### 4 - Power Rule for Differentiation (3)

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

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

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

$=\underset{h\to 0}{\mathrm{lim}}\frac{\left[{\left(x+h\right)}^{\frac{1}{n}}-{x}^{\frac{1}{n}}\right]\left[{\left(x+h\right)}^{\frac{n-1}{n}}+{\left(x+h\right)}^{\frac{n-2}{n}}{x}^{\frac{1}{n}}+{\left(x+h\right)}^{\frac{n-3}{n}}{x}^{\frac{2}{n}}+\cdot \cdot \cdot +{\left(x+h\right)}^{\frac{1}{n}}{x}^{\frac{n-2}{n}}+{x}^{\frac{n-1}{n}}\right]}{h\left[{\left(x+h\right)}^{\frac{n-1}{n}}+{\left(x+h\right)}^{\frac{n-2}{n}}{x}^{\frac{1}{n}}+{\left(x+h\right)}^{\frac{n-3}{n}}{x}^{\frac{2}{n}}+\cdot \cdot \cdot +{\left(x+h\right)}^{\frac{1}{n}}{x}^{\frac{n-2}{n}}+{x}^{\frac{n-1}{n}}\right]}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{\left(x+h\right)-x}{h\left[{\left(x+h\right)}^{\frac{n-1}{n}}+{\left(x+h\right)}^{\frac{n-2}{n}}{x}^{\frac{1}{n}}+{\left(x+h\right)}^{\frac{n-3}{n}}{x}^{\frac{2}{n}}+\cdot \cdot \cdot +{\left(x+h\right)}^{\frac{1}{n}}{x}^{\frac{n-2}{n}}+{x}^{\frac{n-1}{n}}\right]}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{1}{{\left(x+h\right)}^{\frac{n-1}{n}}+{\left(x+h\right)}^{\frac{n-2}{n}}{x}^{\frac{1}{n}}+{\left(x+h\right)}^{\frac{n-3}{n}}{x}^{\frac{2}{n}}+\cdot \cdot \cdot +{\left(x+h\right)}^{\frac{1}{n}}{x}^{\frac{n-2}{n}}+{x}^{\frac{n-1}{n}}}$

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

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

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