5
$\begingroup$

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?

$\endgroup$
9
  • $\begingroup$ 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 $\endgroup$ 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
    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 $x^(1-i)$ $\endgroup$ Jul 27, 2016 at 0:39
  • $\begingroup$ @RossMillikan Thanks for the tip. $\endgroup$
    – Sigma6RPU
    Jul 27, 2016 at 0:40
  • 2
    $\begingroup$ If $x^i=e^{i \ln(x)}$ then its derivative should be $e^{i \ln(x)}\frac{i}{x}=x^{i}\frac{i}{x}=ix^{i-1}$ $\endgroup$ Jul 27, 2016 at 0:42

2 Answers 2

5
$\begingroup$

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:

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

$\endgroup$
4
  • $\begingroup$ So even if the c is an imaginary number, c would simply be that number? $\endgroup$
    – Sigma6RPU
    Jul 27, 2016 at 17:08
  • $\begingroup$ @Sigma6RPU I don't understand your question. $\endgroup$
    – user258700
    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
    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
    Jul 27, 2016 at 20:12
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.