# What is the significance of the Increment Theorem in non-standard analysis?

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 $y = x^3$

$$st( \frac{\Delta y}{\Delta x} ) = st(\frac{(x + \Delta x)^3 - x^3}{\Delta x}) \\ = st(3x^2 + 3x\Delta x + (\Delta x)^2) \\ = 3x^2$$

• The point is we can now write $$\frac{\Delta y}{\Delta x} = f'(x) + \varepsilon$$ where the fraction is legitimately an actual fraction, and doesn't just look like one. – sperners lemma Nov 3 '12 at 15:01
• @spernerslemma It seems that you didn't read the source, since the inference is in the other direction - hence the OP's query as to why this simple rescaling is singled out as a new named theorem. – Bill Dubuque Nov 3 '12 at 15:38
• @spernerslemma I don't understand why I couldn't legitimately write the fraction before, or why this theorem would change that. – Rob N Nov 5 '12 at 1:22
• @RobN Because, $\Delta x and \Delta y$ are individually not numbers, since no number can be infinitely close to 0 but not 0 itself. But they can be thought of as a ratio when they get closer and closer to $0$. Or, as done in nonstandard-analysis, you could say "that's stupid" and declare numbers can be infinitely close to $0$ without being it. – PyRulez May 17 '13 at 20:08

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.