What is the mistake in this proof of product rule of differentiation?

Question

I was trying to derive the product rule of differentiation which states:

If $y=u\cdot v$, then, $y'=u'\cdot v+v'\cdot u$.

So I assumed it like this:

$y=u+u+u+\cdots$ ($v$ number of terms of $u$)

Differentiating

$y'=u'+u'+u'+\cdots$ ($v$ number of terms of $u'$)

Therefore

$y'=u'\cdot v$

Which is clearly wrong...

But I can't see the mistake in this. Please help me point it out.

Note: $y, u, \text{ and } v$ are functions. I know this proof works if either $v$ or $u$ is a constant. We will then get the expected result. Also, I know the proof by method of first principle. I just want to know what's the mistake in this one..

v is a function, so what does "v number of terms of u” mean? — aPaulT, May 14 '14 at 18:36
Well, v(x) is a number for a given value of x. What value of x are you going to choose? — aPaulT, May 14 '14 at 18:47
if $v$ is a continuous function (which it needs to be to be differentiable), what does $$\overbrace{u+u+u+\dots+u}^{v\text{ times}}$$ mean when $v$ is not an integer? If $v$ is integer-valued, then note that all continuous, integer-valued functions on $\mathbb{R}$ are constant. — robjohn, May 14 '14 at 19:22
See Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition) and A contradiction involving derivative and the Winter 2004 issue of North Carolina Association of Advanced Placement Mathematics Teachers. In the last one, scroll down to Dave Renfro posed the following. — Dave L. Renfro, May 14 '14 at 19:28
@DaveL.Renfro You Sir, are great. That should be an answer, an accepted one... — user3459110, May 14 '14 at 19:33
@aPaulT Now I am thinking I am an idiot: you make complete sense... — user3459110, May 18 '14 at 15:50

score 1 · Answer 1 · answered May 14 '14 at 18:26

1

$v$ is a number, not a function (at least this is how I interpret "$v$ terms of $u$), so your product rule is $(uv)^\prime=u^\prime v+uv^\prime=vu^\prime$, since $v^\prime=0$ for a number.

answered May 14 '14 at 18:26

No, v is not a number, its a function.. – user3459110 May 14 '14 at 18:30
Then you can't write $y=u+u+\ldots u$ with "$v$ terms of $u$". – May 14 '14 at 18:31
But why? Even if it is a function, it evaluates to a number which is variable.. isn't it? (Sorry I am not good at maths, I am a physics kinda guy.) – user3459110 May 14 '14 at 18:32
It's right that it evaluates to a number, but all functions do. The point is that you have to "differentiate the dependency" of the number of terms, since they may vary with respect to the argument of $v$. – May 14 '14 at 18:39
I am sorry if I am over doing things.., but I didn't understand how would I "differentiate the dependency".. Whats that? – user3459110 May 14 '14 at 18:42
That was just sloppy language. What I mean is that for any given $x$ you can write $y(x)=v(x)u(x)$ and interpret this as something like "$y(x)=u(x)+\ldots+u(x)$ with $v(x)$ terms of $u(x)$". If you want to differentiate that equation with respect to $x$, you have to differentiate $u(x)v(x)$ as a product, not as a (somewhat constant) multiple of $u(x)$. – May 14 '14 at 18:49
let us continue this discussion in chat – user3459110 May 14 '14 at 18:54

score 1 · Accepted Answer · edited Feb 09 '23 at 18:13

I'm using a revised version of one of my comments as an answer.

One day, when I'm not as busy as I am now, I'll try to write this in a format that works here (copying and pasting doesn't work), but for now I'll just cite something I wrote back in 2003. At the time I was having trouble posting to a certain discussion group where it came up, so I emailed my comments to the original poster, who then posted my comments. A few months later my short essay was reposted in the Winter 2004 issue of North Carolina Association of Advanced Placement Mathematics Teachers.

See also these two Math StackExchange questions:

Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition)

A contradiction involving derivative

(ADDED NEXT DAY) I had a few minutes free just now . . .

Some have said that the issue has to do with $v$ not being integer-valued. However, that's not the essential point, since we could consider the situation where $u$ and $v$ have integer values in the trillions (or $10^{100}$'s, or $10^{1000}$'s, etc.) and increment $u$ and $v$ by small integer amounts, which should provide very close approximations to the behavior of the corresponding differentials, but Awal Garg's calculations do not give close approximations. In fact, Awal Garg's version still shows up in the limit by allowing $u$ and $v$ to be arbitrarily large integers and incrementing them by integer amounts.

robjohn's comment above gives me an idea of how to show what's going one without drawing the rectangles and stars that were used in my 2003 explanation. $$uv \;\; = \;\; \overbrace{u + u + \dots + u}^{v\;\text{times}}$$ $$uv + {\Delta}(uv) \;\; = \;\; \overbrace{(u + {\Delta}u) + (u + {\Delta}u) + \dots + (u + {\Delta}u)}^{v + {\Delta}v\;\text{times}}$$ $$= \;\;\overbrace{u + u + \dots + u}^{v + {\Delta}v\;\text{times}} \;\; + \;\; \overbrace{{\Delta}u + {\Delta}u + \dots + {\Delta}u}^{v + {\Delta}v\;\text{times}}$$ $$= \overbrace{u + u + \dots + u}^{v\;\text{times}} + \overbrace{u + u + \dots + u}^{{\Delta}v\;\text{times}} + \overbrace{{\Delta}u + {\Delta}u + \dots + {\Delta}u}^{v\;\text{times}} + \overbrace{{\Delta}u + {\Delta}u + \dots + {\Delta}u}^{{\Delta}v\;\text{times}}$$ $$ = \;\; uv \; + \; u({\Delta}v) \; + \; ({\Delta u})v \; + \; ({\Delta}u)({\Delta}v) $$ Therefore, $$\left[\, uv + {\Delta}(uv) \, \right] \; - \; uv \;\; = \;\; u({\Delta}v) \; + \; ({\Delta u})v \; + \; ({\Delta}u)({\Delta}v) $$ At this point Awal Garg got $\;{\Delta}(uv) \; = \; ({\Delta u})v,\;$ and we're getting two additional terms. One of these additional terms is not a "first order smallness term", so we can drop it, leaving us with $${\Delta}(u \cdot v) \;\; \approx \;\; u \cdot {\Delta}v \; + \; {\Delta u} \cdot v \;\;\; \text{(approximate version for positive increments)}$$ $$d(u \cdot v) \;\; = \;\; u \cdot dv \; + \; du \cdot v \;\;\; \text{(exact version for infinitesimal increments)}$$

What is the mistake in this proof of product rule of differentiation?

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