What is the 'formal' definition for Total Derivative?

[SebastianRM]

A total derivative dU = (dU/dx)dx + (dU/dy)dy + (dU/dz)dz. I am unsure of how to use latex in the text boxes; so the terms in parenthesis should describe partial differentiations.
My question is, where does this equation comes from?

 

[stevendaryl]

There are several slightly different ways to interpret differentials such as $dU$. In differential geometry, one meaning of $dU$ is that it is an operator that takes a vector $V$ and returns the rate of change of $U$ as you travel along direction $V$.

A little more detail: A parametrized path through space is a function that given a value of the path parameter $s$ returns a point in space. With coordinates $x, y, z$, this path can be described by three functions, $x(s), y(s), z(s)$. The "tangent vector" to the path is what would be called "velocity" if the parameter $s$ were time: $\frac{d\mathcal{P}}{ds} = V$ where $V$ has components $V^x = \frac{dx}{ds}$, $V^y = \frac{dy}{ds}$, $V^z = \frac{dz}{ds}$. The rate of change of $U$ as you follow path is given by:

$\frac{dU}{ds} = \frac{\partial U}{\partial x} V^x + \frac{\partial U}{\partial y} V^y + \frac{\partial U}{\partial z} V^z$

which is also called the "directional derivative" along $V$.

The total derivative $dU$ is an operator which given a vector $V$ returns the directional derivative of $U$ along $V$:

$dU(V) = \frac{\partial U}{\partial x} V^x + \frac{\partial U}{\partial y} V^y + \frac{\partial U}{\partial z} V^z$

The basis one-forms $dx, dy, dz$ in differential geometry don't mean what they mean in calculus, which would be an infinitesimal change in $x, y, z$/. Instead, they are viewed as operators themselves, with the sort of trivial definition:

$dx(V) = V^x$
$dy(V) = V^y$
$dz(V) = V^z$

So $dx$ for example, is the special case of $dU$ where $U(x,y,z) = x$