Editing Power rule (section)

===Proofs for integer exponents===
====Proof by [[mathematical induction|induction]] (natural numbers)====
Let <math>n\in\N</math>. It is required to prove that <math>\frac{d}{dx} x^n = nx^{n-1}.</math> The base case may be when <math>n=0</math> or <math>1</math>, depending on how the set of [[Natural number|natural numbers]] is defined.

When <math>n=0</math>, <math>\frac{d}{dx} x^0 = \frac{d}{dx} (1) = \lim_{h \to 0}\frac{1-1}{h} = \lim_{h \to 0}\frac{0}{h} = 0 = 0x^{0-1}.</math>

When <math>n=1</math>, <math>\frac{d}{dx} x^1 = \lim_{h \to 0}\frac{(x+h)-x}{h} = \lim_{h \to 0}\frac{h}{h} = 1 = 1x^{1-1}.</math> 

Therefore, the base case holds either way.

Suppose the statement holds for some natural number ''k'', i.e. <math>\frac{d}{dx}x^k = kx^{k-1}.</math>

When <math>n=k+1</math>,<math display="block">\frac{d}{dx}x^{k+1} = \frac{d}{dx}(x^k \cdot x) = x^k \cdot \frac{d}{dx}x + x \cdot \frac{d}{dx}x^k = x^k + x \cdot kx^{k-1} = x^k + kx^k = (k+1)x^k = (k+1)x^{(k+1)-1}</math>By the principle of mathematical induction, the statement is true for all natural numbers ''n''.

====Proof by [[binomial theorem]] (natural number)====
Let <math>y=x^n</math>, where <math>n\in \mathbb{N} </math>.

Then,<math display="block">\begin{align}
\frac{dy}{dx}
&=\lim_{h\to 0}\frac{(x+h)^n-x^n}h\\[4pt]
&=\lim_{h\to 0}\frac{1}{h} \left[x^n+\binom n1 x^{n-1}h+\binom n2 x^{n-2}h^2+\dots+\binom nn h^n-x^n \right]\\[4pt]
&=\lim_{h\to 0}\left[\binom n 1 x^{n-1} + \binom n2 x^{n-2}h+ \dots+\binom nn h^{n-1}\right]\\[4pt]
&=nx^{n-1}
\end{align}</math>

Since n choose 1 is equal to n, and the rest of the terms all contain h, which is 0, the rest of the terms cancel. This proof only works for natural numbers as the binomial theorem only works for natural numbers.

====Generalization to negative integer exponents====
For a negative integer ''n'', let <math>n=-m</math> so that ''m'' is a positive integer.
Using the [[reciprocal rule]],<math display="block">\frac{d}{dx}x^n = \frac{d}{dx} \left(\frac{1}{x^m}\right) = \frac{-\frac{d}{dx}x^m}{(x^m)^2} = -\frac{mx^{m-1}}{x^{2m}} = -mx^{-m-1} = nx^{n-1}.</math>In conclusion, for any integer <math>n</math>, <math>\frac{d}{dx}x^n = nx^{n-1}.</math>