The Wronskian of $(n-1)$ times differentiable functions $f_1, \ldots, f_n$ is defined as the determinant $$ W(f_1, \ldots, f_n)(x) = \begin{vmatrix} f_1(x) & f_2(x) & \cdots & f_n(x) \\ f_1'(x) & f_2'(x) & \cdots & f_n'(x) \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)}(x) & f_2^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{vmatrix} $$ and used e.g. in the context of linear differential equations.
While working on Wronskian of functions $\sin(nx), n=1,2,...,k$. I “discovered” the following chain rule for Wronskians:
Let $I, J \subset \Bbb R$ be intervals, $g:I \to J$ and $f_1, \ldots, f_n: J \to \Bbb R$ be $(n-1)$ times differentiable functions. Then $$ W(f_1 \circ g, \ldots, f_n \circ g)(x) = W(f_1, \ldots f_n)(g(x)) \cdot (g'(x))^{n(n-1)/2} \, . $$
It may be surprising (it was to me!) that only the first derivative of $g$ occurs on the right hand side. That is a consequence of Faà di Bruno's formula for the derivatives of a composite function.
This is surely a known identity, but I haven't found a reference so far. I searched for “Wronskian” in connection with ”chain rule”, “Faà di Bruno's formula“, or “Bell polynomials” and checked the Wikipedia and Wolfram Mathworld pages about those topics.
So what I am asking for is a reference for that formula. Or perhaps it is a direct consequence of some other well-known identity for Wronskians?
For the sake of completeness I'll provide a sketch of my proof of the above identity. Faà di Bruno's formula states that $$ \frac{d^k}{dx^k}f_l(g(x)) = \sum_{j=1}^k f_l^{(j)}(g(x)) B_{k, j}(g'(x), g''(x), \ldots, g^{(k-j+1)}(x)) $$ where $B_{k,j}$ are the Bell polynomials. This can be written as a matrix product $$ \Bigl( (f_l \circ g)^{(k)}(x)\Bigr)_{k, l} = B(x) \cdot \Bigl( f_l^{(j)}(g(x))\Bigr)_{j, l} $$
where $B(x)$ is the triangular matrix $$ \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & b_{1, 1}(x) & 0 &\cdots & 0 \\ 0 & b_{2, 1}(x) & b_{2, 2}(x)& \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & b_{n-1, 1}(x) & b_{n-1, 2}(x) & \cdots & b_{n-1, n-1}(x) \end{pmatrix} $$ with $$ b_{k, j}(x) = B_{k, j}(g'(x), g''(x), \ldots, g^{(k-j+1)}(x)) \, . $$
It follows that $$ W(f_1 \circ g, \ldots, f_n \circ g)(x) = \det(B(x)) \cdot W(f_1, \ldots ,f_n)(g(x)) \, . $$ The diagonal entries of $B(x)$ are $B_{k,k}(g'(x)) = (g'(x))^k $, so that $$ \det(B(x)) = \prod_{k=1}^{n-1} (g'(x))^k = (g'(x))^{n(n-1)/2} $$ and that gives exactly the desired result. (It looks easy once you have a proof, but it took me a while to figure this out :)