Author Question: What is the derivative of the imaginary number i? (Read 1705 times)

coco

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I thought of using the definition of the derivative to find it: f'(x)=lim (h->0) (f(x+h)-f(x))/h = lim ((i+h)-i)/h = lim h/h = lim 1 = 1. Is this right? Are there other proofs?



Jesse_J

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Actually i acts just as any other constant does. So the derivative of i with respect to some variable is 0.

Proof:
Let f(x) = i
Square both sides:
[f(x)]² = i²
[f(x)]² = -1
Implicitly differentiate:
2 f(x) f'(x) = 0
f(x) f'(x) = 0
i f'(x) = 0
f'(x) = 0 / i
f'(x) = 0



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Joesy

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The number i is a constant, so its derivative is zero.  I think?



 

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