What is the derivative of $x^i$? What would the derivative be of $x^i$? Would it simply be $ix^{(1-i)} $? I tried running the Power rule, and I got that is that right?
 A: Let $c \in \Bbb C$ be any complex number. By definition, the function $f(x) = x^c$ is given by $\exp(c \log x)$, where, typically, we take $\log$ to be the principal branch of logarithm. This function is not differentiable at any point in $(-\infty, 0]$ (though, it is differentiable everywhere else). So, we have to assume that $x \in \Bbb C \setminus (-\infty, 0]$. If this is the case,
\begin{align*}
f'(x) &=c \left( \frac{d}{dx} \log x \right) \exp( c \log x)\\
 &= c \frac1{x} \exp(c \log x)\\
 &=  c \exp(-\log x) \exp(c \log x)\\
 &= c \exp( (c -1) \log x) \\
 &= c x^{c -1}
\end{align*}
In the particular case where the complex number $c$ is equal to $i$, we have:
$$\frac{d}{dx} x^i = ix^{i-1}$$
A: I will assume that $x \in \mathbb{R}$ and then $f(x) = x^{i}$ is a complex valued function of a real variable. A typical definition of symbol $x^{i}$ for real $x$ is $$x^{i} = \exp(i\log x)$$ where the symbol $\exp(a + ib)$ for real $a, b$ is defined to be $$\exp(a + ib) = \exp(a)(\cos b + i\sin b)$$ Thus $f(x)$ is defined only for $x > 0$ and we have $$f(x) = \cos\log x + i\sin\log x$$ and hence
\begin{align}
f'(x) &= -\frac{\sin \log x}{x} + i\frac{\cos \log x}{x}\notag\\
&= \frac{i}{x}\{\cos \log x + i\sin\log x\}\notag\\
&= \frac{i}{x}\exp(i\log x)\notag\\
&= \frac{i}{x}x^{i}\notag\\
&= ix^{i - 1}\notag
\end{align}
When $x \in \mathbb{C}$ then it is best to use symbol $z$ instead of $x$ and then also the derivative is same, but this requires a knowledge of theory of elementary analytic functions like $\exp(z),\log(z)$ for $z \in \mathbb{C}$. An answer based on this knowledge is already supplied by Ahmed Hussein.
