$V$ vector field, $\omega$ one-form, $V(\omega(V))$=? 
(1-forms) Let $X$ be a manifold and $\omega \in \Omega^1(X)$ be a smooth 1-form,
  and $V, W \in V^{\infty}(X)$ smooth vector fields on $X$. Then, $\omega(V ), \omega(W ) \in C^{\infty}(X)$ are well-defined smooth functions, giving rise to the identity:
$$d\omega(V, W ) = V (\omega(V )) − W \omega(V ) − \omega([V, W ]).$$

I would like to try to solve the problem myself, but I cannot deduce the types of terms in the stated identity. Possibly, there may be some typos in the statement of a problem or an implicit notation of which I am not aware.
My reasoning:
$V$ as a vector field has a type $X \to TX$ where $TX$ is a tangent bundle on $X$. Similarly $[V,W]:X \to TX$. A 1-form $\omega$ is a covector field, hence the type $\omega:X \to T^*X$ where $T^*X$ is a cotangent bundle. So at a particular point $x \in X$, $(\omega([V,W]))(x)=\omega(x)([V,W](x))=$covector $.$ vector=$c \in \mathbb{R}$. So $\omega([V,W]):X \to \mathbb{R}$.
The problem is a calculation of a type of $V(\omega(V))$ since it has to have a type $X \to \mathbb{R}$ since we are adding it with another object of that type, but $\omega(V):X \to \mathbb{R}$ and $V:X \to TX$.
 A: Here's a way to prove this expression that is rather light on computation. The idea is to express everything in a local chart (so that without loss of generality, the manifold is $\Bbb R^n$) and to check how both sides of the equation behave when $\omega, X$, or $Y$ are multiplied by a smooth function $f$. Once this is understood, it remains to check the above formula for differential forms $\omega=dx^k$, and vector fields $X=\frac{\partial}{\partial x^i}$ and $Y=\frac{\partial}{\partial x^j}$. So let $\omega$ be a smooth $1$ form on $\Bbb R^n$, $X,Y$ smooth vector fields on $\Bbb R^n$ and let $f\in C^{\infty}(\Bbb R^n)$ be a smooth map.

What happens when we replace $\omega$ by $f\omega$ : then $d(f\omega)=df\wedge \omega + (-1)^{0}fd\omega=df\wedge \omega + fd\omega$ and
$$\begin{align} d(f\omega)(X,Y)&=df(X)\omega(Y)-df(Y)\omega(X)+fd\omega(X,Y)\\&=(X\cdot f)\omega(Y)-(Y\cdot f)\omega(X)+fd\omega(X,Y)\\&=A+fd\omega(X,Y)\end{align}$$
while
$$\begin{align} X\cdot(f\omega(Y))-Y\cdot(f\omega(X))-f\omega([X,Y])&=(X\cdot f)\omega(Y)+fX\cdot\omega(Y)-((Y\cdot f)\omega(X)+fX\cdot\omega(Y))-f\omega([X,Y])\\&=(X\cdot f)\omega(Y)-(Y\cdot f)\omega(X)+f(X\cdot\omega(Y)-Y\cdot\omega(X)-\omega([X,Y]))\\&=A+f\Big(X\cdot\omega(Y)-Y\cdot\omega(X)-\omega([X,Y])\Big)\end{align}$$
The passage from the first line to the second is by virtue of $X$ and $Y$ begin derivations of $C^{\infty}(\Bbb R^n)$)

What happens when we replace $X$ by $fX$ Then 
$$d\omega(fX,Y)=fd\omega(X,Y)$$
and
$$\begin{align}
(fX)\cdot\omega(Y)-Y\cdot\omega(fX)-\omega([fX,Y])&=f(X\cdot\omega(Y))-Y(f\omega(Y))-\omega(f[X,Y]-(Y\cdot f)X)\\
&=f(X\cdot\omega(Y))-(fY\cdot\omega(Y)+(Y\cdot f)\omega(Y))-f\omega([X,Y])+(Y\cdot f)\omega(X)\\
&=f\Big(X\cdot\omega(Y)-Y\cdot\omega(X)-\omega([X,Y])\Big)
\end{align}$$
Essentially the same calculation work when replacing $Y$ by $fY$.

Thus, we have seen that woth expressions $d\omega(X,Y)$ and $X\cdot\omega(Y)-Y\cdot\omega(X)-\omega([X,Y])$ transform the same way when multiplying $\omega,X$ or $Y$ by smooth functions. Since vector fields are $C^{\infty}(\Bbb R^n)$-linear combinations of the vector fields $\frac{\partial}{\partial x^i}$, $i\in\lbrace 1,\dots,n\rbrace$, and one forms are $C^{\infty}(\Bbb R^n)$-linear combinations of the one forms $dx^j$, $j\in\lbrace 1,\dots,n\rbrace$, respectively, the result will follow if the two expressions coincide for these elementary examples. It is immediate to verify that for all choices $i,j,k\in\lbrace 1,\dots,n$
$$\underbrace{d(dx^k)}_{=0}\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)=0=\frac{\partial}{\partial x^i}\cdot \underbrace{dx^{k}\left(\frac{\partial}{\partial x^j}\right)}_{=\delta_{k,j}}-\frac{\partial}{\partial x^j}\cdot \underbrace{dx^{k}\left(\frac{\partial}{\partial x^i}\right)}_{=\delta_{k,i}}-\omega\left(\underbrace{\left[\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^i}\right]}_{=0}\right)$$
Where $\delta_{a,b}$ is the constant function on $\Bbb R^n$ equal to $1$ if $a=b$ and equal to $0$ otherwise. 

This same method of proof will establish the more general formula 
$$\begin{align}
d\omega(X_0,\dots,X_p)&=\sum_{i=0}^{p}(-1)^i X_i\cdot\omega(X_0,\dots,\widehat{X_i},\dots,X_p)\\ &~~~~+\sum_{0\leq i<j\leq p}(-1)^{i+j}\omega([X_i,X_j],X_0,\dots,\widehat{X_i},\dots,\widehat{X_j},\dots,X_p)\end{align}$$
valid for any $p$-form $\omega$ and vector fields $X_0,\dots,X_p$.
