# Understanding domains mentioned in differentiation of nth root

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.

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