Suppose we have $\frac{dx}{dt} = \dot{x} = f(x)$ with equilibrium point $x_e$ such that $f(x_e) = 0$. Then for the linearized approximation of the differential equation near $x_e$ we hope to use the solution of $\dot{\delta x} = Df(x_e)\delta x$ where $\delta x = x(t) - x_e$.
I'm having trouble figuring out how to set up $Df(x_e)$. Consider a concrete case of the pendulum equation $ml^2\ddot{\theta} = -lmg\sin\theta$ which can be rewritten using $$ x = \left[ \begin{array}{c} \theta \\ \dot{\theta} \end{array} \right] $$ as $$ \dot{x} = \left[ \begin{array}{c} x_2 \\ -(g/l)\sin x_1 \end{array} \right] $$ with equilibrium point $x=0$. In this case, I know that $Df(x_e)$ should be $$ Df(0) = \left[ \begin{array}{c c} 0 & 1 \\ -g/l & 0 \end{array} \right]$$
but I can't quite figure out the various variables that make up $Df$. I think I am having trouble figuring out all the indirect notation, and seeing this example worked out would really help.
My confusion stemmed from mixing up $x$ and $\theta$ as well as not remembering to take $Df(x)$ as $x = x_e = 0$.
Worked out:
From the expression for $\dot{x}$ it is clear that $$ f([x_1 \ x_2]^T) = \left[ \begin{array}{c} x_2 \\ -\frac{g}{\ell}\sin x_1 \end{array} \right] $$ so
$$\begin{align*} Df(x_e) &= \left[ \begin{array}{cc} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{array} \right] \\ &= \left[ \begin{array}{cc} 0 & 1 \\ -\frac{g}{\ell}\cos x_1 & 0 \end{array} \right] \\ &= \left[ \begin{array}{cc} 0 & 1 \\ -\frac{g}{\ell} & 0 \end{array} \right] \quad \textrm{substituting in the value for $x_{e}$} \end{align*} $$
as desired.