The textbook(Ordinary Differential Equations and Dynamical Systems, Gerald Teschl) says:
For the differential equation
$\dot x$ = $f(x)$, $\;x(0) = x_0$ where $f \in C^k(M, R^n)$ where $M$ is open subset of $R^n$,
Let $\phi(t, x_0)$ be solution of the equation.
Then, linearized equation of $\phi(t,x_0)$ is $\dot \phi(t,x_0) = A(t)\,\phi(t,x_0)$ where $A(t) = df_{\phi(t,x_0)}$.
However, I can't understand the notation of $df_{\phi(t,x_0)}$ and confused when calculating example.
For example, when $\phi(t, (1,0)) = (\cos(t), \sin(t))$ and $\dot x_1 = -x_2 + x_1(1-x_1^2-x_2^2),\; \dot x_2 = x_1 + x_2(1-x_1^2-x_2^2)$
Then, $f(\phi(t,(1,0)) = (-\sin(t), \cos(t))$.
Textbook says, in this case, $A(t) = \begin{bmatrix} -2\cos^2(t) &-1+\sin^2(2t) \\ 1-\sin(2t) & \cos(t)\end{bmatrix}$
How do I derive this equation? I thought $df_{\phi(t,x_0)}$ means the gradient of $f(\phi(t,x_0))$, but in this case, $\nabla f(\phi(t,x_0)) = \begin{bmatrix} -\cos(t)\frac{\partial t}{x_1} &-\cos(t)\frac{\partial t}{x_2} \\ -\sin(t)\frac{\partial t}{x_1} & -\sin(t)\frac{\partial t}{x_2} \end{bmatrix} = \begin{bmatrix} \cot(t) &-1 \\ 1 & -\tan(t)\end{bmatrix}$
Is it, $df_{\phi(t,x_0)} \neq \nabla f(\phi(t,x_0))$? or I calculated wrong?
I have very poor understanding about general topic about vector calculus and having trouble with this. Help me please...