as well as a few fundamental properties of determinants.

[*3.3. Differentiation rules and related formulas*]{}

In terms of the notation (3.4), the tau functions $f$ and $g$ can be written as $$f=|D|, \eqno(3.8a)$$ $$\quad g=|D|+{1\over\rho^2}|D({\bf z}_t^*;{\bf z})|. \eqno(3.8b)$$ The differentiation rules of the tau functions with respect to $t$ and $x$ are given by the following formulas:

[**Lemma 3.1.**]{}

$$f_t={\rm i}|D({\bf z}_t^*;{\bf z})| - {1\over \rho^2}|D({\bf z}_t^*;{\bf z}_t)|, \eqno(3.9)$$ $$f_x=-\kappa |D({\bf z}^*;{\bf z})|+{\rm i}|D({\bf z}^*;{\bf z}_x)|, \eqno(3.10)$$ $$f_{xt}={\rm i}\kappa\rho^2|D({\bf z}^*;{\bf z})|-\kappa|D({\bf z}_t^*;{\bf z})|-\kappa|D({\bf z}^*;{\bf z}_t)|
         +{\rm i}|D({\bf z}_t^*;{\bf z}_x)|$$ $$-|D({\bf z}^*,{\bf z}_t^*;{\bf z}_x,{\bf z})|+{\kappa\over \rho^2}|D({\bf z}^*,{\bf z}_t^*;{\bf z},{\bf z}_t)|
-{\rm i\over \rho^2}|D({\bf z}^*,{\bf z}_t^*;{\bf z}_x,{\bf z}_t)|, \eqno(3.11)$$ $$g_t={\rm i}|D({\bf z}_t^*;{\bf z})|+{1\over \rho^2}|D({\bf z}_{tt}^*;{\bf z})|, \eqno(3.12)$$ $$g_x={\rm i}|D({\bf z}_t^*;{\bf z})|+{1\over \rho^2}|D({\bf z}_{t}^*;{\bf z}_x)|
+{\rm i\over \rho^2}|D({\bf z}_t^*,{\bf z}^*;{\bf z},{\bf z}_x)|. \eqno(3.13)$$ [**Proof.**]{} We prove (3.9). Applying formula (3.5) to $f$ given by (3.1) with (3.2a), one obtains $$\begin{aligned}
f_t &=\kappa\rho^2\sum_{j,k=1}^ND_{jk}{\kappa-{\rm i}p_j\over p_jp_k^*}z_jz_k^* \notag \\
    &=-{\rm i}\sum_{j,k=1}^ND_{jk}z_{j}z_{k,t}^* +{1\over \rho^2}\sum_{j,k=1}^ND_{jk}z_{j,t}z_{k,t}^*, \notag\end{aligned}$$ where in passing to the second line, use has been made of the relation $z_{j,t}=(\kappa\rho^2/p_j)z_j$. Referring to formula (3.6) with $z=0$ and taking into account the notation (3.4), the above expression is equal to the right-hand side of (3.9). Formulas (3.10)-(3.13) can 