# Infinitesimals in the denominator

I was reading H. Jerome Keisler's Elementary Calculus: An Infinitesimal Approach, chapter 1.4 ("Slope and Velocity: The Hyperreal Line") (PDF here). Page 23 explains getting the derivative of a function by finding Δy/Δx. It explains that, at the end, we "neglect the term Δx because it is very small."

However, when I was deriving $y=x^2+4$, I ended up with $\Delta y/\Delta x = 2x_0 + \Delta x + 4/\Delta x$. If $\Delta x$ is "very small," then instead of neglecting $4/\Delta x$, shouldn't it be infinity (as 4/0.000001 is 4 million, etc.)?

If it matters, I haven't even taken Pre-calc yet (I'm in Algebra II/Trig right now).

Thanks!

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Surely you've made a mistake in the derivation. The $4$'s should cancel out when you subtract $y(x_0)$ from $y(x_0+\Delta x)$. –  Rahul May 7 '12 at 20:13

You must have made a mistake in algebraic manipulations: $$\begin{eqnarray} \frac{\Delta y}{\Delta x} &=& \frac{y(x+\Delta x)-y(x)}{(x+\Delta x)-x} = \frac{((x+\Delta x)^2 + 4)-(x^2+4)}{\Delta x} \\ &=& \frac{(x+\Delta x)^2 + 4-x^2-4}{\Delta x} = \frac{(x+\Delta x)^2 -x^2}{\Delta x} \\ &\stackrel{\text{use }\color\maroon{a^2-b^2=(a-b)(a+b)}}{=}& \frac{((x+\Delta x)-x)((x+\Delta x)+x)}{\Delta x} = \frac{\Delta x (2x + \Delta x)}{\Delta x} \\ &=& 2x + \Delta x \end{eqnarray}$$
Ah - you're correct. I substituted $y_0$ for $x_0^2$ instead of $x_0^2+4$. Thanks! –  WChargin May 7 '12 at 21:23