Linear connection on a 1-form Let $M$ be a manifold with linear connection $\nabla$ and let $X$ be a vector field on $M$. Given a 1-form $\alpha \in \Omega^{1}(M)$, define $\nabla_{X} \alpha : \scr{X}$ $(M) \to C^{\infty}(M)$ by $$[\nabla_{X} \alpha] (Z) = X(\alpha(Z)) - \alpha(\nabla_{X}Z)$$ for $Z \in \scr{X}$$(M)$. Show that $\nabla_{X} \alpha$  is a 1-form on $M$.
I tried writing everything in coordinate form. $X =\sum X^{i} \frac{\partial}{\partial x^{i}}$ and $\alpha = \sum f_{i} dx^{i}$ and then taking $Z$ to be the basis $\frac{\partial}{\partial x^{i}}$ (for each $i$ separately to see what it does to the basis) but I'm stuck. Not sure where to go from here. Thanks in advance.

$[\nabla_{X} \alpha] (fY) = X (\alpha(f Y)) - \alpha(\nabla_{X} fY)$
$= X(f \alpha Y) - \alpha ((Xf)Y + f(\nabla_{X}Y)$
$= fX(\alpha Y) + (Xf)\alpha Y - (Xf) \alpha Y + f \alpha(\nabla_{X} Y)$
$= f(X(\alpha Y) - \alpha(\nabla_{X} Y)) = f [\nabla_{X} \alpha](Y)$
