# How to rationalize multiple terms with fractional exponents

I'm trying to derive the derivative of $$f(x) = x^{2/3}$$ using the limit definition:

$$f'(x)=\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

$$=\lim_{h \to 0} \frac{(x+h)^{2/3} - x^{2/3}}{h}$$

I suspect I have to rationalize the numerator in order to cancel an $$h$$ from the numerator and denominator, but I'm not sure how to rationalize the numerator. I've tried multiplying by the conjugate and even tried to render the numerator a difference of cubes and then using $$A^3 -B^3 = (A - B)(A^2 +AB + B^2)$$ to rationalize, but to no avail.

My two questions are:

1. How do I rationalize the numerator?
2. Is there a general formula for rationalizing multiple terms with rational exponents? Is there something I can read or study to learn more about this?

The numerator is a difference of cube-roots, not a difference of cubes. So rationalizing the numerator: $$\frac{(x+h)^{2/3}-x^{2/3}}{h}= \frac{(x+h)^2-x^2}{h[(x+h)^{4/3}+x^{4/3}(x+h)^{2/3}+x^{4/3}]}$$ gives the factor of $$h$$ you want in the numerator. Hence \begin{align*} \require{cancel} \lim_{h\to 0} \frac{(x+h)^{2/3}-x^{2/3}}{h} &=\lim_{h\to 0} \frac{\cancel{h}(2x+h)}{\cancel{h}[(x+h)^{4/3}+x^{4/3}(x+h)^{2/3}+x^{4/3}]}\\ &=\frac{2x}{3x^{4/3}}=\frac23 x^{-1/3}. \end{align*}

In general, $$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\dots+B^{n-1})$$, so $$\alpha^{m/n}-\beta^{m/n}=\frac{\alpha^m-\beta^m}{\alpha^{(n-1)m/n}+\alpha^{(n-2)m/n}\beta^{m/n}+\dots+\beta^{(n-1)m/n}}$$ which allows you to deal with the derivative of $$x^{m/n}$$.

• Thank you. I'm curious what did you read or study to come up with the general formula for $A^n -B^n$? Is this something normally taught in university or was this something extra you picked up on your own? What area of math would this fall under? Aug 6 '20 at 0:47
• The formula is a very obvious generalization of $A^2-B^2=(A-B)(A+B)$, $A^3-B^3=(A-B)(A^2+AB+B^2)$, ... and you can come up with it very naturally from summing an $n$ term geometric progression $1+r+r^2+\dots+r^{n-1}=(1-r^n)/(1-r)$ letting $r=B/A$ and clear some denominators. It is definitely secondary school level --- back in the days it was in the "formula book" for O-Level and A-Level exams, not sure if it is still the case now. Aug 6 '20 at 1:04
• Unfortunately, the general formula was never taught (at least not in my classes) and only the expansions for $A^2 - B^2$ and $A^3 - B^3$ were given. Perhaps a quirk of schooling in the U.S. Aug 6 '20 at 1:56
• For this problem, $A^3-B^3=(A-B)(A^2+AB+B^2)$ is sufficient, and that was given. The formula $A^n-B^n=(A-B)\left(A^{n-1}+A^{n-2}B+A^{n-3}B^2+\cdots+B^{n-1}\right)$ is valid and follows from the formula for the sum of a geometric series; however, it is not needed for this problem (but it is quite useful and should be remembered).
– robjohn
Aug 6 '20 at 17:04

Idea

You have $$\frac{A-B}{h}=\frac{(A-B)}{h}\frac{(A^2+AB+B^2)}{(A^2+AB+B^2)}=\frac{A^3-B^3}{h(A^2+AB+B^2)}.$$ Now let $$A=(x+h)^{2/3}$$ and $$B=x^{2/3}$$