>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.

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.


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