A (bi)alternant formula for Wronskian

Question

We know that there exists similarities between power functions and derivative of a function (in particular, Newton binomial formula and Leibniz rule for derivation of a product can be deduced from each other). Thanks to the alternant formula, we can express the determinant of a Vandermonde matrix with a missing power (e.g. see Computing an almost Vandermonde matrix) via a power function of its coefficients.

Question: Does there exist a similar formula for the Wronskian of arbitrary functions :

$$ W_k(f_1,\dotsc, f_{n}) = \begin{vmatrix} f_1&\cdots&f_1^{(k-1)}&f_1^{(k+1)}&\cdots &f_1^{(n)}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ f_n&\cdots&f_n^{(k-1)}&f_n^{(k+1)}&\cdots &f_n^{(n)} \end{vmatrix}? $$

Answer 1

There is an instance where the two formulas are very close: if $f_1,\ldots,f_n$ are fundamental solutions of a linear ODE with indeterminate (constant) coefficients. This is explained in the elegant paper

L. Gatto, I. Shcherbak: "On one Property of one Solution of one Equation" or Linear ODEs, Wronskians and Schubert Calculus

Moscow Mathematical Journal, 2012, Volume 12, Number 2, Pages 275–291

(the link is to the freely available arXiv version)

I doubt that much further can be done. Already the first nontrivial case of $$W_1(f,g,h)=\det\begin{pmatrix}f&f''&f'''\\ g&g''&g'''\\ h&h''&h'''\\ \end{pmatrix}$$ looks somewhat mysterious: the second derivative of the Wronskian is
$$ \det\begin{pmatrix}f&f''&f'''\\ g&g''&g'''\\ h&h''&h'''\\ \end{pmatrix} + \det\begin{pmatrix}f&f'&f''''\\ g&g'&g''''\\ h&h'&h''''\\ \end{pmatrix}, $$ and I do not see any way to use other formulas to split out one of these summands!