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 🙂
Best Answer
Computing the cells of the matrix row by row, starting at the top, for the first three rows we have with $y=g(x)$ for brevity, the matrix $M(f_1\circ g,...,f_n\circ g)$ involved in the Wronskian is $$ M(f_1, \ldots, f_n)(g(x)) = \begin{pmatrix} f_1(y) & f_2(y) & \cdots & f_n(y) \\ f_1'(y)g'(x) & f_2'(y)g'(x) & \cdots & f_n'(y)g'(x) \\ f_1^{(2)}(y)g'^2(x) & f_2^{(2)}(y)g'^2(x) & \cdots & f_n^{(2)}(y)g'^2(x) \\\vdots & \vdots & \vdots & \vdots \end{pmatrix} + \begin{pmatrix} f_1(y) & f_2(y) & \cdots & f_n(y) \\ f_1'(y)g'(x) & f_2'(y)g'(x) & \cdots & f_n'(y)g'(x) \\ f_1'(y)g''(x) & f_2'(y)g''(x) & \cdots & f_n'(y)g''(x) \\\vdots & \vdots & \vdots & \vdots \end{pmatrix}. $$ The second and third rows are linearly dependent (e.g., take the weighted sum of the second row weighted with $g''(x)$ and the third weighted with $-g'(x)$). So for the first three rows, by properties of the determinant we can remove the second term above without changing the value of the determinant. Only the first term survives, which corresponds to the formula you mention, with powers of $g'(x)$ but no higher derivatives of $g$.
For the next rows (4th, 5th etc), say the $k$-th row, only the term in $g'(x)^{k-1}$ survives as all others lead to rows that are linearly dependent with the previous rows (that we have "cleaned" one by one, as for the third row above).
The rest is a combinatorial problem. The second row has $g'(x)$, third row has $g'^2(x)$, fourth row has $g'^3(x)$ etc. As multiplying a row by $C$ multiplies the determinant by $C$, we find a total of $g'(x)^{1+2+...+n-1}$ that can be factored out of the determinant. Since $1+2+...+n-1=n(n-1)/2$, I very much agree with your nice formula!