Definition of derived Hom when both inputs are chain complexes Let $\mathcal{A}$ be an abelian category and let $A \in obj(\mathcal{A})$, and let $D(\mathcal{A})$ denote the derived category of $\mathcal{A}$.
Given a left-exact functor $F: \mathcal{A} \to \mathcal{B}$, I understand how to define its derived functor $RF: D(\mathcal{A}) \to D(\mathcal{B})$.
For example,  $R \operatorname{Hom}_{\mathcal{A}}(R, \cdot): D(\mathcal{A}) \to D(\mathcal{A})$ is the derived functor of $\operatorname{Hom}_{\mathcal{A}}(A, \cdot): \mathcal{A} \to \operatorname{AbGrp}$ .
On the other hand, I have seen written $R \operatorname{Hom}_{\mathcal{A}}(A^{\bullet}, B^{\bullet})$, where $B^{\bullet}, A^{\bullet} \in D(\mathcal{A})$.  However, $\operatorname{Hom}_{\mathcal{A}}(A^{\bullet}, \cdot)$ is not a functor on $\mathcal{A}$ since $A^{\bullet}$ is not an object of $\mathcal{A}$.
How is $R \operatorname{Hom}_{\mathcal{A}}(A^{\bullet}, B^{\bullet})$ defined?
 A: $\operatorname{Hom}_{\mathcal{A}}(A^\bullet,-)$ can be regarded as a functor from $K(\mathcal{A})$ (the chain homotopy category of complexes) to $K(\mathcal{A})$: applying $\operatorname{Hom}_{\mathcal{A}}(-,-)$ to the terms of $A^\bullet$ and $B^\bullet$ gives a double complex, and $\operatorname{Hom}_{\mathcal{A}}(A^\bullet,B^\bullet)$ is the direct product version of the total complex of this double complex (see for example 1.2.6 in Weibel's book on homological algebra).
For a triangulated functor between chain homotopy categories, there is a general definition of a (right or left) derived functor, which will be a triangulated functor between derived categories (see for example 10.5.1 of Weibel's book for an abstract nonsense definition).
More concretely, if $B^\bullet$ is bounded below (for simplicity), and $\mathcal{A}$ has enough injectives, then this can be calculated in the following way, that generalizes the construction for $A^\bullet$ an object of $\mathcal{A}$:
Let $I^\bullet$ be an injective resolution of $B^\bullet$ (i.e., a bounded below complex of injectives with a quasi-isomorphism $B^\bullet\to I^\bullet$). Then $\operatorname{\mathbf{R}Hom}_{\mathcal{A}}(A^\bullet,B^\bullet)= \operatorname{Hom}_{\mathcal{A}}(A^\bullet,I^\bullet)$.
