I'm came across this theorem in a calculus book:

Suppose $n$ is an integer, $n \geq 2$ and $f(x) = x^\dfrac{1}{n}$. Then $f'(x) = (1/n)x^{\dfrac{1}{n} - 1}$. If $n$ is even then the domain of $f'$ is $(0, \infty)$, and if $n$ is odd then it is $(-\infty, 0) \cup (0, \infty)$

I'm having a diffcult time understanding how the domain of $f'$ is satisfied in the above theorem.

Let's take the example of when $n$ is even,

When $n$ is odd, let's consider a case. When $n=3$, the $f'(x)$ is $\dfrac{1}{3*x^{2/3}} $

Having negative numbers of $x$ will lead to $\sqrt{-1}$ etc. But the domain seems to include negative numbers.

  • $\begingroup$ It should be $f'(x)=\frac{1}{n}x^{\frac{1}{n}-1}$, there seems to be a typo in your book $\endgroup$ – Peter Melech Aug 1 at 13:02
  • $\begingroup$ @PeterMelech I think that's what I have written in the question too. I guess the (1/n) is confusing for the readers. $\endgroup$ – Sibi Aug 1 at 13:05
  • $\begingroup$ Ah, okay. I see what you mean. $\endgroup$ – Sibi Aug 1 at 13:07
  • $\begingroup$ No, the exponent is $\frac{1}{n}-1=\frac{1-n}{n}\neq \frac{1}{n-1}$ $\endgroup$ – Peter Melech Aug 1 at 13:08
  • 1
    $\begingroup$ I'm sorry, I have to leave so that I cannot continue this discussion, but I'll be back here $\endgroup$ – Peter Melech Aug 1 at 13:45

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