Determine this

Define the function $g(x_1,x_2,x_3,x_4)=\det[f_i(x_j)]$ , a polynomial of degree 6; the diagonal contributes the term $x_2x_3^2x_4^3$ , for example. If we let $x_p=x_q$ for some $p<q$ then $g(x_1,x_2,x_3,x_4)=0$ since two columns of the matrix are equal.Thus $g(x_1,x_2,x_3,x_4)$ is divisible by all $x_q-x_p$ for $q>p$ , and we have $g(x_1,x_2,x_3,x_4)=C\prod_{q>p}(x_q-x_p)$ for some constant $C$ since the degrees are equal. Since both $g(x_1,x_2,x_3,x_4)$ and $\prod_{q>p}(x_q-x_p)$ contain the term $x_2x_3^2x_4^3$ (multiply all the positive terms in the product), we must have $C=1$ . Now $g(1,2,3,4)=(4-3)(4-2)(4-1)(3-2)(3-1)(2-1)=\boxed{12}$ .

This is just a slight generalisation of the theory of Vandermonde determinants, of course.