problem on matrix derivatives 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
 A: I think (not sure about this one) that it is saying the Kronecker Product of $R2^T$ and a generalized Identity Matrix. I say this because the symbol after $\otimes$ is $\mathbb{I}$ (not $\Pi$), or $\mathbb{I}$
A: $\def\E{{\cal E}}\def\pcolor#1#2#3{\frac{\partial #1}{\color{#3}{\partial #2}}}$
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
$$\eqalign{
X^TX = I \quad\implies\quad dX &= -X\;dX^TX \\
}$$
