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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
1 year, 8 months 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.

- 1 year, 8 months ago

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

- 1 year, 8 months ago

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- 1 year, 8 months ago