Differential forms and partial derivatives If I have a vector valued function $f(\vec R)$ and I imagine that $R$ is also a function of time, so that $R(t)$ Then I'm struggling to prove that:
$$ df \left({\partial R \over \partial t}\right) = {df \over dt}$$
Intuitively, it makes some sense, that if df is a differential form, and a covector; and $\partial R \over \partial t$ is the velocity vector, then a covector and vector product will give a scalar. Since $df \over dt $ is also a scalar output function differentiated in terms of a scalar, it too would be a scalar.
What I'm struggling to see is why this should be true.

The function $f$ takes in a point $(x,y,z)$ in space and outputs a real number $f(x,y,z)$; provided that it is regular enough, you can write by the total differential theorem that
$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz \tag1$$
On the other hand, $\mathbf R$ is a function that takes in a real number (the time $t$) and outputs a point $\mathbf R(t)$ in space; if it is regular enough, its derivative is well-defined and we may say that the vector $\mathbf R$ defines a trajectory in space. The derivative of $\mathbf R$ (total or partial is the same) is the vector
$$\frac{d\mathbf R}{dt} = \left(\frac{dR_x}{dt}, \frac{dR_y}{dt}, \frac{dR_z}{dt} \right)\tag2$$
If you imagine formally dividing $(1)$ by $dt$, you get
$$\frac{df}{dt} =  \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}  \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}, $$
and you can think of “evaluating” this expression on the trajectory of $\mathbf R$ by plugging in the components of $(2)$ into the total derivatives of $x$,$y$, and $z$. I think that is what is meant by the LHS of your formula.
