If we let $\psi:\text{GL}(n,\Bbb R)\times\text{GL}(n,\Bbb R)\rightarrow \text{GL}(n,\Bbb R)$, by $\psi:(A,B)\mapsto AB$, then $\psi$ can be seen as a differentiable map, letting $(A)_{ij}={a}_{ij}$ and $(B)_{ij}={b}_{ij}$, the components of $\psi(A,B)$ are:
$\psi(A,B)_{ij}=\sum\limits_{k=1}^n a_{ik}\;b_{kj}$.
which is a degree two polynomial in $2n$ variables and so is smooth.
I am just wondering if the following is correct for $d_{id,id} \psi\;$, the derivative at the identity in each entry.
We look at $\psi(id+A,id+B)_{ij}=\sum\limits_{k=1}^n (id+ A)_{ik}\;(id+B)_{kj}=\sum\limits_{k=1}^n (\delta_{ik}+ a_{ik})\;(\delta_{kj}+b_{kj})$
$=\sum\limits_{k=1}^n (\delta_{ik}\delta_{kj}+ a_{ik}\delta_{kj}+\delta_{ik}b_{kj}+a_{ik}b_{kj})=1+a_{ij}+b_{ij}+\sum\limits_{k=1}^n a_{ik}b_{kj}$.
To find the derivative we take the linear terms in $a$'s and $b$'s.
So $d_{id,id}\psi(A,B)=A+B$ ?
You're working too hard. You can make this computation much easier for yourself if you just systematically refuse to pick bases. At $(1, 1)$, matrix multiplication looks like
$$(1 + A)(1 + B) = 1 + A + B + AB$$
and after taking linear terms you get $A + B$. This argument works essentially without modification in any Banach algebra.