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n^2 = n + n + n ... (n times)

derivative of n^2 = derivative of n + n + n + n .. (n times)

thus, 2n = 1 + 1 + ... (n times)

thus, 2n = n

and 2 = 1

now i know that the mistake lies in taking the derivative of both sides, because all other steps are correct. So I tried the same thing with n^3, where i wrote n^3 = n^2 into n. Following the same steps as above i got 3 = 2.

derivative of n^3 = derivative of n^2 + n^2 + .. (n times)

thus, 3n^2 = 2n + 2n + .... (n times)

thus, 3n^2 = 2n^2

anyway thats why i think its incorrect to take the derivative like that? but why? is it cause by writing 'n times' is basically just mulitplication and should be written in the correct way? or is it cause when we say n times we are treating n as a constant, when its actually a variable and can assume the value we want? can someone please help Im really confused

anomaly
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Vanessa
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    Possible duplicate of [Where is the flaw in this "proof" that 1=2? (Derivative of repeated addition)](https://math.stackexchange.com/questions/1096/where-is-the-flaw-in-this-proof-that-1-2-derivative-of-repeated-addition), and also https://math.stackexchange.com/questions/1473005/what-is-wrong-with-this-derivation-of-fracddxx2/ – Arthur Mar 28 '19 at 14:41
  • @DietrichBurde The implication $1=0\implies 0=0$ by taking derivatives is perfectly correct. – user647486 Mar 28 '19 at 14:43
  • Cf. also [this question](https://math.stackexchange.com/questions/3160946/why-we-cant-differentiate-both-sides-of-a-polynomial-equation) – J. W. Tanner Mar 28 '19 at 16:43

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