How to relate exactly the notion of differential in math to the notion of elementary variation in physics ( e.g. elementary work).

Question

• I recently encountered the notion of " differential of a function $f$ ", defined as :

$ dy= d[f(x)] = f'(x)\Delta x = f'(x)dx$.

• Out of this I infered ( maybe wrongly) that a differential always requires an original function.

• If I am correct, the notion of " elementary variation " in physics is an application of the mathematical notion of " differential".

• However, I cannot recognize the mathematical form of the differential in the definition of particular elementary variations in physics. For example, in the definition of " elementary work" :

$dW = Fds$

( where $dW$ is elementary work and $ds$ is elementary displacement)

what is the original function?

• Of what function ( if any) is $dW$ a differential? Is it a function having $s$ as independent variable?

• What am missing? Does my question originate in a conceptual confusion?

Answer 1

When you write $dW = F ds$ you are assuming that $W : \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function of the position $s$ and that $W' = \frac{dW}{ds} = F$. So the answer to your question is yes, $dW$ is the differential of $W$ wrt to the independent variable $s$. However in physics the symbol $"ds"$ is often used to indicate a "small displacement" even if the mathematical meaning is different. To be mathematically precise you could rewrite the "equation" of elementary work $dW = F ds$ as a first order Taylor approximation: $$\Delta W = W(s + \Delta s) - W(s) = F \Delta s + o(\Delta s)$$

Answer 2

Physicists assume that their functions are differentiable and that the increments are small enough that the linear approximation is exact (to the experimental errors). In that sense, small increments and differentials coincide.

See Differential of the identity function: how to show that $\Delta x = dx$, in order to explain the general definition of the differential $dy$.