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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?

Note by Choi Chakfung
6 months, 1 week ago

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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. Deeparaj Bhat · 6 months, 1 week ago

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2x = 1 +1+1+1+1+1......(2x times) 2x =2x Not 2x=x So 2 not same with 1 Ivan Tanuwijaya · 5 months, 3 weeks ago

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A link to solution

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Joel Yip · 6 months ago

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