derived functor through acylic resolution 
Let $C,D$ be abelian categoriesy with enough injectives and $G:C \rightarrow D$ left exact functor. How can I show that if $0 \rightarrow X \rightarrow B^{0} \rightarrow B^{1} \rightarrow \cdots$ is a long exact sequence  and for all $i$, $R^{j}(G(B^{i}))=0$ for $j=1,2,3,...$ then we can calculate $R^{j}(G)(X)$ throuh above complex? 

Frankly, the only way to prove it I can come up with is using diagram with first row $0 \rightarrow X \rightarrow B^{0} \rightarrow B^{1} \rightarrow ...$ and columns : $0 \rightarrow B^{i} \rightarrow I_{B^{i}}^{*}$ ($I_{B^{i}}^{*}$ injective resolution) then we can connect columns with chain morphisms and apply functor G everywhere. the only problem is compositions of two neighbouring row morphism are homotopic to zero morphism but they are not necessarily zero. Thank you for all your answers.
 A: Let $X\rightarrowtail B^\bullet$ be a $G$-acyclic resolution of $X$. We may pick an injective resolution of the bounded below complex $B^\bullet$, which is by definition a quasi-isomorphism
$$f\colon B^\bullet \to I^\bullet.$$
In particular, $I^\bullet$ also gives an injective resolution of $X$, so
$$R^n G (X) \cong H^n (G (I^\bullet)).$$
We would like to see that
$$G (f)\colon G (B^\bullet) \to G (I^\bullet)$$
is also a quasi-isomorphism, which means that it induces isomorphisms in cohomology:
$$H^n (G (B^\bullet)) \cong H^n (G (I^\bullet)),$$
and then we are done. (Note that in general, there's no reason for an additive functor to preserve quasi-isomorphisms, so this requires a bit of work.)
Recall the following

Useful fact. A morphism of complexes $f\colon C^\bullet \to D^\bullet$ is a quasi-isomorphism if and only if its cone $\operatorname{Cone} (f)$ is acyclic (i.e. exact, has trivial cohomology).

(To see this, recall that the cone gives a short exact sequence of complexes $D^\bullet \rightarrowtail \operatorname{Cone} (f) \twoheadrightarrow C^\bullet [1]$ and the associated long exact sequence in cohomology has precisely $H^n (f)$ as connecting morphisms.)
So we want to show that the cone of $G (f)$ is acyclic, but the cone construction commutes with any additive functor:
$$\operatorname{Cone} (G (f)) = G (\operatorname{Cone} (f)).$$
The complex $\operatorname{Cone} (f)$ is acyclic, as $f$ is a quasi-isomorphism; it is bounded below and consists of $G$-acyclic objects, as by definition
$$\operatorname{Cone} (f)^n = I^n\oplus B^{n+1},$$
where $I^\bullet$ and $B^\bullet$ are bounded below, and each $B^n$ is $G$-acyclic by our assumption, while each $I^n$ is $G$-acyclic, because any injective object is acyclic with respect to any left exact functor.
To finish the proof, we may apply to $\operatorname{Cone} (f)$ the following

Key lemma. Let $G$ be a left exact functor. If $C^\bullet$ is a bounded below acyclic complex which consists of $G$-acyclic objects, then $G (C^\bullet)$ is acyclic as well.

The proof the Key lemma should go as follows. The acyclicity of $(C^\bullet,d^\bullet)$ means that $\operatorname{im} d^n \xrightarrow{\cong} \ker d^{n+1}$, and we have short exact sequences
$$\tag{*} 0 \to \ker d^n \to C^n \to \ker d^{n+1} \to 0$$
Now I claim that every $\ker d^n$ is $G$-acyclic. This follows by induction, as $\ker d^n = 0$ for $n \ll 0$, and (*) provides the induction step, because of the following

Observation. If $A\rightarrowtail B$ is a monomorphism between two $G$-acyclic objects, then its cokernel $K$ is again $G$-acyclic.

(Indeed, it's evident from the corresponding long exact sequence $\cdots \to R^n G (A) \to R^n G (B) \to R^n G (K) \to R^{n+1} G (A) \to \cdots$)
Now because we know that $\ker d^n$ is $G$-acyclic for all $n$, if we apply our left exact exact functor $G$ to (*), we get short exact sequences
$$0 \to G(\ker d^n) \to G(C^n) \to G
(\ker d^{n+1}) \to 0$$
But this means that the complex $(G (C^\bullet), G (d^\bullet))$ is acyclic.
