## The derivative of x^2

A commenter on Scott Aaronson’s blog, in a discussion of K-12 mathematics education in the U.S., brought up the following (I would imagine fairly common) Algebra I fallacy:

The slope of the line y = mx + b is equal to m. Consider the equation y = x*x + 0. Around x = c, this reduces to y = c*x, and so the slope of the parabola y = x^2 at the point (x, x^2) is x.

Of course, anyone who has taken freshman calculus knows that the slope of the parabola at that point is 2x, not x, and therefore the above reasoning is flawed in some way. But from the perspective of a beginning Algebra I student, there’s nothing at all obviously wrong with it!

So how would one go about deriving (no pun intended) the correct slope of a parabola using only basic Algebra I knowledge? You might imagine that you could just calculate the “rise over run,” and that’s a good start — ((x+h)^2 – x^2)/h = 2x + h. But (without the crucial “take the limit as h approaches 0”) this is ill-defined!

It’s clear that if you could prove the product rule for derivatives, then the correct expression for the slope would follow. Still, though, it seems impossible to do this with such a basic level of knowledge, without resorting to the concept of a limit or an infinitesimal.

The most promising path would likely be to show that y = x^2 actually looks locally like y = 2cx – c^2, rather than y = cx. Actually, it isn’t that difficult to show that y = cx is wrong; if you draw “tangent lines” to the parabola for large enough values of c, it’s clear that those lines don’t come anywhere near the origin. Still, 2cx – c^2 seems incredibly arbitrary without foreknowledge of the calculus.

So, here’s the challenge: How do you convince a reasonably bright Algebra I student, without getting into calculus, that the slope of the parabola is not x but is, in fact, 2x?