Direct calculation of $H_1$ of the cotangent complex

Let $k$ be a commutative ring and let $A$ be a commutative $k$-algebra. The cotangent complex $\mathbf{L}_{A \mid k}$ can be computed using a simplicial resolution of $A$ as follows: choose a simplicial commutative $k$-algebra $P$ and an augmentation $\epsilon : P_0 \to A$ (i.e. a $k$-algebra homomorphism such that $\epsilon \circ d_0 = \epsilon \circ d_1$ as homomorphisms $P_1 \to A$) such that each $P_n$ is a free commutative $k$-algebra and $\epsilon$ considered as a simplicial map $P \to A$ is a weak homotopy equivalence; then $\mathbf{L}_{A \mid k}$ is (the quasi-isomorphism class of the chain complex corresponding to) the simplicial $A$-module $A \otimes_P \Omega_{P \mid k}$, where $\Omega$ denotes Kähler differentials.

By general abstract nonsense, up to quasi-isomorphism, $\mathbf{L}_{A \mid k}$ does not depend on the choice of $P$ or $\epsilon$, buf if it is necessary to be explicit, there is a canonical choice of $P$ and $\epsilon$: take $P_0$ to be the free commutative $k$-algebra generated by $A$ as an abstract set and $\epsilon : P_0 \to A$ to be the unique $k$-algebra homomorphism sending each generator of $P_0$ to the corresponding element of $A$; then iterate, taking $P_{n+1}$ to be the free commutative $k$-algebra generated by $P_n$ as an abstract set. (This is the so-called standard resolution induced by a comonad.)

It is known that the cotangent complex truncated above homological degree $1$ is quasi-isomorphic to the naïve cotangent complex (e.g. Stacks 08RB). Hence, in particular, $\mathrm{H}_1 (\mathbf{L}_{A \mid k}) \cong \ker (\mathrm{d} : A \otimes_{P_0} I \to A \otimes_{P_0} \Omega_{P_0 \mid k})$, where $I = \ker \epsilon \subseteq P_0$.

Question. Can we prove the above isomorphism directly, perhaps by making a particular choice of $P$ and $\epsilon$?

It is not hard to believe that $\mathrm{H}_0 (\mathbf{L}_{A \mid k}) \cong \Omega_{A \mid k}$. This has essentially the same content as the conormal exact sequence, $$A \otimes_{P_0} I \longrightarrow A \otimes_{P_0} \Omega_{P_0 \mid k} \longrightarrow \Omega_{A \mid k} \longrightarrow 0$$ basically because $I$ is the image of the $k$-module homomorphism $d_0 - d_1 : P_1 \to P_0$. It is not even necessary to choose $P$ and $\epsilon$ carefully here – the arguments are general enough. I can also see that $d_0 - d_1 : P_1 \to I / I^2$ is a surjective $P_0$-linear derivation over $\epsilon : P_1 \to A$ (where we think of $P_1$ as a $P_0$-algebra using the degeneracy $s_0 : P_0 \to P_1$), so we get a surjective $A$-module homomorphism $A \otimes_{P_1} \Omega_{P_1 \mid k} \to I / I^2$. But how to proceed?