Differentiation show 2=1?

Given that $6\times6 = 6+6+6+6+6+6$ , or equivalently
$x\cdot x = x + x + x +x +x+x...$ , where $x= 6$ .
Differentiate both sides with respect to $x$ , we get
$\dfrac{d}{dx} (x^2) = \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x + \dfrac d{dx} x...$
$2x = 1 + 1 + 1 + 1 + 1 + 1 ...= 1\cdot x$
$2x = x$
$2 = 1$ . why?

#Calculus

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Comments

You can't differentiate an equation until it's an identity. Also, both sides must be defined in the neighbourhood of the point at which you're differentiating. As these statements don't hold for the above equation, the proof is flawed.

A Former Brilliant Member - 5 years, 2 months ago

A link to solution

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Joel Yip - 5 years, 2 months ago

2x = 1 +1+1+1+1+1......(2x times) 2x =2x Not 2x=x So 2 not same with 1

Ivan Tanuwijaya - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$