If $n$ is an integer I can evaluate the limit in the "difference quotient" to see that the derivative of $x^n$ is $nx^{n-1}$. If $n=p/q$ is rational then I can write x as a $q$th root of $x^p$ and since $p$ is an integer I can evaluate the limit in the difference quotient.
But what if $n$ is an irrational number?
That depends on your definition of $x^n$ for irrational $n$. For instance, if your definition is $x^n=e^{n\ln x}$ then it can be solved by applying the chain rule, the definition of the exponential function as the function that satisfies $f'(x)=f(x)$, and lastly the derivative of the logarithm.
When $n$ is irrational then you need to have a definition of $x^{n}$. However, before we can discuss some definition of $x^{n}$ when $n$ is irrational, it is important to ask the following: Why don't we discuss this problem of definition of $x^{n}$ when $n$ is rational?? Simply because this is within the domain (to some extent) of algebra when $n$ is rational and students are aware of algebraic definition of $x^{n} = x^{p/q}$ as the $p^{\text{th}}$ power of $q^{\text{th}}$ root of $x$. The existence of $q^{\text{th}}$ root of $x$ is the non-algebraic part which students take for granted. A proper proof of the existence of $x^{1/q}$ requires completeness property of real numbers in some form.
As mentioned by Ian in comments, the most intuitive approach for definition of $x^{n}$ when $n$ is irrational (and $x > 0$, this is essential) comes from approximating $x^{n}$ via $x^{q}$ where $q$ is a rational approximation of $n$. Such an approach has its challenges but still students find it easier to digest. Thus let's define $x^{n}$ via limit $$x^{n} = \lim_{k \to \infty}x^{n_{k}}$$ where $n_{k}$ is any sequence of rational numbers with $\lim_{k \to \infty}n_{k} = n$. With this definition it is easy to establish all the usual laws of exponents and we will take them for granted.
Now comes the tricky part of calculating the derivative of $x^{n}$. Since the definition of $x^{n}$ is itself based on limit, the calculation of its derivative requires us to deal with iterated limits and this is a slightly hard problem in general which requires ideas of uniform convergence. I present here one approach which is based on inequalities and does not require uniform convergence. From this answer we can see that $$rx^{r - 1}(x - y) > x^{r} - y^{r} > ry^{r - 1}(x - y)\tag{1}$$ and $$sx^{s - 1}(x - y) < x^{s} - y^{s} < sy^{s - 1}(x - y)\tag{2}$$ for $x > y > 0$ and rational number $r, s$ are rational numbers with $r > 1 > s > 0$.
Now let's assume $n > 1$ and we will use inequality $(1)$ in what follows (if $0 < n < 1$ then we need to use inequality $(2)$ in similar manner, if $n < 0$ then we can use $x^{n} = 1/x^{-n}$). Then there is a sequence of rational numbers $n_{k}$ with $n_{k} > 1$ for all positive integers $k$ and $n_{k} \to n$ as $k \to \infty$. Replacing $y$ by $a$ and $r$ by $n_{k}$ in $(1)$ we get $$n_{k}x^{n_{k} - 1}(x - a) > x^{n_{k}} - a^{n_{k}} > n_{k}a^{n_{k} - 1}(x - a)$$ for $x > a > 0$. Taking limits as $k \to \infty$ we get $$nx^{n - 1}(x - a) \geq x^{n} - a^{n} \geq na^{n - 1}(x - a)$$ From the above inequality (and monotone nature of $x^{n}$) it follows that $\lim_{x \to a}x^{n} = a^{n}$ or in other words $x^{n}$ is continuous for all $x > 0$. Now dividing the above inequality by $(x - a) > 0$ and taking limits as $x \to a^{+}$ we get $$\lim_{x \to a^{+}}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ It should be easy to establish the same limit when $x \to a^{-}$ by replacing $x$ by $a$ and $y$ by $x$ in $(1)$. Thus we have established that $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ where $a > 0$ and $n$ is irrational. This shows that derivative of $x^{n}$ for irrational $n$ is also given by the same formula which holds for rational $n$ i.e. $nx^{n - 1}$.
Note: The inequalities $(1)$ and $(2)$ are established purely using algebra and they are used to prove that derivative of $x^{n}$ is $nx^{n - 1}$ for rational $n$.
Hint:
You can see $n$ as the limit of a converging sequence of rationals,
$$n_k=2^{-k}\left\lfloor2^kn\right\rfloor,$$
and define
$$x^n:=\lim_{k\to\infty}x^{n_k}.$$
Then
$$(x^n)'=(\lim_{k\to\infty}x^{n_k})'=\lim_{k\to\infty}(x^{n_k})'=\lim_{k\to\infty}n_kx^{n_k-1}=nx^{n-1}.$$
The tricky part is to prove that the derivative of the limit is the limit of the derivatives, which requires uniform convergence, I guess.
If necessary, you can also use squeezing
$$2^{-k}\left\lfloor2^kn\right\rfloor\le n\le2^{-k}\left\lceil2^kn\right\rceil.$$
I think it's worth seeing a proof that is based, more or less, on first principles.
Let $\alpha$ be an irrational number, and let $f(x)=x^\alpha$, for $x\gt0$. By the definition of derivative,
$$f'(x)=\lim_{t\to x}{f(t)-f(x)\over t-x}=\lim_{t\to x}{t^\alpha-x^\alpha\over t-x}=\lim_{u\to1}{(xu)^\alpha-x^\alpha\over xu- x}=x^{\alpha-1}\lim_{u\to1}{u^\alpha-1\over u-1}$$
so it suffices to show
$$\lim_{u\to1}{u^\alpha-1\over u-1}=\alpha$$
It will turn out to be helpful to reduce things to computing a one-sided limit. To do so, note that
$$\lim_{u\to1^-}{u^\alpha-1\over u-1}=\lim_{v\to1^+}{v^{-\alpha}-1\over v^{-1}-1}=\lim_{v\to1^+}\left(v^{1-\alpha}{v^\alpha-1\over v-1}\right)=\lim_{v\to1^+}{v^\alpha-1\over v-1}$$
since $\lim_{v\to1}v^p=1$ for any power $p$. In sum, changing notation slightly, we need to show
$$\lim_{h\to0^+}{(1+h)^\alpha-1\over h}=\alpha$$
That is, we need to show that for any $\epsilon\gt0$ there exists $\delta\gt0$ such that
$$0\lt h\lt\delta\implies\left|{(1+h)^\alpha-1\over h}-\alpha\right|\lt\epsilon$$
Now let's assume we know that the derivative of $x^r$ is $rx^{r-1}$ when $r$ is rational. Furthermore, let's assume we know the generalized Bernoulli inequality, $(1+h)^p\le1+ph$ for $0\lt p\lt1$. With this in hand, let $r$ be a rational number slightly larger than $\alpha$ (how slight, we'll see momentarily). Then
$$\begin{align} {(1+h)^\alpha-1\over h}-\alpha &=\left({(1+h)^\alpha-1\over h}-{(1+h)^r-1\over h}\right)+\left({(1+h)^r-1\over h}-r\right)+(r-\alpha)\\ &=-(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)+\left({(1+h)^r-1\over h}-r\right)+(r-\alpha)\\ \end{align}$$
Given $\epsilon\gt0$, we can clearly choose $r$ so that $|r-\alpha|\lt{\epsilon\over3}$. Once we fix that choice of $r$, the known derivative formula for $x^r$ means we can choose a $\delta_r$ such that
$$0\lt h\lt\delta_r\implies\left|{(1+h)^r-1\over h}-r\right|\lt{\epsilon\over3}$$
Finally, we see that, if $0\lt h\lt1$ and $0\lt\alpha\lt r$, then
$$0\lt(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)\le(1+h)^\alpha\left({(1+(r-\alpha)h)-1\over h}\right)=2^\alpha(\alpha-r)$$
So let's choose $r\lt\alpha$ so that $2^\alpha(\alpha-r)\lt{\epsilon\over3}$ and then let $\delta=\min\{\delta_r,1\}$. Then we have
$$\begin{align} 0\lt h\lt\delta\implies\left|{(1+h)^\alpha-1\over h}-\alpha\right| &=\left|-(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)+\left({(1+h)^r-1\over h}-r\right)+(r-\alpha)\right|\\ &\le\left|(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)\right|+\left|{(1+h)^r-1\over h}-r\right|+\left|r-\alpha\right|\\ &\lt{\epsilon\over3}+{\epsilon\over3}+{\epsilon\over3}=\epsilon \end{align}$$
as desired.
Remark: Bernoulli's inequality, $(1+h)^p\le1+ph$ if $0\lt p\lt 1$ and $h\gt0$ can be proved by differentiating the function $f(h)=(1+h)^p-ph$ and showing that $f$ is strictly decreasing from its value $f(0)=1$. This may seem circular, since we're applying the inequality to the irrational power $r-\alpha$. But we really only need the derivative-based proof for rational exponents $p$. Then, if $0\lt\beta\lt1$ is an irrational number, we have
$$(1+h)^\beta\lt(1+h)^p\le1+ph$$
for any rational $p\in(\beta,1)$, after which it suffices to let $p\to\beta$.
