# Geometric representation of product rule?

At time 1:06 of this video by minutephysics, there is a geometric representation of the product rule:

However, I don't understand how the sums of the areas of those thin strips represent $d(u\cdot v)$. The only way I can see it is that $d(u\cdot v)$ is a small change in the area of the square, and those thin strips do represent that; however, I'm not sure if this is correct and if it is, how formal of a proof is this? Thanks!

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How can a Youtube video be considered a formal proof? – Hagen von Eitzen May 20 '13 at 19:53
You can link to a specific time in a Youtube video. Also, always take screenshots or copy down the information yourself into the question, so that your question will still make sense if external content is removed. – Zev Chonoles May 20 '13 at 19:56
@Hagen von Eitzen: I'm talking about the diagram, just like the phytagorean theorem was proved with a diagram by Bhaskara. I really don't know if that was considered a formal proof, but I think it's pretty convincing. – Ovi May 20 '13 at 19:58
@Zev Chonoles: Ok thanks I'll do that next time. – Ovi May 20 '13 at 19:59
Also, it is not a square, but a rectangle. It will only confuse you if you think of it as a square. – Thomas Andrews May 20 '13 at 19:59

I use the picture of the rectangle in my own teaching (without the differential notation) and show it to grad students who are starting their teaching careers. It is far superior to the usual tricky addition-of-$0$ argument found in most textbooks.

Here is the argument in greater detail:

\begin{align*} \frac{\Delta(uv)}{\Delta x} &= \frac{(u+\Delta u)(v+\Delta v) - uv}{\Delta x} \\ &= \frac{u\Delta v + v\Delta u + \Delta u\Delta v}{\Delta x} = u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \Delta u\cdot\frac{\Delta v}{\Delta x}\,. \end{align*} Taking $\lim\limits_{\Delta x\to 0}$ gives the product rule.

This can all be written out with the usual $f(x+h)g(x+h)$ notation, if so desired.

By the way, this same picture can be used to give a more motivated proof of the product theorem for limits, as well.

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One tiny little tweak I'd make is to replace the $\Delta u\cdot\frac{\Delta v}{\Delta x}$ at the end of the last line with a $\Delta x\cdot\frac{\Delta u}{\Delta x}\cdot\frac{\Delta v}{\Delta x}$ so it's immediately clear that that quantity goes to zero (as long as $u'$ and $v'$ are bounded, of course), as opposed to needing to argue that $\Delta u\to 0$ which can sometimes throw a wrench in the works. – Steven Stadnicki May 20 '13 at 20:15
Well, of course, that is the argument. But it doesn't hurt to remind folks that differentiability implies continuity, precisely by that argument :) – Ted Shifrin May 20 '13 at 20:18