Cyclic Homology of a mixed complex as derived tensor product. Let $k$ be a commutative ring and $A$ a $k$-algebra. Denote by $C_{\bullet}(A)$ the Hochschild chain complex, with $C(A)_n = A^{\otimes n+1}$. Let $B$ denote Connes' operator
$$B_n \colon A^{\otimes n+1} \to A^{\otimes n+2}\, .$$
This induces a bicomplex $\mathcal{B}_{\bullet\bullet}(A)$ with $\mathcal{B}(A)_{pq} = A^{\otimes q-p+1}$ if $q \geq p$ and $0$ otherwise, where the vertical differential $b$ is the Hochschild differential and the horizontal differential is given by $B$. Define the cyclic homology of $A$ as
$$HC_{\bullet}(A) := H_n(\text{Tot}\, \mathcal{B}_{\bullet\bullet}(A)) \, .$$
This definition makes sense for any mixed complex $(M, b, B)$ (i.e. $M$ is a graded $k$-module together with a homological differential $b$ and a cohomological differential $B$ such that $bB + Bb = 0$). Any mixed complex can be identified with a dg $\Lambda$-module over the dg-algebra $\Lambda$ given by
$$\Lambda_0 = k \, , \; \Lambda_1 = k[\epsilon]/\epsilon^2 \, \; \Lambda_i = 0 \quad \text{for $i >1$}\, ,$$
where $\epsilon$ has degree $1$. An alternative definition of cyclic homology that works for any bicomplex $(M, b, B)$ is
$$HC_{\bullet}(M) = H^n(k \otimes_{\Lambda}^{\mathbf{L}} M) \, ,$$
where $-\otimes_{\Lambda}^{\mathbf{L}}-$ stands for the derived tensor product. My question is

Why are these two definitions equivalent?

If I wanted to compute $HC_{\bullet}(M)$ as in the second definition, I need a resolution of $k$ by dg $\Lambda$-modules, and then somehow compute the derived tensor product. I don't know how to proceed, probably because my comprehension of the "derived world" is very limited. Any help would be very much appreciated.
 A: What you want is explained in this paper of Kassel, in Proposition 1.3.
A mixed module is a (non-negatively graded) dg-module $(M,b)$ for the algebra $\Lambda = k[t]/(t^2)$ where $t$ has degree $1$, so it comes equipped with an extra differential $B$ of degree $1$ that (anti)commutes with $b$.
The trivial module $k$, seen as a (dg) module over $\Lambda$ admits a very simple free resolution over $\Lambda$, where the maps are multiplication by $t$, where I use $s$ as a formal symbol to remember the degree where $\Lambda$ sits on:
$$C:\cdots\longrightarrow s^4\Lambda \longrightarrow s^3\Lambda \longrightarrow s^2\Lambda \longrightarrow s\Lambda \longrightarrow 0$$
If you tensor this with $M$ over $\Lambda$, you get a complex $C\otimes_\Lambda M$ representing the derived tensor product
such that $$(C\otimes_\Lambda M)_n = \bigoplus_{i+j=n} s^iM_j = M_n\oplus M_{n-1} \oplus \cdots$$
and where we both have an internal differential $b$ that preserves this homological grading (and simply goes from $M_n$ to the next summand, etc) and the external differential $B$ going from one "row of $M_i$s" to the next one above. This is precisely what your cyclic bar (double) complex looks like:
$$\begin{matrix} 
\uparrow & & \uparrow &  &  \uparrow\\
M_2 &\to& M_1& \to &M_0 \\
\uparrow & &\uparrow \\
M_1 &\to &M_0 \\
\uparrow \\
M_0\end{matrix}$$
right? (You need to totalize at the end and look at diagonals, and this introduces the funny situation that as in Kassel's paper the total complex now goes $M_n\oplus M_{n-2}\oplus\cdots$.)
