Why $2x$? Can't it be $x$?

Question

So today in my school our neighbor class monitors were complaining to that few of our students were yelling and making noise. Actually the case was that we were having very aggressive debate over a question. The question is quite simple but there is lot of confusion regarding it's solution because according to different students there are two very different approach toward this question. The question is,

Question: Find the derivative of $$f(x)=x+x+x+x+x+...\quad x \mathrm\quad {times}$$

So here are two approach that my classmates are pretending to do.

First one (that I think is the correct approach) is that we can simplify the above function as $x$ times $x$ ( i.e. $f(x)=x^2$**)** , so that equation becomes,

$$f(x)=x^2$$

and simply we can write

$$f'(x)=2x$$

But my classmates are pretending to say that there is another way ( that I think is wrong ) to do this. In this method without simplifying the $f(x)$ we can differentiate individual terms in the sum. i.e.

$$f'(x)=\frac{d(x)}{dx}+\frac{d(x)}{dx}+\frac{d(x)}{dx}+\frac{d(x)}{dx}+...\quad x \quad\mathrm{times}$$ $$\implies f'(x)=1+1+1+1+1...\quad x \quad \mathrm{times}$$

$$\implies f'(x)=x$$

So in this way we are having two different answers (i.e $2x$ and $x$). However I know that $2x$ is the correct answer but honestly speaking I am really tiered to explain them how. However I tried to tell them that in second method (that I think is wrong method) we unintentionally ignore that $x$ times term and didn't involve in the differentiation while we do take care of that in the first method (that I think is correct).

Some are saying that $x$ is right while some are pretending to $2x$ to be right while some are saying that that both are wrong. I need a perfect answer that can satisfy my classmates.

Thanks!

What about $x=2.5$ or $x=\pi$? Addition in $\mathbb R$ cannot be defined via a sum of $n$ summands easily. — CBenni, Jul 03 '14 at 16:24
This is not particularly well-defined; however, if you want to restrict to cases wherein $x$ is an integer, and the find the derivative of your function's extention to the reals, then you have to recognize that "$x$ times" is part of your function's definition, and cannot be excluded from the differentiation process. — Emily, Jul 03 '14 at 16:28
To show your classmates they are wrong try simply to show how their method fails with $$x^2+x^2+x^2+x^2+\dots\quad x\ \mbox{times}\ .$$ — Dario, Jul 03 '14 at 16:34
Also virtually the same: What is the mistake in this proof of product rule of differentiation? — Dave L. Renfro, Jul 03 '14 at 17:19

score 11 · Accepted Answer · edited Apr 20 '19 at 14:30

11

You have to use the product rule:

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} (\underbrace{x + x + x + \cdots + x}_{x\text{ times}}) &= (\underbrace{1 + 1 + 1 + \cdots + 1}_{x\text{ times}}) + (\underbrace{x + x + x + \cdots + x}_{1\text{ time}}) \\ &= x + x \\ &= 2x. \end{align*}

edited Apr 20 '19 at 14:30

L. F.

1,940

answered Jul 03 '14 at 16:25

Umberto P.

52,165

score 3 · Answer 2 · answered Jul 03 '14 at 16:33

3

The second evaluation is a function on the naturals, and therefore no derivative exists.

answered Jul 03 '14 at 16:33

This doesn't really get at the issue, as I explained in my answer to What is the mistake in this proof of product rule of differentiation?. – Dave L. Renfro Jul 03 '14 at 17:21

score 2 · Answer 3 · answered Jul 03 '14 at 16:29

You are correct. The second "method" has a variable number of terms and so the distribution of the derivative sign is incorrect, you can't distribute over the variables if you don't know how many there are. Take a look at the definition of the derivative, it has an infinitesimal (usually called h) which approaches zero. This would mean that the number of terms is not a whole number.

Why $2x$? Can't it be $x$?

3 Answers3