### 6 - Power Rule for Differentiation (5)

Let $y=f\left(x\right)={x}^{\frac{m}{n}}$ ,where $m$ is a positive integer and $n$ is any integer. Then

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

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

$=\underset{h\to 0}{\mathrm{lim}}\frac{{\left({\left(x+h\right)}^{\frac{1}{n}}\right)}^{m}-{\left({x}^{\frac{1}{n}}\right)}^{m}}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{\left[{\left(x+h\right)}^{\frac{1}{n}}-{x}^{\frac{1}{n}}\right]\left[{\left({\left(x+h\right)}^{\frac{1}{n}}\right)}^{m-1}+{\left({\left(x+h\right)}^{\frac{1}{n}}\right)}^{m-2}{x}^{\frac{1}{n}}+\cdot \cdot \cdot +{\left(x+h\right)}^{\frac{1}{n}}{\left({x}^{\frac{1}{n}}\right)}^{m-2}+{\left({x}^{\frac{1}{n}}\right)}^{m-1}\right]}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{\left(x+h{\right)}^{\frac{1}{n}}-{x}^{\frac{1}{n}}}{h}\left[{\left({\left(x+h\right)}^{\frac{1}{n}}\right)}^{m-1}+{\left({\left(x+h\right)}^{\frac{1}{n}}\right)}^{m-2}{x}^{\frac{1}{n}}+\cdot \cdot \cdot +{\left(x+h\right)}^{\frac{1}{n}}{\left({x}^{\frac{1}{n}}\right)}^{m-2}+{\left({x}^{\frac{1}{n}}\right)}^{m-1}\right]$

$=\frac{1}{n}{x}^{\frac{1}{n}-1}\left[m{x}^{\frac{m-1}{n}}\right]$

$=\frac{m}{n}{x}^{\frac{m}{n}-1}$