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We know $x.x=x^2$(Consider $x\ne 0$)

$x.x$ is adding $x$ $x$ times. So we have

$x+x+\dots+x$($x$ times $x$ is added)=$x^2$......(1)

Differentiating both sides of (1) we get,

$1+1+\dots+1$(x times)=$2.x$

$\Rightarrow x=2x$

$1=2$ Proved.

Find out the mistake in this proof.

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How does something occur $x$ times if $x\notin\mathbb{N}$? –  Ben West Feb 15 '13 at 7:01
You forgot the chain rule; you need to account for the derivative of the operation "... (x times)" (which, of course, doesn't exist since it's restricted to $x$ being an integer) –  Hurkyl Feb 15 '13 at 7:04
there is one evident mistake: x times is valid when x is integer. when it is integer differentiation is not valid. –  user59671 Feb 15 '13 at 7:06
The same reasoning shows that differentiating $x^2=\underbrace{1+\cdots+1}_{x^2\text{ times}}$ we get $0$. –  Stefan Hansen Feb 15 '13 at 7:13
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marked as duplicate by MJD, Stefan Hansen, Hurkyl, Michael Albanese, Jesse Madnick Feb 15 '13 at 7:09

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1 Answer

The equation $$x^2 = x\cdot x = \underbrace{x + \dots + x}_{\text{$x$ times}}$$ only holds when $x$ is a positive integer (what does it mean to add $\pi$ lots of $\pi$?), so the left hand side of the equation following (1) is not a valid expression for the derivative of $x^2$.

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Wait... This really doesn't seem like the issue to me. I don't think the problem is that the equation only holds when $x \in \mathbb{N}$, but rather that the number of times that addition is repeated is itself a variable quantity! In other words, the $x$ in "$x$ times" needs to be accounted for in the differentiation (if that were somehow possible computationally). –  Jesse Madnick Feb 15 '13 at 7:11
It's the same flaw as saying that the derivative of $f(x) = x^x$ is $f'(x) = x\cdot x^{x-1}$. –  Jesse Madnick Feb 15 '13 at 7:13
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