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:
- How do I rationalize the numerator?
- 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?
Best Answer
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 $x^{m/n}$.