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](http://www.claymath.org/library/monographs/cmim04.pdf), 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?
> 
> 
> 



Different definitions of derived functors
The total right derived functor ${\bf R}F(-)$ contains a bit more information than just its individual cohomologies ${\bf R}^iF(-) = H^i({\bf R}F(-))$. This information can indeed be described as a kind of gluing data, and can be encoded as suitable $k$-invariants. For example, if $C\_{\bullet}$ is a cochain complex in ${\cal A}$ of length $2$ then $C\_\bullet$ is completely determined by the data of $H^0(C\_\bullet), H^1(C\_\bullet)$ and a suitable $k$-invariant $\alpha\_{C} \in {\rm Ext}^2\_{{\cal A}}(H^1(C\_\bullet),H^0(C\_\bullet))$. 


An example where this additional gluing information is important is when one is trying to compute ${\bf R}^iF(A)$ for a functor $F = G \circ H$ which is a composition of two left exact functors. In this case one has the Grothendieck spectral sequence 
$${\bf R}^pG({\bf R}^qH(A)) \Rightarrow {\bf R}^{p+q}F(A) .$$ 
While the $E\_2$-page of this spectral sequence depends only on the individual derived functors $\{{\bf R}^pG(-)\}$ and $\{{\bf R}^qH(-)\}$, the **differentials** in this spectral sequence depend on the precise gluing data, or $k$-invariants, of the total derived functor ${\bf R}H(A)$.


