### 9 - Power Rule for Differentiation (6)

Suppose $y=f\left(x\right)={x}^{q}$  where $q$  is an irrational number. And, suppose ${\left\{{r}_{k}\right\}}_{k=1}^{k=\infty }$  is an infinite sequence of rational numbers such that $\underset{k\to \infty }{\mathrm{lim}}{r}_{k}=q$. Then

$\frac{dy}{dx}=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(x+h\right)}^{q}-{x}^{q}}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(x+h\right)}^{\underset{k\to \infty }{\mathrm{lim}}{r}_{k}}-{x}^{\underset{k\to \infty }{\mathrm{lim}}{r}_{k}}}{h}$

$=\underset{k\to \infty }{\mathrm{lim}}\underset{h\to 0}{\mathrm{lim}}\frac{{\left(x+h\right)}^{{r}_{k}}-{x}^{{r}_{k}}}{h}$

$=\underset{k\to \infty }{\mathrm{lim}}{r}_{k}{x}^{{r}_{k}-1}$

$=\underset{k\to \infty }{\mathrm{lim}}{r}_{k}\underset{k\to \infty }{\mathrm{lim}}{x}^{{r}_{k}-1}$

$=q{x}^{q-1}$