If $M$, $N$ are smooth manifolds, $p \in M$ and $X_p \in T_p M$ and $F$ a smooth map from $M$ to $N$ the definition of the differential of $F$ at $p$ is given by the equality
$$ F_{*,p}(X_p) f = X_p(f \circ F) , \;\; \forall f\in C^{\infty}(N) $$
It's easy to show that $F_{*,p}$ is a linear map.
I was trying to figure what the differential is in the following example... Let $R^z(\phi)$ defined as
$$ R^z(\phi) = \begin{pmatrix} \cos\phi & -\sin(\phi) & 0 \\ \sin\phi & \cos(\phi) & 0 \\ 0 & 0 & 1 \end{pmatrix} $$
We can regard this rotation matrix as a map $R^z(\cdot) : [0,2\pi] \to SO(3)$.
My question is... what's the differential in this case, in local coordinates at least, I was trying to use the definition
$$ R^z_{*,\phi_0}(X_{\phi_0})f = X_{\phi_0}(f \circ R^z) $$
but I got stuck, and the reason is because I'm dealing with a matrix, I think what I should end up with is the componentwise derivative of the matrix, but I'd like to show that.
Also I'd like to generalize this to general element of $SO(3)$, using euler angles representation.
My background is Tu's Introduction to manifolds.
Can you help?
You can see, $SO(3)$ has a subset of the vector space of $3\times 3$ matrices. The differential of $R(\phi)$ is $\pmatrix{-sin(\phi)&-cos(\phi)&0\cr cos(\phi)&-sin(\phi)&0\cr 0&0&0}$