Editing Power rule (section)

==Proofs==

===Proof for real exponents===
Let {{nowrap|<math>f(x) = x^r</math>,}} where <math>r</math> is any real number. 

If {{nowrap|<math>f(x) = e^x</math>,}} then {{nowrap|<math>\ln (f(x)) = x</math>,}} where <math>\ln</math> is the [[natural logarithm]] function, 
or {{nowrap|<math>f'(x) = f(x) = e^x</math>,}} as was required. 
Therefore, applying the chain rule to {{nowrap|<math>f(x) = e^{r\ln x}</math>,}} we see that 
<math display="block">f'(x)=\frac{r}{x} e^{r\ln x}= \frac{r}{x}x^r</math>
which simplifies to {{nowrap|<math>rx^{r-1}</math>.}}

When {{nowrap|<math>x < 0</math>,}} we may use the same definition with {{nowrap|<math>x^r = ((-1)(-x))^r = (-1)^r(-x)^r</math>,}} where we now have {{nowrap|<math>-x > 0</math>.}} This necessarily leads to the same result. Note that because <math>(-1)^r</math> does not have a conventional definition when <math>r</math> is not a rational number, irrational power functions are not well defined for negative bases. In addition, as rational powers of −1 with even denominators (in lowest terms) are not real numbers, these expressions are only real valued for rational powers with odd denominators (in lowest terms).

Finally, whenever the function is differentiable at {{nowrap|<math>x = 0</math>,}} the defining limit for the derivative is:
<math display="block">\lim_{h\to 0} \frac{h^r - 0^r}{h}</math>
which yields 0 only when <math>r</math> is a rational number with odd denominator (in lowest terms) and {{nowrap|<math>r > 1</math>,}} and 1 when {{nowrap|<math>r = 1</math>.}} For all other values of {{nowrap|<math>r</math>,}} the expression <math>h^r</math> is not well-defined for {{nowrap|<math>h < 0</math>,}} as was covered above, or is not a real number, so the limit does not exist as a real-valued derivative. For the two cases that do exist, the values agree with the value of the existing power rule at 0, so no exception need be made.

The exclusion of [[Zero to the power of zero|the expression <math>0^0</math>]] (the case {{nowrap|<math>x = 0</math>)}} from our scheme of exponentiation is due to the fact that the function <math>f(x, y) = x^y</math> has no limit at (0,0), since <math>x^0</math> approaches 1 as x approaches 0, while <math>0^y</math> approaches 0 as y approaches 0. Thus, it would be problematic to ascribe any particular value to it, as the value would contradict one of the two cases, dependent on the application. It is traditionally left undefined.

===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>

===Generalization to rational exponents=== 
Upon proving that the power rule holds for integer exponents, the rule can be extended to rational exponents.
====Proof by [[chain rule]]====
This proof is composed of two steps that involve the use of the chain rule for differentiation.
# Let <math>y=x^r=x^\frac1n</math>, where <math>n\in\N^+</math>. Then <math>y^n=x</math>. By the [[chain rule]], <math>ny^{n-1}\cdot\frac{dy}{dx}=1</math>. Solving for <math>\frac{dy}{dx}</math>, <math display="block">\frac{dy}{dx}
=\frac{1}{ny^{n-1}}
=\frac{1}{n\left(x^\frac1n\right)^{n-1}}
=\frac{1}{nx^{1-\frac1n}}
=\frac{1}{n}x^{\frac1n-1}
=rx^{r-1}</math>Thus, the power rule applies for rational exponents of the form <math>1/n</math>, where <math>n</math> is a nonzero natural number. This can be generalized to rational exponents of the form <math>p/q</math> by applying the power rule for integer exponents using the chain rule, as shown in the next step.
# Let <math>y=x^r=x^{p/q}</math>, where <math>p\in\Z, q\in\N^+,</math> so that <math>r\in\Q</math>. By the [[chain rule]], <math display="block">\frac{dy}{dx}
=\frac{d}{dx}\left(x^\frac1q\right)^p
=p\left(x^\frac1q\right)^{p-1}\cdot\frac{1}{q}x^{\frac1q-1}
=\frac{p}{q}x^{p/q-1}=rx^{r-1}</math> 
From the above results, we can conclude that when <math>r</math> is a [[rational number]], <math>\frac{d}{dx} x^r=rx^{r-1}.</math>

====Proof by [[implicit differentiation]]====
A more straightforward generalization of the power rule to rational exponents makes use of implicit differentiation.

Let <math> y=x^r=x^{p/q}</math>, where <math>p, q \in \mathbb{Z}</math> so that <math>r \in \mathbb{Q}</math>.

Then,<math display="block">y^q=x^p</math>Differentiating both sides of the equation with respect to <math>x</math>,<math display="block">qy^{q-1}\cdot\frac{dy}{dx} = px^{p-1}</math>Solving for <math>\frac{dy}{dx}</math>,<math display="block">\frac{dy}{dx} = \frac{px^{p-1}}{qy^{q-1}}.</math>Since <math>y=x^{p/q}</math>,<math display="block">\frac d{dx}x^{p/q} = \frac{px^{p-1}}{qx^{p-p/q}}.</math>Applying laws of exponents,<math display="block">\frac d{dx}x^{p/q} = \frac{p}{q}x^{p-1}x^{-p+p/q} = \frac{p}{q}x^{p/q-1}.</math>Thus, letting <math>r=\frac{p}{q}</math>, we can conclude that <math>\frac d{dx}x^r = rx^{r-1}</math> when <math>r</math> is a rational number.