Let $f_0,\ldots,f_N$ be smooth functions over an interval $I\subset{\mathbb R}$. Let $x_0,\ldots,x_N\in I$ be given, and form the Vandermonde-like determinant $$\Delta_N=\det((f_i(x_j)))_{0\le i,j\le N}.$$ I should bet that there exists an $a\in I$ such that $$\Delta_N=\frac1{0!1!\cdots N!}\,V_N(\vec x)W_N(\vec f)(a),$$ where $V_N$ is the Vandermonde determinant $\prod_{i<j}(x_j-x_i)$ and $W_N$ is the Wronskian $\det((f_i^{(j)}))_{0\le i,j\le N}$.

This must be a well-known Taylor-like formula. I should appreciate an accurate reference, or a sketch of proof.

**Edit**. After Christian R.'s answer, who shows by a counter-example that the above claim is wrong, let me suggest a weaker property, that
$$\left|\Delta_N\right|\le\frac1{0!1!\cdots N!}\,\left|V_N(\vec x)\right|\cdot\|W_N(\vec f)\|_{L^\infty(I)}.$$
After all, if $W_N(\vec f)\equiv0$, then the functions $f_j$ are linearly dependent and $\Delta_N$ vanishes too.

2more comments