TITLE: Product rule for matrix-valued functions and Differentiability of matrix multiplication
QUESTION [5 upvotes]: Let $F, G: \mathbb{R}^{n \text{ x } n} \to \mathbb{R}^{n \text{ x } n}$ be two differentiable functions. Defining $(F \cdot\ G)(A):=F(A)G(A)$ using the usual matrix multiplication, prove that $F\cdot\ G$ is differentiable and express $(F\cdot\ G)'$ in terms of $F$, $F'$, $G$, $G'$.

What I thought of doing was to write $A=[a_{ij}]_{i, j \in {1, ..., n}}$ and express $F$ as a matrix $[f_{ij}]_{i, j \in {1, ..., n}}$ of differentiable functions $f_{ij}:\mathbb{R} \to \mathbb{R}$ (same idea for $G$), and I imagine $(F\cdot\ G)'$ will look something like $F'\cdot\ G+F\cdot\ G'$, but I don't know how to formalize it.
[obs.: the matrix norm considered here is $||A||=\sup_{|v|=1}|A(v)|$, where $v \in \mathbb{R^n}$ and $|\cdot\ |$ is the usual euclidean norm]

REPLY [2 votes]: $E = \mathbb{R}^{n \times n}$ is a Banach space (which, with the given norm, is isomorphic to the space of continuous linear maps $\mathscr{L}(\mathbb{R}^n; \mathbb{R}^n)$).
The usual matrix multiplication $\mu \colon E \times E \to E$, $(B, C) \mapsto BC$ is bilinear; and it is continuous, because $\|BC\| \leqslant \|B\|\|C\|$ for all $B, C \in E$. Continuity follows from e.g. J. Dieudonne, Foundations of Modern Analysis (1969), proposition (5.5.1).
By e.g. Dieudonne's proposition (8.1.4), $\mu$ is differentiable at every point $(B, C)$ in $E \times E$, and its derivative at $(B, C)$ is the continuous linear mapping
\begin{equation}
\tag{1}\label{eq:mu}
\mu'(B, C) \colon E \times E \to E, \ (U, V) \mapsto BV + UC.
\end{equation}
By e.g. Dieudonne's proposition (8.1.5), the function
$$
H \colon E \to E \times E, \ A \mapsto (F(A), G(A))
$$
is differentiable at every point $A \in E$, and - under the natural linear isometry between $\mathscr{L}(E; E \times E)$ and $\mathscr{L}(E; E) \times \mathscr{L}(E; E)$, where both products have the $\sup$ norm - the derivative at $A$ is identified with the continuous linear mapping
$$
H'(A) = (F'(A), G'(A)).
$$
(I had better now spell that out in more detail - especially as I haven't properly studied differential calculus in Banach spaces! It took some brass nerve to post this answer. Partly it's in order to learn some of this stuff myself, but it is also to show that an answer requires virtually no calculation.)
The natural linear isometry in question sends $L : E \to E \times E$ to $(\operatorname{pr}_1 \circ L, \operatorname{pr}_2 \circ L)$, where $\operatorname{pr}_i$ ($i = 1, 2$) are the projection mappings $E \times E \to E$. So we have
\begin{equation}
\tag{2}\label{eq:H}
H'(A) \colon E \to E \times E, \ T \mapsto (F'(A)(T), G'(A)(T)).
\end{equation}
By the Chain Rule (which is Dieudonne's proposition (8.1.2)),
\begin{align*}
(F \cdot G)'(A) & = (\mu \circ H)'(A) \\ & = \mu'(H(A)) \circ H'(A) \\ & = \mu'(F(A), G(A)) \circ H'(A).
\end{align*}
Therefore, by \eqref{eq:mu} and \eqref{eq:H},
\begin{align*}
(F \cdot G)'(A)(T) & = \mu'(F(A), G(A))(H'(A)(T)) \\ & = \mu'(F(A), G(A))(F'(A)(T), G'(A)(T)) \\ & = \boxed{F(A)G'(A)(T) + F'(A)(T)G(A)}
\end{align*}
- which agrees with rmdmc89's answer.
By the same argument, the result holds generally for differentiable functions $F: W \to \mathbb{R}^{n \times p}$, $G: W \to \mathbb{R}^{m \times n}$, where $W$ is an open subset of a Banach space and $m, n, p$ are positive integers.