Differentials for algebraic stacks Let $S$ be a base scheme. For which algebraic stacks $X$ over $S$ can we define a sheaf of differentials $\Omega^1_{X/S}$ (classifying derivations)? Probably it works when $X$ is Deligne Mumford because $\Omega^1$ is stable under pullbacks of étale morphisms? But for example for the classifying stack $B \mathbb{G}_m$ there is no $\Omega^1$ because deformations of invertible sheaves have non-trivial automorphisms, right?
Even if there is no $\Omega^1_{X/S}$ in general, I would like to know if there is a "tangent bundle" $T(X/S)$ which satisfies the adjunction $\hom_S(Y[\varepsilon]/\varepsilon^2,X) \simeq \hom_S(Y,T(X/S))$ for algebraic stacks $Y$ over $S$. If $\Omega^1_{X/S}$ exists then one may take $T(X/S) = \mathrm{Spec} \mathrm{Sym} \Omega^1_{X/S}$, but perhaps this definition is "too discrete" for algebraic stacks. Perhaps one can encode the cotangent complex into $T(X/S)$?
Edit: At this moment I am not interested in derived stacks. Algebraic stacks mean Artin stacks in the usual sense.
 A: I might be "too derived" for the kind of framework you're looking for, but in general, if you think of a stack as a functor from CDGAs to spaces, the replacement for $\Omega^1$ is the cotangent complex for the stack. As you point out, you have to prove the existence for such a thing, but when a stack $X$ is n-geometric, a cotangent complex exists. In particular, it exists for $BG$. In that case, a sheaf on $BG$ is the same thing as giving a representation of $G$, and the cotangent complex can be written as
$$
\mathfrak{g}^\vee[1]
$$
i.e., the dual of the Lie algebra, shifted in degree. The representation is the dual of the adjoint representation.
In this framework, the adjunction property you seek won't hold, because $Y[\epsilon]/\epsilon^2$ isn't semi-free. (Semi-free resolutions are the usual ways you get cofibrant objects in CDGAs.)
Regardless, the cotangent complex still "classifies" derivations in the following sense:
If $A$ is an affine scheme (i.e., a CDGA), then for every $A$-point
$$f: Spec(A) \to X$$
there exists an $A$-module $\mathbb{L}_{X,f}$ such that
\begin{equation}
(*) \qquad
Map_{AMod}(\mathbb{L}_{X,f},M)
\simeq
hofib(X(A\oplus M) \to X(A))
\end{equation}
for every $A$-module $M$ with cohomology concentrated in non-positive degree. (Note that $A \oplus M$ is an algebra as the square-zero extension.)
This must be functorial in the sense that for all commutative diagrams
$$
Spec(B) \to Spec(A) \to X
$$
you have that pullback preserves the cotangent complex---i.e.,
$$
\mathbb{L}_{X,f_A} \otimes_A^{ho} B \simeq \mathbb{L}_{F,f_B}.
$$
It might help to note that (*) is just a homotopical way to write a universal property from the non-derived setting: Whenever you have a map of rings $f: B \to A$, then
$$
Hom_A(f^* \Omega_B, M)
\cong
\{ \text{maps $B \to A \oplus M$ factoring the map $B \to A$} \}.
$$
A: There is a first-order jet stack, given by the Hom stack construction $\underline{\operatorname{Hom}}_S(S \times \operatorname{Spec} \mathbb{Z}[x]/(x^2), X)$.  When $X/S$ is representable, this describes the total space of the relative tangent sheaf.
It does not encode the cotangent complex, and is insensitive to inertia - you would need a derived jet stack to capture more data.
