Let y = x^(a/b) for integers "a" and "b". Raise both sides to the "bth" power and use implicit differentiation to prove the power rule y' = (a/b)x^(a/b-1).
Hello, erimat89!
I'll do it in LaTex . . .
Let $\displaystyle y \:=\: x^{\frac{a}{b}}$ for integers $\displaystyle a\text{ and }b$
Raise both sides to the power $\displaystyle b$ and use implicit differentiation
to prove the power rule: .$\displaystyle y' \:=\:\frac{a}{b}\!\cdot\!x^{\frac{a}{b}-1}$
We have: .$\displaystyle y \;=\;x^{\frac{a}{b}}$ .[1]
Raise to the $\displaystyle b^{th}$ power: .$\displaystyle y^b \:=\:x^a$ .[2]
Differentiate implicitly: .$\displaystyle by^{b-1}y' \;=\;ax^{a-1} \quad\Rightarrow\quad y' \;=\;\frac{a}{b}\cdot\frac{x^{a-1}}{y^{b-1}}
$ .[3]
Divide [2] by $\displaystyle y\!:\;\;y^{b-1} \:=\:\frac{x^a}{y}$
Substitute [1]: .$\displaystyle y^{b-1} \:=\: \frac{x^a}{x^{\frac{a}{b}}} \:=\:x^{a-\frac{a}{b}}$
Substitute into [3]: .$\displaystyle y' \;=\;\frac{a}{b}\cdot\frac{x^{a-1}}{x^{a-\frac{a}{b}}} \quad\Rightarrow\quad\boxed{ y' \;=\;\frac{a}{b}\!\cdot\!x^{\frac{a}{b}-1}}$