Deriving rigid body motion from Euler-Lagrange equations I am trying to derive the equations of motion of the inertial tumbling of a rigid body. I represent the rigid body's orientation by time-varying axis-angle $\theta$ and write $R_{\theta}$ for the exponential map 
$$R_{\theta} = \exp([\theta]_{\times}) \in SO(3)$$ 
(so that the rotation mapping the time-zero body to the time-$t$ body is $R_{\theta(t)}$).
Then the angular velocity $\omega$ in the body's coordinates is given by the relationship
$$R_{\theta(t)} [\omega(t)]_{\times} = \frac{d}{dt}R_{\theta(t)};$$
if I expand the right-hand side using the derivative of the $\mathfrak{so}(3)$ exponential map described here: https://arxiv.org/pdf/1312.0788.pdf and solve for $\omega$, I get
$$\omega(t) = T(\theta)\dot\theta$$
where
$$T(\theta) = \frac{\theta\theta^T + (R_{-\theta}-I)[\theta]_{\times}}{\|\theta\|^2}.$$
Finally the kinetic energy of the spinning body is simply
$$L(\theta(t), \dot\theta(t)) = \frac{1}{2} \omega(t)^T M_I \omega(t)$$
where $M_I$ is the (constant) inertia tensor in the body's coordinates, and $\omega$ depends implicitly on $\theta$ and $\dot\theta$. The Euler-Lagrange Equations are
\begin{align*}
0 &= \frac{d}{dt}D_2L - D_1L \\
&= \frac{d}{dt}\left[\omega(t)^T M_I T(\theta(t))\right] - \omega(t)^TM_I D_1\omega\\
&= \dot\omega(t)^T M_I T(\theta) + \omega(t)^TM_I\left(\frac{d}{dt}T(\theta) - \frac{\partial T\dot\theta}{\partial \theta}\right),
\end{align*}
where $D_i$ is differentiation with respect to the $i$th parameter.
Now it is tempting to declare $\left(\frac{d}{dt}T(\theta) - \frac{\partial T\dot\theta}{\partial \theta}\right)$ zero by the chain rule, but this is subtly wrong: the first derivative is of the matrix $T(\theta)$ with respect to the scalar $t$, and the second is of the vector $T(\theta)\dot\theta$ with respect to the vector $\theta$, and equality requires symmetry of the rank-three tensor $dT$ that does not seem to hold here. Instead, some precession term is supposed to come out of this expression, but the derivatives seem extremely unpleasant and I'm not able to simplify them into a simple form. What are the next steps?
 A: For those who come later, I figured it out! The key is to use equality of mixed partial on the second derivative
$$d\left(\frac{d}{dt}R_{\theta}v\right)(\delta \theta)$$
where $v$ is an arbitrary test vector, and $d(\cdot)(\delta\theta)$ is the differential applied to the tangent vector $\delta \theta$.
On the one hand, we have
$$d\left(\frac{d}{dt}R_{\theta}v\right)(\delta \theta) = - R_{\theta}[v]_{\times}d\left(T(\theta)\dot\theta\right)(\delta \theta) + R_{\theta}\left[[v]_{\times}T(\theta)\dot\theta\right]_{\times}T(\theta)\delta\theta$$
and on the other,
$$\frac{d}{dt}\left(d\left(R_{\theta}v\right)(\delta\theta)\right) = - R_{\theta}[v]_{\times}\frac{d}{dt}\left(T(\theta)\right)\delta \theta + R_{\theta}\left[[v]_{\times}T(\theta)\delta\theta\right]_{\times}T(\theta)\dot\theta.$$
Equating these expressions and rearranging gives
\begin{align*}
[v]_{\times}\left(\frac{d}{dt}T(0)\delta\theta - d\left(T(\theta)\dot\theta\right)(\delta\theta)\right) &= \left[[v]_{\times}T(\theta)\delta\theta\right]_{\times}T(\theta)\dot\theta - \left[[v]_{\times}T(\theta)\dot\theta\right]_{\times}T(\theta)\delta\theta\\
&= \left[T(\theta)\delta\theta\right]_{\times}[v]_{\times}T(\theta)\dot\theta - \left[T(\theta)\dot\theta\right]_{\times}[v]_{\times}T(\theta)\delta\theta\\
&= -[v]_{\times}\left[T(\theta)\dot\theta\right]_{\times}T(\theta)\delta\theta\\
\left(\frac{d}{dt}T(0)\delta\theta - d\left(T(\theta)\dot\theta\right)(\delta\theta)\right) &= -\left[\omega\right]_\times T(\theta)\delta\theta,
\end{align*}
where the second-to-last equality is by the Jacobi identity. Then the Euler-Lagrange equations in the OP reduce to
$$\dot\omega^T M_I T(\theta) -\omega^TM_I [\omega]_{\times}T(\theta)=0$$
or
$$M_I\dot\omega + [\omega]_{\times}M_I\omega = 0,$$
in agreement with the usual Euler equations.
