be a *derivation*, that means it fulfills a *Leibniz rule* $$\begin{aligned}
 &\text{with $f,g\in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$}& ({\mathcal{A}}^\alpha_\beta (f \cdot g))(x) &= f(x) ({\mathcal{A}}^\alpha_\beta g)(x) + g(x) ({\mathcal{A}}^\alpha_\beta f)(x) \\ &\text{and a \textit{chain rule}, }& ({\mathcal{A}}^\alpha_\beta (f \circ g))(x) &= f'(g(x)) ({\mathcal{A}}^\alpha_\beta g)(x) + \left(\frac{x}{g(x)} \right)^{\beta} e^{\frac{g(x) - x}{\alpha x g(x)}} ({\mathcal{A}}^\alpha_\beta f ) (g(x)), \end{aligned}$$ where $(f \cdot g)(x) = f(x) g(x)$ and $(f \circ g)(x) = f(g(x))$. These statements will be derived from elementary properties of the Gamma function. Note that the chain rule involves a peculiar correction term if $f$ has a non-trivial asymptotic expansion. The fact that the chain rule cannot be simple, that means for general $f,g$: $({\mathcal{A}}^\alpha_\beta (f \circ g))(x) \neq f'(g(x)) ({\mathcal{A}}^\alpha_\beta g)(x)$, is obvious. The reasonable requirement that the function $g(x)=x$ has a trivial asymptotic expansion, $({\mathcal{A}}^\alpha_\beta g)(x) = 0$, would otherwise imply that all $f\in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ have trivial asymptotic expansions. The formalism can be applied to calculate the asymptotic expansions of implicitly defined power series. This procedure is similar to the extraction of the derivative of an implicitly defined function using the implicit function theorem. In sections \[sec:prereq\]-\[sec:comp\] the derivation ring ${{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ will be described and the main Theorem \[thm:chainrule\], which establishes the chain 