Issues with the chain rule in derivatives in chiral perturbation theory I realize that this is a purely math question, which however has arisen in a physics computation. The reason to post it here is that I want a fast dirty answer. Not something cluttered with definitions. If the mods think that it isn't fit for the site feel free to remove it.
I am computing some derivatives of the matrix $U$ in chiral perturbation theory
$$U=e^{it^a\phi_a}$$
where $t^a$ are the Pauli matrices and $\phi_a$ are complex scalar fields. I want to take a derivative of this, for example
$$\partial_{\mu}U=\partial_{\mu}e^{it^a\phi_a}$$
since the exponent contains a matrix a am uncertain when applying the chain rule. Is the proper way to take the derivative
$$e^{it^a\phi_a}\partial_{\mu}(it^b\phi_b)$$
or
$$\partial_{\mu}(it^b\phi_b)e^{it^a\phi_a}$$
 A: I think the answer is neither one sadly. If you do a taylor expansion you find
\begin{equation}
\partial_\mu e^{i t^a \phi_a} = \sum_n \left(\frac{i^n}{n!}\right) \partial_\mu (\phi^a t_a)^n = i \partial_\mu (\phi^a t_a) - \frac{1}{2}\left( \partial_\mu (\phi^a t_a) \phi^b t_b + \phi^a t_a \partial_\mu (\phi^b t_b) \right) + \cdots
\end{equation}
The second term on the right hand side already prevents us from factoring the answer into anything nice, since $\partial \phi $ does not commute with $\phi $.  At higher order you can get even uglier things like $\phi^n \partial \phi \phi^m$.
If we work in $0+1$ dimensions, then what you are basically asking is for the derivative of $e^{X(t)}$ with respect to $t$, where $X$ is a matrix. There is a (terrible) formula for this on wikipedia (https://en.wikipedia.org/wiki/Matrix_exponential)
\begin{equation}
\frac{d}{dt} e^{X(t)} = \int_0^1 d\alpha e^{\alpha X(t)}\frac{dX}{dt}e^{(1-\alpha)X(t)}
\end{equation}
I suspect this is rarely useful in practice (or, at least in chiral perturbation theory).
I'm assuming this comes up when expanding out the effective action in terms of $\phi$. I think probably the only thing you can really do is work perturbatively and keep track of all the different vertices you get. Physically I think what's going on is basically "identical particles are identical," so if you have a vertex like $\partial \phi \partial \phi \phi \phi$ you must also have a corresponding vertex where you exchange some of the $\phi$'s, like $\partial \phi \phi \partial \phi \phi$. 
If you are working with a lagrangian there is presumably a trace, in which case you can use the cyclic property of the trace to simplify things somewhat, for example from ${\rm tr}(\partial_\mu U \partial^\mu U^\dagger)$ you end up with vertices like
\begin{equation}
{\rm tr}(\partial U \partial U^\dagger) \supset {\rm tr}\left( t_a t_b t_c t_d\right)\left( \partial_\mu \phi^a \phi^b + \phi^a \partial_\mu \phi^b \right) \left(\partial^\mu \phi^c \phi^d + \phi^c \partial^\mu \phi^d \right) = 2 {\rm tr}\left(t_a t_b t_c t_d\right) \left(\partial_\mu \phi^a \phi^b \partial^\mu \phi^c \phi^d + \partial_\mu \phi^a \partial^\mu \phi^b \phi^c \phi^d\right)
\end{equation}
where in the last line we used the cyclic property of the trace.
