How do we prove that the derivative of $x^r$ is $r x^{r-1}$ for all $r$? Let the function $f$ be defined as:
$f(x)=x^r, r\in \mathbb R$.
We have all heard that $f'(x)=r x^{r-1}$. And apparently, there exists a proof for that. But in that proof, there is something that doesn't look convincing to me. As you see, we first take this into account:
$$f(x)=x^r=e^{r\ln{x}}$$
and then, use the existing rules for the derivative of exponential functions.
But let's take a little step back. As far as I know, the exponential function is defined as the inverse of the logarithm function. And the logarithm is defined as:
$$\ln(x)=\int_1^x \frac{1}{t} dt,\quad x>0$$
At the first steps, it is proved that this function is well-defined, continuous and anywhere differentiable in its domain. Then we prove that it is injective, and denote its inverse function by $\exp(x)$ and we don't mention anything about the term exponential. Then it is shown that the derivative of this particular function is equal to itself, by using those abstract definitions and some smart-looking workarounds.
But something caught my attention that, in the process of those proofs, it is taken as an assumption that the derivative of the power function is equal to $rx^{r-1}$, and this seems like a loop to me. Also if the other definition of the $\exp$ is used, which is based on the Taylor series, we would still be stuck to the derivative of power function.
So, is there a way to prove the derivative of the power function other than using the exponential function? Or in other words, is there any way to differentiate the so-called $\exp$ without using the power function?
 A: $$\newcommand{\dd}{\frac{\mathrm{d}}{\mathrm{d}x}}$$
I haven't looked at your link, but it's important to note that the rule for taylor series and the rule you want to prove are not the same.
The first one says that if $r\in\mathbb R$, then $\dd x^r = rx^{r-1}$.  For the taylor series rule, we only need that $r\in\mathbb N$, then $\dd x^r = rx^{r-1}$.  This is much easier to prove (namely, we can use the difference quotient).  Calculations with the difference quotient show that:
\begin{align}
\dd x^r & = \lim_{h\to 0}\frac{(x+h)^r-x^r}{h}
\end{align}
We have the binomial formula that $$(a+b)^n = \sum_{i = 0}^n \binom{n}{i}a^ib^{n-i}$$
We can use this to find that:
\begin{align}
\dd x^r & = \lim_{h\to 0}\frac{\sum_{ i  = 0}^r\binom{r}{i}x^ih^{r-i}-x^r}{h} \\
&=\lim_{h\to 0}\frac{h\binom{r}{r-1}x^{r-1}+h^2\sum_{i = 0}^{r-2}\binom{r}{i}x^{i}h^{r-2-i}}{h} \\
&=\binom{r}{r-1}x^{r-1}=\frac{r!}{(r-1)!}x^{r-1} = rx^{r-1}
\end{align}
As we used the binomial formula, we assume that $r\in\mathbb N$.  In this way, it wasn't a strong enough proof to establish the result you want.  It WAS strong enough to establish the power rule for differentiation taylor series.  In this way, the proof isn't circular.
A: I see no such circular logic in the link you provided. The proof there only uses the Chain Rule and the derivative of $\ln(x)$ as established facts outside the scope of the proof. It reaches a point where it asks the reader to compute $$\frac{d}{dx}e^{r\ln(x)}$$ and then uses the Chain Rule, the derivative of $\ln(x)$, and the derivative of $e^x$ (established earlier in the proof) to calculate this derivative is $$r x^{r-1}\text{.}$$
A: The linked pdf (see comments to question by OP) gives a theory of exponential, logarithmic and general power functions of a real variable using the simplest possible approach (here simple means short and concise proofs and we assume that the theory of Riemann integrals is already developed and available for use) and this approach of defining logarithm via integral is very common in many textbooks of calculus/analysis.
There is no circularity involved in the proof of the derivative formula $$\frac{d}{dx}(x^{r}) = rx^{r - 1}\tag{1}$$ for $x > 0$ and $r \in \mathbb{R}$. It should however be noted that when $r$ is rational then the function $f(x) = x^{r}$ can be defined without any reference to $\log$ or $\exp$ and a proof of derivative formula then follows from algebraic inequalities. Moreover it is possible to extend this definition to define $x^{r}$ for any irrational $r$ and then the derivative formula $(1)$ also holds according to this extended definition. See this answer for more details on this approach which avoids logarithmic/exponential functions as far as the definition of $x^{r}$ and calculation of its derivative are concerned.
A: Here's a way to prove it at least for rational exponents without resorting to the exponential function or logarithms. Extending to real coefficients probably can be done by using a sequence of rational numbers $q_n$ converging to the $r$, but then one has to show that the limits for $n\to\infty$ and $h\to 0$ can be exchanged, which I don't know how to do.
In the following, for simplicity I assume that $x>0$; the case $x\le 0$ needs additional considerations of whether the functions are defined, and when the power laws can be used.
First, use induction and the product rule (which I assume has already been proven before) to prove it for $n\in\mathbb N$.
Obviously, it's true for $n=0$: $(x^0)' = 1' = 0 = 0x^{-1}$
Assume it is true for $x^n$ Then we find:
$$(x^{n+1})' = (x\cdot x^n)' = x'\cdot x^n + x\cdot (x^n)'
= x^n + x\cdot (nx^{n-1}) = (n+1)x^n$$
So it's true at least for $x\in\mathbb N$.
We can also determine that it is true for $x^{-1}$ by explicitly calculating the limit:
$$\lim_{h\to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}
= \lim_{h\to 0}\frac{-1}{x(x+h)}= -\frac{1}{x^2}$$
Now we can use the chain rule (which I also assume to have been proven) to extend the rule to negative integers, and thus to all of $\mathbb Z$:
$$(x^{-n})' = ((x^n)^{-1})' = -\frac{1}{(x^n)^2}\cdot (x^n)' = -n\frac{x^{n-1}}{x^{2n}} = -nx^{-n-1}$$
Next, we can use the fact that $x^{1/n}$ is the inverse function of $x^n$ and use the rule of the derivative of inverse functions (also assumed to already have been proved) to get
$$(x^{\frac{1}{n}})' = \frac{1}{n(x^{1/n})^{n-1}} = \frac{1}{n}x^{\frac{1}{n}-1}$$
Now we can use the chain rule again for $x^{m/n} = (x^m)^{1/n}$ to prove the rule for arbitrary rational exponents.
