Power Rule for Derivatives

Theorem: If $$f(x)=x^n$$, $$\forall x\in R$$ and $$\forall n \in N$$, then $$\frac{d}{dx}f(x) = nx^{n-1}$$

Proof
Let $$f(x)=x^n$$,$$x\in R$$ and $$n\in N$$.

By the definition of the derivative, $$\frac{d}{dx}f(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0} \frac{(x+h)^n-x^n}{h}$$.

Using the binomial theorem this simplifies to:

$$\lim_{h\rightarrow 0}\frac{({n\choose 0}x^n+{n\choose 1}x^{n-1}h+{n\choose 2}x^{n-2}h^2+\dots+{n\choose n-1}xh^{n-1}+{n\choose n}h^n)-x^n}{h}$$

$$=\lim_{h\rightarrow 0}\frac{{n\choose 1}x^{n-1}h+{n\choose 2}x^{n-2}h^2+\dots+{n\choose n-1}xh^{n-1}+{n\choose n}h^n}{h}$$

$$=\lim_{h\rightarrow 0} {n\choose 1}x^{n-1}+{n\choose 2}x^{n-2}h^1+\dots+{n\choose n-1}xh^{n-2}+{n\choose n}h^{n-1}$$

Evaluating the limit gives,

$$={n\choose 1}x^{n-1} = nx^{n-1}$$

Q.E.D.