### 2 - Power Rule for Differentiation (1)

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

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

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

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

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

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

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

If $y=f\left(x\right)=k$  , where $k$  is a constant, then

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

$=\underset{h\to 0}{\mathrm{lim}}\frac{k-k}{h}$

$=\underset{h\to 0}{\mathrm{lim}}\frac{0}{h}$

$=\underset{h\to 0}{\mathrm{lim}}0$

$=0$

In particular, if $y=f\left(x\right)=1={x}^{0}$  , then

$\frac{dy}{dx}=0=0{x}^{0-1}$

Thus we see that if $y=f\left(x\right)={x}^{n}$  , where $n$  is a nonnegative integer, then

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