Let $\theta$ and $\psi$ be two vectors in $\mathbb{R}^3$. I want to compute $$\nabla_{\psi} \log \left( e^{[\theta]_\times}e^{[\psi]_\times}\right)$$
Where $[v]_\times$ is the skew-symmetric cross-product matrix $[v]_\times w = v\times w.$
If rotations commuted I would just get $\nabla_\psi (\theta+\psi) = I$, but, of course, they don't. Is there still a nice formula for the derivative?
Using the formula for the derivative of the exponential map in this paper and the chain rule, I get that
$$\nabla_\psi \log\left(e^{[\theta]_\times}e^{[\psi]_\times}\right) = T\left(\log\left(e^{[\theta]_\times}e^{[\psi]_\times}\right)\right)^{-1}T(\psi),$$ where the matrix $T(v)$ is given by $$T(v) = \begin{cases}I, & v = 0\\ \frac{vv^T+\left(e^{-[v]_\times}-I\right)[v]_\times}{\|v\|^2}, & v \neq 0\end{cases}$$ which is tractable for computation, if unpleasant.