$0$-th cohomology of derived tensor product.

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I want to show the following lemma:

Lemma 4.8. Let $M$ and $N$ be two complexes of $A$-module whose homology is concentrated in negative degrees. Then there is an isomorphism $$ H^0(M \otimes^{\mathbb{L}}_A N) \simeq H^0(M) \otimes_A H^0(N) $$

Since $H^0(M)\simeq \operatorname{Hom}_{\mathcal{D}(A)} (A,M)$, we only need to show $$ \operatorname{Hom}_{\mathcal{D}(A)} (A,M\otimes^{\mathbb{L}}_A N) \simeq \operatorname{Hom}_{\mathcal{D}(A)} (A,M) \otimes_A \operatorname{Hom}_{\mathcal{D}(A)} (A,N) $$ Is this isomorphism true?

This may seem trivial to you, but the relationship between derived Hom functors and derived tensor product functors does confuse me. Thank you.