Can $n$ & $a$ be any real number in the power rule? $$\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}\tag{1}$$
$$\frac{d}{da}(a^n)=na^{n-1}\tag{2}$$
In $(1)$ & $(2)$, can $a$ & $n$ be any real number?
 A: The differentiation rule in question applies to all these cases:

*

*the base is positive and the exponent is real;

*the base is real and the exponent is a positive integer;

*the base is nonzero and the exponent is an integer.

A: Certainly $a$ cannot be negative if $n$ is irrational or rational with even denominator, since the function $a^n$ is not even defined for negative values of $a$ in those cases. Otherwise the answer is yes; one can prove this for positive $a$ by writing $a^n = e^{n\log a}$ and differentiating using the Chain Rule.
A: I think this is elementary as it requires:
1-Definition of continuity and derivative
2-Binomial Theorem ( which I think is quite elementary)
3-Geometric series
4-Series convergence test by comparison
5-limit of Uniformly Convergent series is continuous
It suffice to prove $\frac{d}{dx}\log{x}=\frac{1}{x}$ using elementary results
$x > 0$ , $n \in N$
define $e=\lim_{n \to \infty}(1+\frac{1}{n})^n$ which exists according to binomial theorem and comparison with geometric series
define $g(x)=\lim_{n \to \infty}(1+\frac{x}{n})^n$  which exists as shown below
$h(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$ which exists by comparison with geometric series and is uniformly convergent
using binomial theorem $g(x)=h(x)$
because $x>0$, $g(x)$ is increasing on $[0,\infty)$
$k,m,j \in N$
$ y \in Q$
$y=\frac{k}{m}$
$g(y)=\lim_{n \to \infty}((1+\frac{y}{n})^{\frac{n}{y}})^y$
set $n=kj$
$g(y)=\lim_{j \to \infty}((1+\frac{1}{mj})^{mj})^y=e^y$
because $g(x)$ is increasing for every $x>0$ we can find two rational numbers $p,q$ and $q>p$ such that $g(p)\le g(x) \le g(q)$
since $h(x)$ is continuous the limit $\lim_{r_n \to r}e^{r_n}=\lim_{r_n \to r}h(r_n)=h(r)$ , where $\{r_n\}$ is a sequence of rational numbers converging to $r$
So it makes sense to define $e^r=\lim_{r_n \to r}e^{r_n}$
since we can let $ p \to q$ and $e^x$ is continuous ,this implies $e^x=h(x)$
$\frac{dh}{dx}=\lim_{t \to 0} e^x\frac{e^t-1}{t}=\lim_{t \to 0} e^x\frac{\sum_{n=1}^\infty \frac{t^n}{n!}}{t} =  e^x$
$x=e^{\log{x}}$
taking derivative of both sides and using chain rule
$1=e^{\log{x}}\frac{d}{dx}\log{x}$
$\frac{d}{dx}\log{x}=\frac{1}{e^{\log{x}}}=\frac{1}{x}$
