Is there a natural Lie bracket for $\mathfrak X (M) \times C^\infty(M)$ (pairs of vector fields and smooth functions)? Space of smooth vector fields $\mathfrak X(M)$ on a manifold $M$ has a structure of Lie algebra with the bracket being a commutator of two vector fields.
Does cartesian product $\mathfrak X (M) \times C^\infty(M)$ of a space of smooth vector fields $\mathfrak X(M)$ with a space of smooth functions $C^\infty(M)$ on the same manifold naturally form a Lie algebra?
What about $\mathfrak X (M) \times C^\infty(M) \times \ldots \times C^\infty(M)$ ?
 A: If $\mathfrak g$ is a Lie algebra, and $V$ a representation of $\mathfrak g$, then you can define a Lie algebra structure on $\mathfrak g\oplus V$ by the formula
$$[(X,u),(Y,v)]=([X,Y],X\cdot v-Y\cdot u)$$
This applies to your example via the standard action of vector fields on functions by derivation.

Here are the details. The bracket is clearly antisymmetric. We only need to check the Jocaobi identity. If $X,X',X''\in\mathfrak g$ and $v,v',v''\in V$, then
$$\begin{align}
[(X,v),[(X',v'),(X'',v'')]]&=[(X,v),([X',X''],X'\cdot v''-X''\cdot v')\big]\\
&=([X,[X',X'']],X\cdot (X'\cdot v''- X''\cdot v')-[X',X'']\cdot v)\\
&=([X,[X',X'']],X\cdot X'\cdot v''- X\cdot X''\cdot v'\\
&\qquad\qquad -X'\cdot X''\cdot v+X''\cdot X'\cdot v)
\end{align}$$
So for $X,Y,Z\in\mathfrak g$ and $u,v,w\in V$. Then
$$\begin{align}
[(X,u),[(Y,v),(Z,w)]]\\
+\:[(Y,v),[(Z,w),(X,u)]]\\
+\:[(Z,w),[(X,u),(Y,v)]]
\end{align}$$
has its first coordinate equal to
$$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$
and its second coordinate equal to
$$\begin{align}
&\phantom{+}
  X\cdot Y\cdot w- X\cdot Z\cdot v -Y\cdot Z\cdot u+Z\cdot Y\cdot u\\
&+Y\cdot Z\cdot u- Y\cdot X\cdot w -Z\cdot X\cdot v+X\cdot Z\cdot v\\
&+Z\cdot X\cdot v- Z\cdot Y\cdot u -X\cdot Y\cdot w+Y\cdot X\cdot w
\end{align}$$
which equals $0$, as every term appears along with its opposite.
