Functors which are non-isomorphic but whose derived functors are isomorphic I was wondering if there is a good/interesting examples of functors that are non-isomorphic as functors but whose derived functors are isomorphic? The example I have encountered so far, the derived Nakayama functor $$\nu = \mathbb{R}\text{Hom}(\text{Hom}(\_,A), K) : \mathcal{D}^b(\text{mod}A) \rightarrow \mathcal{D}^b(\text{mod}A)$$ and the derived functor $$\mathbb{L}(\_ \otimes DA) : \mathcal{D}^b(\text{mod}A) \rightarrow \mathcal{D}^b(\text{mod}A)$$ are isomorphic, but this holds already for the non-derived functors. 
Thank you very much for any help. 
 A: A typical situation when this happens is when you left-derive (for example) a functor $F$ that is not right exact: then its zeroth left derived functor $L_0F$ is actually right exact (in particular, it is not isomorphic to $F$), and the left derived functors for $F$ and $L_0F$ are the same.
(In fact, this is in some sense what always happens (assumig enough projectives etc.): if two functors $F \not\simeq F'$ have $\mathbb{L}F \simeq \mathbb{L}F'$, then at least one of them, say $F$, cannot be right exact, otherwise they are both identified with $L_0F$. Then taking $F$ and $L_0F$ instead of $F, F'$ works too.) 
The one example of this that actually comes up is when $F=\widehat{(-)}_I,$ the functor of $I$-adic completion for a finitely generated ideal (over, generally, non-Noetherian ring). In nice cases (when $I$ is "weakly proregular"), $L_0F$ can be described in a nice intrinsic way (it agrees with the so-called "derived $I$-completion"/"analytic $I$-completion").
So, for example, when $I=(x_1, \dots, x_n)$ is generated by a regular sequence, the two functors that share their derived functor are the classical $I$-completion $M\mapsto \varprojlim_k M/I^kM$ and the "derived $I$-completion", described by $$M \mapsto M[[X_1, \dots, X_n]]/(X_1-x_1, \dots, X_n-x_n)M[[X_1, \dots, X_n]].$$
