Here is the equation I want to solve:
$$\frac{d\vec{A}}{d\vec{C}}=\frac{d\mathbf{M}(\vec{C})\vec{B}}{d\vec{C}}$$
where
$\vec{A} = \mathbf{M}(\vec{C})\vec{B}$
$\vec{B}$ is a constant vector
$\vec{C}$ is a vector
$\mathbf{M}(\vec{c}) = e^{[\vec{c}]_\times}$
$[\vec{c}]_\times = \begin{bmatrix} 0&-c_z&c_y\\\ c_z&0&-c_x\\\ -c_y&c_x&0 \end{bmatrix}$ is to transform a vector into a skew matrix
I have asked a similar question to differentiate a scalar related to a matrix w.r.t. vector.
I try to use Frobenius product notation to solve the problem. Like this: (I'm not sure if it's correct or not)
$$d\vec{A}=d\mathbf{M}\vec{B}$$ $$\ \quad\quad=d\mathbf{M}：\vec{B}$$ then I stuck
I searched some materials online. But I can't find some handy information may be due to the incorrect keywords.
Please help me to figure out this question.
It would be great if you could provide me with the correct keywords or relevant information for a systematic study.
I appreciate your answer and suggestion.
$ \def\LR#1{\left(#1\right)} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\B{\m{b}_\times} \def\C{\m{c}_\times} $For consistency, use lowercase letters for vectors and uppercase for matrices $$M=\exp\LR{\:\C}$$ Then an amazing paper by Gallego & Yezzi directly addresses your question.
It's challenging to follow their derivation, but here is what they call $\,\sf Result\:1$ $$\eqalign{ \grad{\LR{Mb}}{c} &= -M\,\B Y \\ }$$ I call $Y$ the Yezzi Matrix$.\,$ Like $M$, it is also a function of the vector $c$ $$\eqalign{ Y &= \frac{cc^T+\LR{M^T-I}\C}{c^Tc} \\ }$$