2 - Power Rule for Differentiation (1)

Let $y=f(x)=x^n$ , where $n$ is a positive integer. Then

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

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

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

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

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

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

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

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

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

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

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

$=0$

In particular, if $y=f(x)=1=x^0$  , then

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

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

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