What is the derivative of the imaginary number i?

[coco]

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]

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

[Joesy]

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