4
$\begingroup$

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?
$\endgroup$
1
6
$\begingroup$

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 $x^{m/n}$.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – Slecker
    Aug 6 '20 at 0:47
  • $\begingroup$ 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. $\endgroup$ Aug 6 '20 at 1:04
  • $\begingroup$ 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. $\endgroup$
    – Slecker
    Aug 6 '20 at 1:56
  • 1
    $\begingroup$ 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). $\endgroup$
    – robjohn
    Aug 6 '20 at 17:04
6
$\begingroup$

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 $B=x^{2/3}$

$\endgroup$

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.