I was curious about one thing: Let $A$ and $B$ be abelian categories with enough projectives, let $X$, $Y$ be objects in $A$ and let $P_{\bullet} \rightarrow X$, $P'_{\bullet} \rightarrow Y$ be projective resolutions of $X$ and $Y$ respectively, this already well-known theorem (http://bit.ly/1t0y2Rm, theorem 6.9), says that any morphism $f: X \rightarrow Y$ defines a chain map $f_{\bullet}: P_{\bullet} \rightarrow P'_{\bullet}$ and any two chain maps $f_{\bullet},g_{\bullet}: P_{\bullet} \rightarrow P'_{\bullet}$ are chain homotopic. So that means that any two morphisms $f,g: X \rightarrow Y$ define a chain homotopy being that they both define chain maps $f_{\bullet}$, $g_{\bullet}: P_{\bullet} \rightarrow P'_{\bullet}$?
Let's say I want to compute the derived functor of some right exact functor $F: A \rightarrow B$ for object $Y$ and resolution $P'_{\bullet} \rightarrow Y$, since chain homotopies define quasi-isomorphisms which are preserved by additive functors then if I have a resolution $P_{\bullet} \rightarrow X$ and a morphism $f: X \rightarrow Y$ I can just use the zero morphism $0_{XY}: X \rightarrow Y$ and get a chain homotopy between $f_{\bullet}$ and $0_{XY_{\bullet}}$ and there'll be a chain homotopy equivalence between $P_{\bullet}$ and $P'_{\bullet}$ and I can compute $L_nF(Y)$? So any morphism that is not trivial defines a chain homotopy since I can always use the zero morphism to define a chain homotopy? Does that mean that objects and their subobjects have the same derived functor groups for any right (left) exact functor? For any subobject $A' \subseteq A$ I can always define a chain homotopy between the chain maps induced by the monomorphism $A' \rightarrow A$ and the zero morphism to define a chain homotopy between their resolutions and compute derived functors?
It looks like you're misreading the theorem in a critical detail. It's not the case that any two maps $f_\bullet,g_\bullet:P_\bullet\to P'_\bullet$ are chain homotopic, but that any two such maps induced by the same map $f:X\to Y$ are. (For a counterexample, if $X$ and $Y$ are themselves projective, then two chain maps between $X$ and $Y$ are chain homotopic if and only if they're equal.) This shows that your procedure for computing the left derived functor won't work out. And that's good news, because if it did the whole theory would be rather trivial!