>Prove that $\frac d {dx} x^n=nx^{n-1}$ for all $n \in \mathbb R$. 
Prove that $\frac d {dx} x^n=nx^{n-1}$ for all $n \in \mathbb R$.

I saw some proof of $\frac d {dx} x^n=nx^{n-1}$ using binomial theorem, which is only available for $n \in\mathbb N$. Do anyone have the proof of $\frac d {dx} x^n=nx^{n-1}$ for all real $n$? Thank you.
 A: Writing $x^r=e^{r\ln x}:$
$$\begin{align*}(x^r)'&=\lim_{h\to 0}\frac{e^{r\ln (x+h)}-e^{r\ln x}}{h}\\&=\lim_{h\to 0}\left(\frac{e^{r\ln (x+h)}-e^{r\ln x}}{r\ln(x+h)-r\ln x}\right)\left(\frac{r\ln (x+h)-r\ln x}{h}\right)\end{align*}$$
$(\ln $ is continuous and injective)
Let $r\ln x =w,\;\;r\ln (x+h)=w+i:$
$$\begin{align*}\cdots &=\left(\lim_{i\to 0}\frac{e^{w+i}-e^w}{i}\right)\cdot \lim_{h\to 0}r\ln\left(1+\frac{h}{x}\right)^{\frac{1}{h}}\quad\qquad\quad\\&=\left(\lim_{i\to 0}\frac{e^{w+i}-e^w}{i}\right)\left(\lim_{h\to 0}\frac{r}{x}\ln\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right)\\&=\displaystyle\lim_{i\to 0}e^w\left(\frac{e^{i}-1}{i}\right)\cdot \frac{r}{x}\end{align*}$$
Let $i=\ln \left(1+\frac{1}{a}\right):\qquad (i\to0 \Rightarrow a\to \infty)$
$$\begin{align*}\cdots &=\lim_{a\to \infty}e^w\left(\frac{1}{a \ln \left(1+\frac{1}{a}\right)}\right)\cdot \frac{r}{x}\quad\qquad\qquad\qquad\qquad\\&=\lim_{a\to \infty}e^w\left(\frac{1}{\ln \left(1+\frac{1}{a}\right)^{a}}\right)\cdot \frac{r}{x}\\&=\dfrac{re^w}{x}\\&=rx^{r-1}\end{align*}$$
A: Assuming you know the derivatives of $e^x$ and $\log(x)$,
$$\dfrac{dx^n}{dx}=\dfrac{de^{n\log(x)}}{dx}=n\dfrac{x^n}{x}=nx^{n-1}$$
Here's a proof for $n\in Q$.
Let n=$\frac pq$ and $y=x^{\frac pq}$.
$y^q=x^p$
With implicit differentiation we get the result.
For irrational n,
$x^n=\lim_{r\to n}x^r \quad r \in Q$
$\dfrac{dx^n}{dx}=\dfrac{d}{dx}\lim_{r\to n}x^r$
Not sure if we can interchange the limit and the differentiation, though.
A: From the definition of derivatives, we have
$$ f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} $$
Let's assume here that $f(x) = x^n$. Then,


*

*Case 1, $n \in \mathbb{N}$


$$
\begin{align}
f'(x) &= \lim_{h \to 0} \frac{(x + h)^n - x^n}{h} \\
&= \lim_{h \to 0} \frac{x^n + nhx^{n-1} + \frac{n(n-1)}{2!}h^2x^{n-2} + \cdots - x^n}{h}\\
&= \lim_{h \to 0} \text{ } h \text{ } \frac{nx^{n-1} + \frac{n(n-1)}{2!}hx^{n-2} + \cdots \text{ higher powers in h}}{h} \\
&= nx^{n-1} + \lim_{h \to 0} \text{ } h \cdot \left( \frac{n(n-1)}{2!}x^{n-2} + \cdots \text{multiples of h}\right)\\
&= nx^n + 0\\
&= nx^n
\end{align}
$$


*

*Case 2, $r \in \mathbb{R}$: A similar expansion of binomials exists for real powers, as given by Issac Newton.


$$
\begin{align}
f'(x) &= \lim_{h \to 0} \frac{(x + h)^r - x^r}{h} \\
&= \lim_{h \to 0} \frac{x^r + hrx^{r-1} + \frac{r(r-1)}{2!}h^2x^{r-2} + \cdots - x^r}{h}\\
&= \lim_{h \to 0} \text{ } h \text{ } \frac{rx^{r-1} + \frac{r(r-1)}{2!}hx^{r-2} + \cdots \text{ higher powers in h}}{h} \\
&= rx^{r-1} + \lim_{h \to 0} \text{ } h \cdot \left( \frac{r(r-1)}{2!}x^{r-2} + \cdots \text{multiples of h}\right)\\
&= rx^r + 0\\
&= rx^r
\end{align}
$$
