TITLE: What is the significance of the Increment Theorem in non-standard analysis?
QUESTION [9 upvotes]: A bit of background: I'm an engineer, not a mathematician, and I need to review and improve my calculus.  In college, I never liked how they said $dy/dx$ was a single symbol, not a ratio; and then proceeded to write things like $dy = f(x) dx$ and integrate.  So I'm trying a different angle this time, and reading the textbook Elementary Calculus: An Infinitesimal Approach, by H. Jerome Keisler, which is available online.
I'm at the part (p.55) that discusses the Increment Theorem.  It says: 

Let $y = f(x)$. Suppose $f'(x)$ exists at a certain point x, and $\Delta x$ 
  is infinitesimal.  Then $\Delta y$ is infinitesimal, and
$\Delta y = f'(x)\Delta x + \epsilon\Delta x$
for some infinitesimal $\epsilon$, which depends on $x$ and $\Delta x$.

And then he works some examples, finding $\epsilon$. For example, with $y = x^3$...
$$
y' = 3x^2 \\
\Delta y = (x + \Delta x)^3 - x^3 \\
\epsilon = \Delta y / \Delta x - y' \\
...\\
\epsilon = 3x \Delta x + (\Delta x)^2
$$
I'm left wondering... what is the point?  Where are we going with this? We seem to be revisiting the definition of the derivative, where $\epsilon$ is the part of the equation that we were able to discard because it was infinitesimal.  For example, to get the derivative of $y = x^3$
$$
st( \frac{\Delta y}{\Delta x} ) = st(\frac{(x + \Delta x)^3 - x^3}{\Delta x}) \\
= st(3x^2 + 3x\Delta x + (\Delta x)^2) \\ 
= 3x^2
$$

REPLY [4 votes]: The increment theorem is sort of a restatement of the usual procedure for derivatives, but it is more convenient because $\Delta x$ can be 0.  Keisler uses it on page 86 to prove the chain rule.