How do we relate the exterior derivative of a scalar function to the "differential?" So Munkres' Analysis on Manifolds says that the exterior derivative of a $0$-form (scalar function) $f$ is a $1$-form, 
$$df_x (v) = \nabla f \cdot v\tag{1}$$
which is just a functional, and the domain is that $v$ must be in the tangent space about $x$.
But back in the regular multivariate calculus world, I see people define the "differential" of $f$ by $$df = \frac{df}{dx}dx + \frac{df}{dy}dy = \nabla f \cdot \left<dx,dy\right>\tag{2}$$
I'm having trouble understanding the relationship between (1) and (2). In (1), $v$ can be any element of the tangent space, whereas in (2) it seems that people are restricting $v$ to be some kind of differential step...?
Any thoughts?
Thanks!
 A: I don't think you want to think of $df$ as the dot product of $\nabla f$ with a vector of differentials $\left<dx,dy\right>$.  Instead, think of $dx$ and $dy$ as the dual basis to the standard basis vectors $\mathbf{i}$ and $\mathbf{j}$.  In other words,
\begin{align*}
    dx(a\mathbf{i} + b\mathbf{j}) &= a \\
    dy(a\mathbf{i} + b\mathbf{j}) &= b
\end{align*}
Then $df_{(x,y)}$ is a dual vector whose coefficients are $\frac{\partial f}{\partial x}(x,y)$ and $\frac{\partial f}{\partial y}(x,y)$.  It acts on vectors $v=a\mathbf{i} + b\mathbf{j}$ as
$$
    df(v) = \left(\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy\right)(a\mathbf{i} + b\mathbf{j}) = \frac{\partial f}{\partial x}a + \frac{\partial f}{\partial y}b
$$
So you see that this is really the same thing as $\nabla f \cdot v$.  
The differential one is the more “natural” of the two because it uses dual vectors rather than the dot product.  On a general Riemannian manifold with inner product $g$, the gradient operator can be defined by the equation
$$
    df(v) = g(\nabla f,v)
$$
