How to derive the derivative of the r-determinant? Given a matrix $A=(a_{ij})\in M_n(\mathbb{C})$ and $r\in\mathbb{N}$, define the r-determinant of $A$ as $$det_r[A]:=\sum_{\sigma\in S_n}\prod_{i=1}^na_{i\sigma(i)}sign(\sigma)r^{c(\sigma)},$$
where $c(\sigma)$ is the number of cycles in $\sigma$.
In particular, when $r=  1$, it specializes to the determinant.
How to prove the following result?
$$
det_r[Z_i-A_i] = \frac{1}{r}\frac{\partial}{\partial z_i}det_r[Z-A],
$$
where $Z = diag(z_1,\ldots,z_n)$ and $Z_i, A_i$ denote the matrix that is deleted $i$th row and $j$th row in $Z$, $A$, respectively.
 A: Take $i=n$ without loss of generality. Note that the product $\prod_{i=1}^n(A-Z)_{i\sigma(i)}\operatorname{sgn}(\sigma)r^{c(\sigma)}$ will include $z_n$ if and only if $\sigma(n) = n$; all other terms in the sum become zero under $\frac{\partial }{\partial z_n}$.
With that, we have
\begin{align}
\frac{\partial }{\partial z_n} {\det}_r(Z-A) &= 
\frac{\partial }{\partial z_n}\sum_{\sigma\in S_n}\prod_{i=1}^n(Z-A)_{i\sigma(i)}\operatorname{sgn}(\sigma)r^{c(\sigma)}
\\ & = 
\frac{\partial }{\partial z_n}\sum_{\color{red}{\sigma\in S_n, \sigma(n) = n}}
\prod_{i=1}^n(Z-A)_{i\sigma(i)}\operatorname{sgn}(\sigma)r^{c(\sigma)}
\\ & = 
\frac{\partial }{\partial z_n}\sum_{\sigma\in S_n, \sigma(n) = n}
(a_{nn} - z_n)r\cdot \prod_{i=1}^{n-1}(Z-A)_{i\sigma(i)}\operatorname{sgn}(\sigma)r^{c(\sigma)-1}
\\ & = 
\frac{\partial }{\partial z_n}\sum_{\color{red}{\tau\in S_{n-1}}}
(a_{nn} - z_n)r\cdot \prod_{i=1}^{n-1}(Z-A)_{i\tau(i)}\operatorname{sgn}(\tau)r^{c(\tau)}
\\ & = 
\frac{\partial }{\partial z_n}\left[(z_n - a_{nn})r\sum_{\tau\in S_{n-1}}
\prod_{i=1}^{n-1}(Z-A)_{i\tau(i)}\operatorname{sgn}(\tau)r^{c(\tau)}\right]
\\ & = 
\frac{\partial }{\partial z_n}[r(z_n - a_{nn}){\det}_r(Z_n - A_n)] = r\,{\det}_r(Z_n - A_n).
\end{align}
