Consider the split monic $f=\prod_{i=1}^n(x-x_i)\in \mathbb Z[x_1 ,\dots ,x_n,x]$. Its discriminant is usually defined as $$(-1)^{n(n-1)/2}\prod_{i=1}^nf^\prime(x_i)=\prod_{1\leq i<j\leq n}(x_i-x_j)^2.$$
What is the reason for taking this definition as opposed to $\prod_{i=1}^nf^\prime(x_i)$? The product of the derivatives at the roots "feels" to me more canonical than the product on the RHS.
The reason is that the formula on the right side should be considered more fundamental, not the formula on the left, when seeking a symmetric expression in the roots. Don't use a product of anything "at" the roots, but a symmetric expression in the roots that vanishes if any pair of roots are equal. That explains the factors $(x_i-x_j)^2$. Do you consider the simplest polynomial with a double root at 0 to be $x^2$ or $-x^2$? The product on the left (without the sign) is also interesting and has a name: it is called the resultant of $f(x)$ and $f'(x)$.
The definition of the resultant $\text{res}(f,g)$ of two (monic) polynomials $f,g\in {\mathbb Z}[x]$ as the determinant of the corresponding Sylvester matrix, readily implies that if $\alpha_1,\ldots,\alpha_n$ are the roots of $f(x)$ (say, in ${\mathbb C}$) then $$\text{res}(f,g)=\prod_{i=1}^ng(\alpha_i).$$ Now, if you define the discriminant of the (monic) polynomial $f(x)$ as $$\Delta(f(x))=\text{res}(f(x),f'(x))=\prod_{i=1}^nf'(\alpha_i),$$ writing $f(x)=(x-\alpha_1)\cdots (x-\alpha_n)$ it follows that $$f'(\alpha_j)=\prod_{i\neq j}^n(\alpha_j-\alpha_i).$$ In this equality, once you order the indices, for each pair $i<j$ there are two factors $\pm(\alpha_i-\alpha_j)$ and thus \begin{align*} \Delta(f(x))&=\prod_{i=1}^nf'(\alpha_i)=\prod_{i\neq j}^n(\alpha_j-\alpha_i)\\ &=(-1)^{n(n-1)/2}\prod_{i< j}^n(\alpha_j-\alpha_i)^2 \end{align*} where the sign is because there are precisely $n(n-1)/2$ pairs $\pm(\alpha_i-\alpha_j)$.
