Consider the decomposition of any rotation matrix $R\in SO(3)$ as $$ R=\exp(\theta_3 \hat{e}_3)\exp(\theta_1\hat e_1 + \theta_2 \hat e_2)$$ where the hat designates the hat operator, and the application $\phi$ that "extracts" the angles $(\theta_1,\theta_2,\theta_3)$: \begin{align} \phi: SO(3)&\longrightarrow \mathbb{R}^3 \\ R &\longmapsto (\theta_1,\theta_2,\theta_3) \end{align}
Is it possible to derive a closed-form expression for $d\phi$?
What I have done: I computed $d(\phi^{-1})$: $$\phi^{-1}=\mathrm{comp}\circ \alpha$$ where $\mathrm{comp}$ is the product in $SO(3)$ and $$\alpha:(\theta_1,\theta_2,\theta_3)\longmapsto (\exp(\theta_3 \hat e_3),\exp(\theta_1\hat e_1 + \theta_2 \hat e_2))$$
I find, with $\Theta = (\theta_1,\theta_2,\theta_3)$: \begin{align}d(\phi^{-1})(\Theta)(u)& = d(\mathrm{comp}(\alpha(\Theta))(d\alpha(\Theta)(u)) \\ & = \mathrm{Ad}\big(\mathrm{inv}\circ\exp(\theta_1\hat e_1 + \theta_2 \hat e_2)\big)(d\exp(\theta_3\hat e_3)(u)) + d\exp(\theta_1\hat e_1 + \theta_2 \hat e_2)(u) \end{align}
and then I thought of using the identity: $$d\phi(R)= \big( d(\phi^{-1})(\phi(R))\big)^{-1}$$
This provides a way of computing $d\phi(R)$ numerically. Any better idea?