I want to linearize a system to this form $$\begin{bmatrix} \Delta\dot{x}_1\\ \Delta\dot{x}_2\\ \Delta\dot{x}_3 \end{bmatrix} = A\begin{bmatrix}\Delta x_1\\ \Delta x_2\\ \Delta x_3\end{bmatrix}+B\begin{bmatrix} \Delta u\end{bmatrix}$$ $$\begin{bmatrix} \Delta y \end{bmatrix} = C\begin{bmatrix}\Delta x_1\\ \Delta x_2 \\ \Delta x_3\end{bmatrix}+D\begin{bmatrix} \Delta u\end{bmatrix}$$
The A matrix
$$A = \begin{bmatrix} \frac{\partial \dot x_1}{\partial x_1} & \frac{\partial \dot x_1}{\partial x_2} & \frac{\partial \dot x_1}{\partial x_3}\\[6pt] \frac{\partial\dot x_2}{\partial x_1} & \frac{\partial \dot x_2}{\partial x_2} & \frac{\partial \dot x_2}{\partial x_3}\\[6pt] \frac{\partial\dot x_3}{\partial x_1} & \frac{\partial \dot x_3}{\partial x_2} & \frac{\partial \dot x_3}{\partial x_3}\end{bmatrix}_{|P}$$
$$=\begin{bmatrix} M_1L & \frac{d}{dt} \left[\frac{R_2}{M^2}\ln(x_2(t)+1)\right] & R_1\\[6pt] L_1L_2 & \frac{d}{dt} \left[\frac{L_1}{M^2}\ln(x_2(t)+1)\right]& R_2\\[6pt] 0 &\frac1C & 0\end{bmatrix}_{|P}$$
has in its second comulmn logarithm. Pay attention only there. ($M$, $L$'s and $R$'s are some constants) If I derivate it, I get
$$=\begin{bmatrix} M_1L & \frac{R_2 \frac{dx_2(t)}{dt}}{M^2(x_2(t)+1)} & R_1\\[6pt] L_1L_2 & \frac{L_1 \frac{dx_2(t)}{dt}}{M^2(x_2(t)+1)}& R_2\\[6pt] 0 &\frac1C & 0\end{bmatrix}_{|P}$$
Am I right?
And I'm not sure what to do with this. Since my initial condition $P$ are in equilibrium and $\frac{dx_2(t)}{dt} = 0$, should I put there the $0$? The result would look like
$$=\begin{bmatrix} M_1L & 0 & R_1\\[6pt] L_1L_2 & 0 & R_2\\[6pt] 0 &\frac1C & 0\end{bmatrix}_{|P}$$
But few weeks ago I have been solving similar situation and I didn't put $\frac{dx_2(t)}{dt}$ in the matrix at all. Lecturer told me it was right. According to that, this matrix should look like
$$=\begin{bmatrix} M_1L & \frac{R_2}{M^2(x_2(t)+1)} & R_1\\[6pt] L_1L_2 & \frac{L_1 }{M^2(x_2(t)+1)}& R_2\\[6pt] 0 &\frac1C & 0\end{bmatrix}_{|P}$$
But I'm confused. The mode $x_i(t)$ isn't consider as a function of time?
How to solve this?
