Very confused about directional derivatives as vectors I'm currently reading about GR and how partial derivatives are basis vectors for a manifold, but I'm confused about how $dx^μ$ is the dual basis to $(\partial /\partial x)^μ$. I can't see how $dx(\partial/\partial x)=1$. Maybe this is a dumb question and I'm missing something, but it says this equals $\partial x^μ/\partial x^ν$, which of course is the kroneker delta, but where does the extra $\partial$ on the top go? It just neglects it in the step from $dx^μ(\partial /\partial x^ν)$ to $\partial x^μ/ \partial x^ν$! Why?
 A: This flirts with being an abuse of notation. The idea is that vector fields on a manifold act on functions defined on the manifold as derivatives. More precisely, $X \in TM$ acts on a (smooth) function $f : M \to \mathbb{R}$ by taking it to another function $X(g) : M \to \mathbb{R}$. This definition of the tangent bundle only requires the following facts (for arbitrary $f : M \to \mathbb{R}$, $g : M \to \mathbb{R}$, $a \in \mathbb{R}$:
$$X(f g) = X(f) g + g X(f)$$
$$X(f + g) = X(f) + X(g)$$
$$X(a f) = a X(f)$$
The first you will recognize as the product rule for derivatives, and the other two make $X$ linear. This is actually sufficient to define the derivative, and it agrees with the usual derivative when you apply it to $\mathbb{R}$ (proving this might be instructive).
In a particular chart on the manifold $M$, a natural basis for the tangent bundle is $\frac{\partial}{\partial x_i}$, where $\frac{\partial}{\partial x_i} f = \frac{\partial f}{\partial x_i}$. So to define things accurately, when you write $\frac{\partial}{\partial x_i}$ above you mean the element of $TM$ that acts on a (smooth) function $f : M \to \mathbb{R}$ by taking its derivative with respect to the coordiante $x_i$.
Now when we turn to $dx^i$, it is defined to be an element of the co-tangent bundle $T^* M$, which is generally defined as the dual space to $TM$ (i.e. the vector space of linear functions $\alpha : TM \to \mathbb{R}$). That is the idea behind the notation $dx(\frac{\partial}{\partial x})$.
The reason $dx(\frac{\partial}{\partial x}) = 1$ is that that is usually how the basis for the co-tangent bundle is defined. Basically, just as $\{ \frac{\partial}{\partial x_i} \}$ forms a basis for $TM$, the $\{ dx^i \}$ defined as the unique elements of $T^* M$ s.t. $dx^i(\frac{\partial}{\partial x_j}) = \delta_{ij}$ form a basis for $T^* M$.
The reason this isn't entirely an abuse of notation is that if you apply these to $\mathbb{R}^n$ (along with a notion of taking an integral with respect to $\alpha \in T^* M$), you get the usual derivative and integral from elementary vector calculus!
