Editing Power rule (section)

===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.