\(x\cdot x = x + x + x +x +x+x...\), 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?

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## Comments

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TopNewestYou 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.

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

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