Exterior derivative of a $1-$form is a tensor field I'm confused about this:
If $\theta$ is a one-form its exterior derivative $$(d\theta)(X,Y)=X\theta Y-Y\theta X-\theta[X,Y]$$ is a tensor field. In fact is a two-form.
Why $d\theta$ is a tensor field?
As a comment of my class note says that it's enough to check $(d\theta)(X,fY)=f(d\theta)(X,Y).$
So, $d\theta$ is a tensor field of type (0,2)?
Any kind of help is thanked in advanced.
 A: Suppose we have a continous tensor field $A : M \to T^kT^*M$. Then given $X_1,\dots,X_k \in \mathfrak{X}(M)$, we have a function $A(X_1,\dots,X_k) : M \to \mathbb{R}$ induced by $A$, defined as
$$
A(X_1,\dots,X_k)(p) = A_p(X_1|_p,\dots,X_k|_p).
$$
It can be check that if $A$ is smooth, then so is $A(X_1,\dots,X_k)$. Also it turns out that the converse is also true. That is, if the map 
$$
\mathcal{A} : \mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M) \to C^{\infty}(M)
$$
defined as $\mathcal{A} (X_1,\dots,X_k) := A(X_1,\dots,X_k)$ (induced by a covariant $k$-tensor field $A$) is multilinear over $C^{\infty}(M)$, then the $k$-tensor field $A$ is smooth. You can see the proof in Lee's book p.318, called $\textbf{Tensor Characterization Lemma}$.
The map $\mathcal{A}$ multilinear over $C^{\infty}(M)$ means that for any $X_1,\dots,X_i,X_i',\dots,X_k$, and $f,g \in C^{\infty}(M)$, then
$$
A(X_1,\dots,fX_i+gX_{i}',\dots,X_k) = fA(X_1,\dots,X_i,\dots,X_k) + gA(X_1,\dots,X_i',\dots,X_k).
$$
holds for all $i=1,\dots,k$. This theorem may be used alot , but it is not trivial to prove.
$\textbf{Edit : }$ In your case, $d\theta(X,Y)$ is now a function induced by $d\theta$. So we need to show that $d\theta(X,fY) = fd\theta(X,Y)$. By your formula, property of Lie derivative $[X,fY] =f[X,Y]+ (Xf) Y$, property of vector field $X(fg) = fXg + gXf$, and apply the theorem for the smooth covector field $\theta$ (that is $\theta$ is linear $\theta(fY) = f\theta Y$), we have
\begin{align*}
d\theta(X,fY) &= X(\theta (fY)) +fY(\theta X) - \theta [X,fY] \\
&= X(f\theta Y) + fY(\theta X) - f\theta[X,Y] - (Xf)(\theta Y) \\
&= \require{cancel}\cancel{(Xf)(\theta Y)} + f X(\theta Y)-fY(\theta X)- f\theta [X,Y] - \require{cancel}\cancel{(Xf)(\theta Y)} \\
&=  fX(\theta Y)-fY(\theta X)- f\theta [X,Y]\\
&= f d\theta [X,Y]
\end{align*}
