# How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+\cdots\ (x\text{ times})$ are not same. [duplicate]

$$x^2=\underbrace{{x+x+x+\cdots+x}}_{x \text{ times}}$$

$$\therefore \frac {dy}{ dx}\\ =\frac{d}{dx} \left( \underbrace{{x+x+x+\cdots+x}}_{x \text{ times}} \right) \\ = \underbrace{{1+1+1+\cdots+1}}_{x \text{ times}} \\ =x$$

but we know, $$\frac{dy}{dx} =2x$$ So what is the problem ?

My take :

We cannot derivate both sides because the terms $\underbrace{{x+x+x+\cdots+x}}_{x \text{ times}}$ is not finite and thus $1$ is not equal to $2$.

-

## marked as duplicate by Nate Eldredge, Douglas S. Stones, Rahul, Hans Lundmark, Zev ChonolesJun 29 '12 at 7:17

What do you mean by $x$ times? –  Frank Science Jun 29 '12 at 3:55
Adding x, x times x+x+x+x....(x times x) –  Bazinga Jun 29 '12 at 3:56
Worth reading. (maa.org/devlin/devlin_0708_08.html) –  user17762 Jun 29 '12 at 4:03
Note that the terms will be finite for any finite value of $x$ –  Pedro Tamaroff Jun 29 '12 at 4:05
$x^2 = \lfloor x\rfloor x + (x-\lfloor x\rfloor)x$. The first part, $\lfloor x\rfloor x$, is roughly like adding $x$ "$x$ times" while still adding an integer number of $x$s. The derivative of $\lfloor x\rfloor x$ is $\lfloor x\rfloor$ where it is defined. –  Jonas Meyer Jun 29 '12 at 4:40

Simply because "$x \text{ times}$" is also a "function" of $x$. One mistake is not considering that in the derivation.
You say "$x\text{ times}$". The number of "times" you add it up---the number of terms in the sum---keeps changing as $x$ changes. An what if $x=1.6701$? How do you add up $x$ $1.6701$ times?
The arrangement of math equations in the question is very ugly. Help to edit it, such as $\underbrace{x+x+\cdots+x}_{x\textrm{ times}}$ instead of the one in the question, and displaymath is not necessary. His $dy/dx$ can be arranged with inline math. –  Frank Science Jun 29 '12 at 4:21
Concerning defining repeated addition for the reals, was your question for the OP's author's consideration? I think the problem is not with defining such addition, but with using the definition for $x^2$ that has been used in the OP. –  ThisIsNotAnId Jul 27 '12 at 3:36