Abitrary derivatives of lagrange basis functions The lagrange basis functions are given by
\begin{align}
\phi_k(x) =\prod_{j\not = k} \frac{x-x_j}{x_k-x_j}
\end{align}
I try to reproduce the numerical results of a paper. In this paper, the fractional derivative of $\phi$ has to be computed. This is done, by approximating $\phi$ by its Taylor expansion and then using analytical results for fractional derivatives for polynomials.
The Taylor expansion of $\phi_k$ is given by
\begin{align}
\phi_k(x) \approx \sum_{m=0}^N\phi_k^{(m)}(0)\frac{x^m}{m!}
\end{align} 
Now in the paper, there are no information about the computation of $\phi^{(m)}(0)$. Since the method is told to be of high order, I doubt that finite differences are the way to go. For $N=1,2$ I found analytic results for $\phi^{(m)}(x)$. 
Are there any results for $\phi_k^{(m)}(0)$ in the literature? If it helps, the $x_i$ are the Gauss Lobatto points. Since I perform a convergence analysis and no performance analysis, a slow roboust method is prefered over a fast inaccurate method. 
Thanks!
 A: The monomial basis, $\pmb m(x) = (1,x,x^2,x^3\ldots,x^n)^T$ should be easier to work with when applying the operator $D^{k}_0=\frac{d^k}{dx^k}\mid_{x=0}$ than the Lagrange basis, $\pmb l(x)$. Let us find the transformation between the two bases.
The Lagrange interpolating polynomial $P_n(x)$ is the sum
$$
P_n(x)=\sum_{i=0}^n f_il_i(x)=\pmb f^T\pmb l(x)
$$
If we choose to interpolate functions $f$ that are polynomials of degree $k\leq n$, then obviously $P_n(x)\equiv f(x)$. Let us take $f$ from the monomial list, $\pmb m$. For example, if $f(x)=x^k$, then
$$
\sum_{i=0}^n x^k_il_i(x)=x^k
$$
and so on, we can write it with the Vandermonde matrix $V$ as
$$
V^T \pmb l(x)=\pmb m(x)
$$
Since the Vandermonde matrix is invertible (for distinct nodes), we have,
$$
\pmb l(x)=V^{-T}\pmb m(x).
$$
Using this transformation, which is in fact the Maclaurin series of $\pmb l(x)$, it's easy to see that the k-th derivative at 0 of the column of the Lagrange basis polynomials is proportional to the k-th column of the inverse transposed Vandermonde matrix,
$$
D^{k}_0[\pmb l(x)]=k!\left[V^{-T}\right]_k
$$
For the derivatives $0\leq k\leq n$, we can write for the matrix $[A]_k^i=l_i^{(k)}(x)\mid_{x=0}$
$$
A=V^{-T}\operatorname{diag}(k!)_{k=0}^n.
$$
