Let $A$ be an $n\times n$ matrix over the field $\mathbb{F}$, $V, W$ be vector spaces over $\mathbb{F}$, $f(B):V\to V, g(B):W\to W$ be automorphisms and $L:W\to V$ be a linear mapping. Denote by $\exp(A)$ the matrix exponential of $A$.
Then how would you apply the chain rule to find $\frac{d}{dt}f(\exp(tA))\circ L \circ g(\exp(tA))$ at $t = 0$? What confuses me is that 1.) the outer function depends directly on the parameter $t$ 2.) how to deal with the derivative w.r.t. the linear mapping $L$?
One consequence of the multivariate chain rule is that if you're differentiating with respect to $t$ and multiple $t$'s appear, then you obtain the correct answer by separately differentiating with respect to each instance and adding the results.
Define $\phi(t_1,t_2) = f(\exp(t_1 A))\circ L \circ g(\exp(t_2 A))$. We find that $$ \frac{\partial \phi}{\partial t_1} = df(\exp(t_1 A)) \circ A \exp(t_1 A)\circ L \circ g(\exp(t_2 A))\\ \frac{\partial \phi}{\partial t_2} = f(\exp(t_1 A)) \circ L \circ dg(\exp(t_2 A))\circ A\exp(t_2A) $$ Thus, the chain rule yields $$ \frac{d \phi(t,t)}{dt} = \frac{\partial \phi}{\partial t_1}(t,t) + \frac{\partial \phi}{\partial t_2}(t,t) = \\ df(\exp(t A)) \circ A \exp(t A)\circ L \circ g(\exp(t A)) \\ \quad + f(\exp(t A)) \circ L \circ dg(\exp(t A))\circ A\exp(tA). $$