\nu\le n\atop(\iota,\nu)\ne(i,k)}(x_\iota-x_\nu).\end{align}$$
It may be easier to try with the logarithmic derivative:
$$\frac{\frac\partial{\partial x_k}f}{f}= \frac\partial{\partial x_k}\ln f =\frac\partial{\partial x_k}\sum_{1\le i<j\le n}\ln(x_i-x_j)=\sum_{k<j\le n}\frac1{x_k-x_j}-\sum_{1\le i<k}\frac1{x_i-x_k}$$
although this is formally correct only if $x_1>x_2>\ldots >x_n$.

