This question is perhaps somewhat soft, but I'm hoping that someone could provide a useful heuristic. My interest in this question mainly concerns various derived equivalences arising in geometric representation theory.
Background
For example, Bezrukavnikov, Mirkovic, and Rumynin have proved the following: Let $G$ be a reductive algebraic group over an algebraically closed field of positive characteristic. Then there is an equivalence between the bounded derived category of modules for the sheaf $\cal D$ of crystalline (divided-power) differential operators on the flag variety, and the bounded derived category of modules with certain central character for the enveloping algebra $\cal U$ of Lie($G$). What is interesting is that it is not true that this equivalence holds on the non-derived level: The category of $\cal D$-modules is not equivalent to the category of $\cal U$-modules with the appropriate central character. This is true in characteristic 0 (this is the Beilinson-Bernstein correspondence), but something is broken in positive characteristic: there are certain "bad" sheaves that are $\cal D$-modules which make the correspondence not hold.
There are other results in geometric representation theory of this form. For example, Arkhipov, Bezrukavnikov, and Ginzburg have proved that there is an equivalence (in characteristic 0) between the bounded derived category of a certain block of representations for the quantum group associated to $G$, and the bounded derived category of $G \times \mathbb C^*$-equivariant sheaves on the cotangent bundle of the flag variety of $G$. Again, this equivalence does not hold on the non-derived level.
In general, there are a number of results in geometric representation theory that hold on the derived level, but not the non-derived level.
Question
Here's my question: Why would one be led to expect that a derived equivalence holds, when the non-derived equivalence does not? It seems as though the passage to the derived level in some sense fixes something that was broken on the non-derived level; how does it do that?
Best Answer
One crude answer is that passing to derived functors fixes one obstruction to being an equivalence. Any equivalence of abelian categories certainly is exact (i.e. it preserves short exact sequences), though lots of exact functors are not equivalences (for example, think about representations of a group and forgetting the G-action).
What derived functor does is fix this problem in a canonical way; you have to replace short exact sequences with exact triangles, but you get a functor which is your original "up to zeroth order," exact, and uniquely distinguished by these properties.
So, what BMR do is take a functor which is not even exact (and thus obviously not an equivalence), and show that the lack of exactness is "the only problem" for this being an equivalence.
EDIT: Let me just add, from a more philosophical perspective, that derived equivalences are just a lot more common. There are just more of them out in the world. Given an algebra A, Morita equivalences to A are classified essentially by projective generating A-modules, whereas derived Morita equivalences of dg-algebras are in bijection with all objects in the derived category of $A-mod$ which generate (in the sense that nothing has trivial Ext with them): you look at the dg-Ext algebra of the object with itself. If you have an interesting algebra (say, a finite dimensional one of wild representation type), there are a lot more of the latter than the former in a very precise sense. Of course, the vast majority of these are completely uncomputable an tell you nothing, but there are enough of them in the mix to make things interesting.