Derivative of a rotated vector with respect to the quaternion Let us say we have a right-handed unit quaternion, describing the rotation from frame $a$ to frame $b$: $q_a^b$.  The rotation matrix formed from this quaternion is $R\left( q_a^b \right)$ and describes a passive rotation.  That is, $R\left( q_a^b \right)v$ describes the same object $v$ in the new frame $b$.
The following expression is given in Michael Andre Bloesh's dissertation without explanation link - (unfortunately embargoed until April 2018)
$$\frac{d}{dq_a^b} R\left( q_a^b \right)v = -\left( R\left( q_a^b \right)v  \right)^\times $$
where the $\left( \cdot \right)^\times $ notation is the skew-symmetric matrix.
I played with these expressions numerically to confirm the above and 
 also discovered that the derivative of the active rotation is
$$\frac{d}{dq_a^b} R\left( q_a^b \right)^\top v = R \left( q_a^b \right)^\top \left( v \right)^\times $$
which I guess makes some intuitive sense as well.
While these expressions seem to work, how do I approach this problem in a principled way (i.e. not guessing and checking with numerical differentiation)?

Did you also confirm the second one numerically? I think it is wrong. Here is the right one.
$$\frac{d}{dq_b^a}R(q_b^a)^⊤v=R(q_b^a)^⊤(v)_×$$
