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?

  • $\begingroup$ How are you defining $x^i$? It will be $ix^{i-1}$ in most cases, however. Not sure how you got used the power rule $\endgroup$ Commented Jul 27, 2016 at 0:38
  • $\begingroup$ @ThomasAndrews Sorry I don't know how to raise the x to the power of $i-1$. $\endgroup$
    – Sigma6RPU
    Commented Jul 27, 2016 at 0:39
  • $\begingroup$ MathJax hint; to get multicharacter exponents, enclose them in braces. So x^{(1-i)} gives $x^{(1-i)}$ instead of $x^(1-i)$ $\endgroup$ Commented Jul 27, 2016 at 0:39
  • $\begingroup$ @RossMillikan Thanks for the tip. $\endgroup$
    – Sigma6RPU
    Commented Jul 27, 2016 at 0:40
  • 2
    $\begingroup$ If $x^i=e^{i \ln(x)}$ then its derivative should be $e^{i \ln(x)}\frac{i}{x}=x^{i}\frac{i}{x}=ix^{i-1}$ $\endgroup$ Commented Jul 27, 2016 at 0:42

2 Answers 2


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:

$$\frac{d}{dx} x^i = ix^{i-1}$$

  • $\begingroup$ So even if the c is an imaginary number, c would simply be that number? $\endgroup$
    – Sigma6RPU
    Commented Jul 27, 2016 at 17:08
  • $\begingroup$ @Sigma6RPU I don't understand your question. $\endgroup$
    – user258700
    Commented Jul 27, 2016 at 19:16
  • $\begingroup$ If the constant C were to be an imaginary number would it be simply that number? $\endgroup$
    – Sigma6RPU
    Commented Jul 27, 2016 at 19:50
  • $\begingroup$ @Sigma6RPU I honestly don't understand what you are talking about. I'll try to edit the post to make things more precise. $\endgroup$
    – user258700
    Commented Jul 27, 2016 at 20:12

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.


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