In principle one uses the notion of derived category, and the other doesn't.

Suppose $F: \mathcal A \to \mathcal B$ is a left exact (additive) functor between abelian categories, and suppose the category $\mathcal A$ has enough injective objects. Then we have two kinds of terminology of derived functor:

(1) The standard one follows Hartshorne's book on algebraic geometry. For an object $A\in \mathcal A$ choose an injective resolution $I^\bullet$ of $A$, i.e. an exact sequence $0 \to A \to I^0 \to I^1 \to \cdots$. Then we define a collection $\{R^iF | i \ge 0 \}$ of the so-called *right derived functor* by setting $$R^iF(A) := H^i (F(I^\bullet)) = \frac {\mathrm{ker}\big(F(I^i) \to F(I^{i+1})\big) }{ \mathrm{im} \big( F(I^{i-1}) \to F(I^i) \big)}$$

(2) Alternatively, one can resort to the notion of derived category (cf. Dirichlet Branes and Mirror Symmetry, sec. 4.4.5) For instance, suppose for simplicity that $\mathcal A = \mathcal B = \mathbf{Mod}(R)$ is the category of $R$-modules for a given ring $R$. Fix $P \in \mathcal A$, then it is known that the Hom functor $$F\equiv \mathrm{Hom}(P,- ) : \mathbf{Mod}(R)\to \mathbf{Mod}(R)$$ is a left exact functor. It is not hard to check that $F$ trivially induces a functor $\hat F$ on the category $\mathcal C(\mathcal A)$ of complexes of $R$-modules. Now we want to find a way to get a How functor on the derived category as follows. Given a $R$-module $M\in \mathcal A$ we take a injective resolution: $$0 \to M\to L^0 \to L^1 \to \cdots \to L^n \to \cdots$$ Then the $R$-module $M$, considered as a complex, is quasi-isomorphic to the complex $L^\bullet = \{0 \to L^0 \to L^1 \to \cdots \to L^n \to \cdots \}$. Applying the functor $\hat F$ to $L^\bullet$ yields a complex $$ 0 \to F(L^0) \to F(L^1) \to \cdots \to F(L^n) \to \cdots $$ This defines an object of the derived category $D(\mathcal A)$, which we denote $\mathbf RF(M)$. It is easy to see different choices of $L^\bullet$ yields isomorphic object of $D(\mathcal A)$. Hence this defines a derived functor $$ \mathbf RF : D(\mathcal A) \to D(\mathcal A) $$

By definition we immediately have $H^i(\mathbf RF(A)) =R^iF(A)$. My question is how to recover the derived functor $\mathbf RF(-)$ from the collection $\{ R^iF(-): i\ge 0\}$?

Indeed, it is generally believed that "an object $E$ of $D( \mathcal A)$ consists of its cohomology objects $H^i(E)\in \mathcal A$ together with some "glue" which holds them together. " which I read from the second reference mentioned above.

Hence, it seems that the derived functor $\mathbf RF$ defined in (2) contains more information than $\{ R^iF\}$ defined in (1), right? If so, what should be the additional information here?