# What is the 'formal' definition for Total Derivative?

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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?

Staff Emeritus
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##

• SebastianRM