TITLE: How to evaluate this determinant?
QUESTION [7 upvotes]: Can someone give me a hint how to solve $$\left|\begin{array}{ccccc}
1 & 1 & \ldots &  & 1\\
2x_{1} & 2x_{2} &  &  & 2x_{n}\\
\vdots\\
nx_{1}^{n-1} & nx_{2}^{n-1} & \ldots &  & nx_{n}^{n-1}\\
\\
\end{array}\right|$$ ?
I know that I somehow have to use the Vandermonde determinant to do this, but I can't figure out how to get rid of the coefficients. Can someone give me a hint please ?

REPLY [3 votes]: Because $\det A = \sum \epsilon_{i_1, i_2, \ldots, i_n} a_{1,i_1} \cdot a_{2,i_2} \cdots a_{n,i_n}$. Since $k x_i^{k-1} = \partial_{x_i} x_i^k$, determinant of your matrix $A$ is
$$
   \det A = \partial_{x_1, x_2, \ldots, x_n} \left( x_1 x_2 \cdots x_n W(x_1, x_2, \ldots, x_n) \right)
$$
where $W(x_1, x_2, \ldots, x_n)$ is the determinant of the Vandermonde matrix.
$$
   \det A = \partial_{x_1, x_2, \ldots, x_n} \left( x_1 x_2 \cdots x_n \prod_{i < j} (x_i - x_j) \right)
$$
Added:
It actually looks like the derivative can be found in closed form, with the result of
$$
   \det A = n! \prod_{i < j} (x_i - x_j)
$$