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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
8 months, 3 weeks 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 · 8 months, 3 weeks 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 · 8 months, 1 week ago

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A link to solution

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Joel Yip · 8 months, 2 weeks ago

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