Exterior derivative on the Frechet manifold $C^{\infty}(M)$ It is well known that $C^{\infty}(M)$ is an infinite dimensional Frechet manifold and the tangent space of it can be identified with $C^{\infty}(M)$. Given a one form $\alpha$ which as far as I understand is just  a linear functional on $C^{\infty}(M)$ which varies smoothly. Now I read that in order to check that $\alpha$ is closed we only need to check that $\frac{\partial}{\partial t}|_{t=0} \alpha_{f+tg}(h)$ is symmetric in $h$ and $g$.
The logic is certainly: the exterior derivative of a $1-$form will be a $2-$form which is anti-symmetric, and it is zero iff it is symmetric. Then essentially we are claiming that $d\alpha_g(h) = \frac{\partial}{\partial t}|_{t=0} \alpha_{f+tg}(h)$
Why is this true?
I try to understand this on a finite dimensional vector space $V$ first. Suppose $dimV = n$, and take a basis $v_1, \ldots, v_n$, $\alpha =  \sum_{i} f_i dv_i$, so $\frac{\partial}{\partial t}|_{t=0} \alpha_{p_0+tp_1}(p_2) = \sum_{i} Df_i|_{p_0}(p_1) dv_i(p_2) = \sum_{i} df_i(p_1)dv_i(p_2)$, it seems to be the tensor product instead of the wedge product. And $d\alpha = \sum_{i} df_i \wedge dv_i$ is in terms of the wedge product. What is wrong here?
 A: If $V$ is a vector space and $U \subseteq V$ is an open subset, a $k$-form on $V$ can be identified with a smooth map $\alpha \colon U \rightarrow \operatorname{Hom}(V^{\otimes k}, \mathbb{R})$ (such that $\alpha(p)$ is alternating for all $p \in U$). Given a $k$-form, you can take its Fréchet derivative (the "usual derivative") and get a smooth map $D\alpha \colon U \rightarrow \operatorname{Hom}(V, \operatorname{Hom}(V^{\otimes k}, \mathbb{R}))$ defined by
$$ D\alpha|_{p}(v) = D\alpha(p,v) = \lim_{t \rightarrow 0^{+}} \frac{\alpha(p + tv) - \alpha (p)}{t}. $$
Now you can identify $\operatorname{Hom}(V,\operatorname{Hom}(V^{\otimes k}, \mathbb{R}))$ with $\operatorname{Hom}(V^{\otimes (k+1)}, \mathbb{R})$ and think of $D\alpha$ as a map $D\alpha \colon U \rightarrow \operatorname{Hom}(V^{\otimes (k+1)}, \mathbb{R})$. The basic slogan is that the exterior derivative $d\alpha$ is the anti-symmetrization of $D\alpha$ up to a non-zero universal constant which depends on your conventions (see this answer). In particular, $d\alpha = 0$ if and only if $D\alpha$ is symmetric. 
For example, if $f \colon U \rightarrow \mathbb{R}$ is a smooth function then $Df = df$ and $D^2f \colon U \rightarrow \operatorname{Hom}(V^{\otimes 2}, \mathbb{R})$ is the Hessian of $f$ which is symmetric, hence $d^2f = 0$.
Regarding the relation between the tensor product and the wedge product, this generalizes so that if $\alpha = f_I dx^I$ is a $k$-form then
$$ D\alpha = d(f_I) \otimes dx^I,\,\, d\alpha = d(f_I) \wedge dx^I. $$
The whole thing can be generalized to arbitrary manifolds with a torsion-free connection (which is needed to define the "full derivative" $\nabla \alpha$ of $\alpha$, also known as the full covariant derivative) and this is outlined in the answer I linked above.
