### 3 - Power Rule for Differentiation (2)

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

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

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

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

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

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

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

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

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

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

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