If I have two $3\times3$ rotation matrices $R_1$ and $R_2$, what is the matrix derivatives of $\frac{\partial(R_1 R_2^T)}{\partial R_1}$?
Also, I have found a website which can answer my question
http://www.matrixcalculus.org/ The following is the answer:
I am not sure whether this is the correct answer? Also,I don't understand the meaning of the notation ∏. Can someone explain it for me？Thanks
$\def\E{{\cal E}}\def\pcolor#1#2#3{\frac{\partial #1}{\color{#3}{\partial #2}}}$ Consider the fourth-order tensor $\E$ whose components (in terms of Kronecker deltas) are $$\E_{ijk\ell} = \delta_{ik} \delta_{j\ell}$$ This tensor is useful for rearranging matrix products, e.g. $$ABC = A\E C^T:B$$ where a colon denotes the double-contraction product $$\eqalign{ {\cal C} &= {\cal A}:{\cal B} \\ {\cal C}_{ijpq} &= \sum_{k=1}^m\sum_{\ell=1}^n {\cal A}_{ijk\ell}\,{\cal B}_{k\ell pq} \\ }$$ Then consider the following matrix-valued function and its gradient $$\eqalign{ F &= F(X) &\doteq XA \\ dF &= dX\,A &= \E A^T\color{red}{:dX} \\ \pcolor{F}{X}{red} &=\E A^T&\qquad\big({\rm tensor\,gradient}\big) \\ }$$ Another way to tackle the problem is to use vectorization, which is what the website has implicitly done $$\eqalign{ {\rm vec}(dF) &= (A^T\otimes I)\,{\rm vec}(dX) \\ \pcolor{\big({\rm vec}(F)\big)}{\big({\rm vec}(X)\big)}{black} &= A^T\otimes I \;\qquad\big({\rm matrix\,gradient}\big) \\ }$$ where $\otimes$ denotes the Kronecker product and $I$ is the $3\times 3$ identity matrix.
In either case, setting $$X=R_1 \qquad A=R_2^T$$ yields the desired gradient.
However, neither solution takes into account the orthogonality constraint $$\eqalign{ X^TX = I \quad\implies\quad dX &= -X\;dX^TX \\ }$$