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)?
 A: I may have an answer, taking the derivation straight from "A Primer on the Differential Calculus of 3D Orientations" by Bloesch et al. (Appendix I: Section 3: Derivative of a Coordinate Map)
First, let $\Phi_{BA} \in SO(3)$ be a relative orientation of a coordinate system $B$ w.r.t. a coordinate system $A$. In the paper, they defined a mapping $\boldsymbol{C}: SO(3) \rightarrow \mathbb{R}^{3 \times 3}$ such that $\Phi(\mathbf{r}) \triangleq \boldsymbol{C}(\Phi) \mathbf{r}$ which means $\Phi$ can be a quaternion or euler angle (if I'm not mistaken). $\boldsymbol{e}_{i} \in \mathbb{R}^{3}$ be the standard basis vectors in $\mathbb{R}^{3}$, $\epsilon$ be a small scalar pertubation, and finally, 
$$
\begin{align}
  \boxplus : SO(3) \times \mathbb{R}^{3} \rightarrow SO(3), \\
  \Phi, \boldsymbol{\varphi} \mapsto \exp(\boldsymbol{\varphi}) \circ \Phi
\end{align}
$$
be the box-plus operator that forms the addition operator between $SO(3)$ and $\mathbb{R}^{3}$
Copying from the appendix to here, the map of an orientation applied to a coordinate tuple can be differentiated w.r.t. the orientation itself.
$$
\begin{align}
  \begin{bmatrix}
    \dfrac{\partial}{\partial \Phi} \Phi(\boldsymbol{r})
  \end{bmatrix}_{i}
  &= 
  \lim_{\epsilon \rightarrow 0}
  \dfrac{
    (\Phi \boxplus \boldsymbol{e}_{i} \epsilon)(\boldsymbol{r})
    - \Phi
  }
  {
    \epsilon
  } \\
  &=
  \lim_{\epsilon \rightarrow 0}
  \dfrac{
    \boldsymbol{C}(\boldsymbol{e}_{i} \epsilon)
    \boldsymbol{C}(\Phi)(\boldsymbol{r})
    - \boldsymbol{C}(\Phi)(\boldsymbol{r})
  }
  {
    \epsilon
  } \\
  &=
  \lim_{\epsilon \rightarrow 0}
  \dfrac{
    (\boldsymbol{I} + \boldsymbol{e}_{i}^{\times} \epsilon)
    \boldsymbol{C}(\Phi)(\boldsymbol{r})
    - \boldsymbol{C}(\Phi)(\boldsymbol{r})
  }
  {
    \epsilon
  } \\
  &=
  \lim_{\epsilon \rightarrow 0}
  \dfrac{
    \boldsymbol{e}_{i}^{\times} \epsilon
    \boldsymbol{C}(\Phi)(\boldsymbol{r})
  }
  {
    \epsilon
  } \\
  \dfrac{\partial}{\partial \Phi} \Phi(\boldsymbol{r})
  &= 
  -(\boldsymbol{C}(\Phi) \boldsymbol{r})^{\times}
\end{align}
$$
Highly encouraged to look at the paper to see the identities used for this derivation.
Please let me know if I got this wrong, I'm very new to differential geometry.
A: Did you also confirm the second one numerically? I think it is wrong. Here is the right one.
$$\frac{d}{dq_b^a}R(q_b^a)^⊤v=R(q_b^a)^⊤(v)_×$$
