Derivative of a Rotation Matrix w.r.t. an angle I am struggling with something that is probably really easy.
I have the canonical rotation matrix w.r.t. only to a rotation around the z axis.
I can't write the fornula because I am writing from my smartphone but the matrix is this one:

I want to know how does it work if I want to make the derivative of the rotation matrix in the figure w.r.t. the angle $\alpha$. Is it just the derivative conponent by component ?
Thanks for your help.
 A: Yes. You just take derivatives componentwise.
More can be said. If your group of matrices describes rotations about the axis $\vec{n}\in\Bbb{R}^3$ in the right handed direction, then the derivative evaluated at $\alpha=0$ will be the matrix of the linear transformation corresponding to cross product with $\vec{n}$, i.e. the transformation $\vec{x}\mapsto \vec{n}\times\vec{x}$.
The length of the vector $\vec{n}$ comes into it as follows. If $\vec{n}$ is a unit vector, then $\alpha$ should be exactly the angle of rotation. Otherwise we scale the angle using $||\vec{n}||$ as speed of rotation.
For example differentiating your matrix w.r.t. $\alpha$ and then setting $\alpha=0$ gives the matrix
$$
\left(\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&0\end{array}\right),
$$
which is the matrix of the linear transformation $\vec{x}\mapsto \mathbf{k}\times \vec{x}$. Fittingly, as the rotation is about the axis $\mathbf{k}$.
A: You can See $R_z(a)$ as vector function in $\mathbb{R}^9$, therefore, yes, the derivative with respect to $\alpha$ is, as you say, the derivative component by component. If you know how to differentiate curves, then I think it helps to notice that $R_z$ can be seen as the unit circle $e^{i\theta}$.
