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

\begin{align*} x^2 &= \underbrace{x + x + x + \dots + x}_{x \text{ times}}, \\ \therefore \frac{\mathrm{d}}{\mathrm{d}x} (x^2) &= \frac{\mathrm{d}}{\mathrm{d}x} (\underbrace{x + x + x + \dots + x}_{x \text{ times}}) \\ &= \underbrace{1 + 1 + 1 + \dots + 1}_{x \text{ times}} \\ &= x. \end{align*}

But we know that $$\frac{\mathrm{d}}{\mathrm{d}x} (x^2) = 2x.$$

So what is the problem?

My take is that we cannot differentiate both sides because $$\underbrace{{x+x+x+\cdots+x}}_{x \text{ times}}$$ is not fixed 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? – Yai0Phah 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. – Yai0Phah 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