\[\underbrace{x+x+x+x+\cdots+x}_{x \times x } = x^{2}\]

Differentiating both sides with respect to \(x\)

\[\underbrace{1+1+1+1+\cdots+1}_{x \times 1 } = 2x\]

\[\implies x=2x\]

\[\underbrace{x+x+x+x+\cdots+x}_{x \times x } = x^{2}\]

Differentiating both sides with respect to \(x\)

\[\underbrace{1+1+1+1+\cdots+1}_{x \times 1 } = 2x\]

\[\implies x=2x\]

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## Comments

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TopNewestEquation 1 is only true when \(x\) is an positive integer. It is not continuous and therefore non-differentiable. Therefore, equation 2 does not apply. – Chew-Seong Cheong · 1 year, 2 months ago

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– Hritesh Mourya · 1 year, 2 months ago

Thanks a lotLog in to reply

@Hritesh Mourya here is your answer – Satyabrata Dash · 1 year, 2 months ago

Absolutely perfect.Log in to reply

– Chloe Marshall · 11 months, 2 weeks ago

You have no answerLog in to reply

How many times does this question keep popping up? @Calvin Lin

In one month I've seen two of these. Can't it added to the misconceptions page? – Deeparaj Bhat · 1 year, 2 months ago

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