The twisted tensor product $BA\otimes_{\tau} A$ as the non-unital Hochschild complex The twisting universal morphism $\tau: BA\rightarrow A$ induces a differential $\partial_{\tau}$ on $BA\otimes_{\tau}A$, we have:
$$\partial_{\tau}(x\otimes y)=\partial x\otimes y+(-1)^{\lvert x\rvert}x\otimes\partial y+\displaystyle\sum{(-1)^{\lvert x_{(1)}\rvert}}x_{(1)}\otimes \tau(x_{(2)})y$$
According to some references as Algebraic Operads from Loday-Vallette, the proof of proposition 2.2.13 says that $BA\otimes_{\tau}A$ (as a complex) is identified with a Hochschild complex where the differential of $A$ dissapears and only the product of $A$ remains. I mean, let's consider ($BA,\partial$) with $\partial=d_{1}+d_{2}$. I need to 'kill' $d_{1}$, which is induced by the differential of $A$, but I do not find it evident. Could anybody please give me a hint?. 

An element of $BA\otimes A$ looks like $u = [a_1 \vert \cdots\vert a_n]a$. Let us assume for a minute that $A$ has trivial differential. Then the map $\partial_\tau$ acts as follows on such a basis element: $\tau$ is nonzero only on elements of the form $[a]$, so the only term in the coproduct of $[a_1 \vert \cdots\vert a_n]$ is $[a_1 \vert \cdots\vert a_{n-1}]\otimes [a_n]$, and then what you obtain for $\partial_\tau(u)$ is $[a_1\vert\cdots\vert a_{n-1}]a_na$, so here what you get is the complex which is a reoslution of the trivial module. 
To get the 'cyclic' Hochschild complex you not only need to use $-\frown\tau = \partial_\tau$ (check the definition of the cap products, say in Proute's thesis) but also $\tau\frown -$ so what you are doing is twisting the differential $d_{BA}\otimes 1$ of $BA\otimes A$ by the 'commutator' $[\tau,-]$, giving the terms where $a_n$ and $a_1$ pop out of the bars and act on $a$.
