Let $\mathcal{D},\mathcal{E}$ be abelian categories and $\mathcal{C}$ be a Serre subcategory of $\mathcal{D}$.
Let $D(\mathcal{C}), \, D(\mathcal{D})$ denote the derived categories of $\mathcal{C},\mathcal{D}$. 
Suppose that the natural functor between the derived category $D(\mathcal{C})$ and the category $D\_C(\mathcal{D}):=\{X\in D(\mathcal{D}) \, | \, H^i(X) \in \mathcal{C}\}$ is an equivalence of categories. 


Let $F:\mathcal{D} \rightarrow \mathcal{E}$ be a right exact functor, that has derived functor $DF:D(\mathcal{D}) \to D(\mathcal{E})$. Does this imply that the restriction of $DF$ to $D(\mathcal{C})$ is the derived functor of the restriction of $F$ to $\mathcal{C}$?


The derived categories can bounded unbounded or semi-bounded, Whatever make the answer simpler.


A typical example is when $\mathcal D$ is the category of abelian groups and $\mathcal C$ is the category of finite abelian groups.



When the restriction of a derived functor to a subcategory is the derived functor of the restriction
In the example where $\mathcal{D}$ is the category of abelian groups and $\mathcal{C}$ is the category of finite abelian groups, take $F(X)=X\otimes\_\mathbb{Z}\mathbb{Q}/\mathbb{Z}$. Then the restriction of $F$ to $\mathcal{C}$ is zero, and so has the zero functor as its derived functor. However, the derived functor of $F$ doesn't vanish on $\mathcal{C}$.


