I am reading a paper concerning the action of monoidal category to another category. 
Let $k$ be a commutative ring, $R$ is a k-algebra. $A=R-mod$, $B=R^{e}-mod=R\bigotimes \_{k}R^{o}-mod$.


Consider the action:


$B\times A\rightarrow A,(M,N)\mapsto M\bigotimes \_{R}N$ is an action of monoidal category of $R^{e}-mod=B^{~}=(B,\bigotimes \_{R},R)$ on A. 


The paper said this action induces the action 


$\Phi : D^{-}(B)\times D^{-}(A)\to D^{-}(A)$ of the monoidal derived category $D^{-}(B)$ on $D^{-}(A)$


I know this action should be $(M,N)\mapsto M\bigotimes\_{R}^{L}N$. 


But I do not know how is this action of monoidal derived category on the other derived category **induced** by the action of monoidal abelian category. Is there a canonical way(A natural transformation)to get this action?


Notice that the action of monoidal abelian category is defined as follows


$\Psi:=(\Phi ,\phi ,\phi \_{0})$


$\Phi :B=(B,\bigotimes \_{R},R)\rightarrow End(A)$


$\Phi (V)\cdot \Phi (W)\overset{\phi }{\rightarrow}\Phi (V\bigotimes \_{R}W)$


The back ground of this question is localization of differential operator in derived category, so I added the tag"algebraic geometry"


This paper is "Differential Calculus in Noncommutative algebraic geometry I" which is available in MPIM 



How is this action of monoidal derived category induced?
The original action takes the form of an additive functor $A \times B \to B$ with notation as in the question (and appropriate coherence conditions giving compatibility with the monoidal structure on $A$ presumably). By additivity this extends to the level of homotopy categories giving $K(A)\times K(B) \to K(B)$ where there is an obvious triangulation on the product category. Left deriving this functor gives the desired action $D(A)\times D(B) \to D(B)$ as uniquely as one can expect (and that there are coherent natural isomorphisms making this act as nicely as one could expect).


Note that taking categories of complexes bounded above is unneccesary.


