Suppose you have a rank 4 Tensor $T$ whose coordinates are, in some basis, $T_{ijkl}$. Say the indices take values in some representation of $T$. Now suppose I construct a matrix by dotting this tensor with 2 vectors $u $ and $v$ to get $T(u, v)$ with coordinates $\sum_{k,l} T_{ijkl}u^k v^l$ and I want to study the exponential matrix $M= e^{T(u, v)}$ defined through it's power series (Let's assume everything is convergent).
My question is: what can be said about the gradient with respect to one of the two vectors? Say with resp. to $\vec{v}$
$$\nabla_{\vec{v}}\,\, e^{T(u, v)}=?$$
To be clear this is a rank 3 tensor, in some basis the above would be
$$ \frac{\partial}{\partial v_i} (e^{T(u,v)})_{mn}. $$
I guess my question is if and when some sort of identity like this one holds
\begin{equation}\label{1} \frac{\partial}{\partial v_i} (e^{T(u,v)})_{mn}=\sum_l (e^{T(u,v)})_{ml} ({T(u)})_{lni}, \end{equation}
Which would be sort of reminiscent of $\partial_t e^{At}=A e^{At}$, for some parameter $t$ and matrix $A$. My problem here is that there seem to be ordering issues when I power expand $e^T$. That is at order $n$ in the expansion you differentiate the vector $v$ in a string of $n$ matrices timed into each other, and you cannot just "move" $T(u)$ in front or after the string to get my guessed formula. So the result doesn't seem to exponentiate like the easy case.
But is there still a nice relation or a clever way to rearrange the RHS of $\frac{\partial}{\partial v_i} (e^{T(u,v)})_{mn}$? What if $T$ has some additional property, like it's (anti)symmetric.
Thanks!