### 5 - Power Rule for Differentiation (4)

Let $y=f\left(x\right)={x}^{\frac{1}{n}}$ , where $n$ is a negative integer. Then $n=-m$ , where $m$ is a positive integer. So,

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

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

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

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

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

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

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

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

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

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