Defining derivative of powers of $x$ We know that derivative of $x^n$ is $nx^{(n-1)}$ if $n$ is an integer.
My question is how do we define derivative of $x^r$ is $r$ is an irrational number. For example what is the derivative of $x^\sqrt2$ or $x^\pi$?
 A: We define $x^r$ as $x^r = e^{r \log x}$ so that
$$\begin{aligned}
\frac{d}{dx} x^r &= \frac{d}{dx} e^{r \log x} \\&= \frac{r}{x} e^{r \log x} \\&= \frac{r}{x} x^r \\&= r x^{r-1}.
\end{aligned}$$
A: If you know the generalized binomial theorem, you can get it right out of the definition in first semester calculus:
$$\begin{align*}
\lim_{h\to 0}\frac{(x+h)^r-x^r}h&=\lim_{h\to 0}\frac1h\sum_{k\ge 1}\binom{r}kh^kx^{r-k}\\
&=\lim_{h\to 0}\frac1h\sum_{k\ge 1}\frac{r^{\underline{k}}}{k!}h^kx^{r-k}\\
&=\lim_{h\to 0}\sum_{k\ge 1}\frac{r^{\underline{k}}}{k!}h^{k-1}x^{r-k}\\
&=r^{\underline 1}x^{r-1}+\lim_{h\to 0}\sum_{k\ge 2}\frac{r^{\underline{k}}}{k!}h^{k-1}x^{r-k}\\
&=rx^{r-1}+\lim_{h\to 0}h\sum_{k\ge 2}\frac{r^{\underline{k}}}{k!}h^{k-2}x^{r-k}\\
&=rx^{r-1}+0\\
&=rx^{r-1}
\end{align*}$$
Here $r^{\underline{k}}$ is a falling factorial.
A: If you can differentiate $\ln$, you can argue as follows for any $r$:
$y = x^r; \tag 1$
$\ln y = r \ln x; \tag 2$
$\dfrac{y'}{y} = \dfrac{r}{x}; \tag 3$
$y' = \dfrac{ry}{x} = \dfrac{rx^r}{x} = rx^{r - 1}. \tag 4$
$OE\Delta$.
If you want to argue for $r$ rational but not an integer, you may also proceed as follows:
Set
$r = \dfrac{p}{q}, \; p, q \in \Bbb Z, \; q \ne 0; \tag 5$
then
$y = x^r = x^{p/q}; \tag 6$
so,
$y^q = x^p; \tag 7$
since $p$ and $q$ are integers, we may use the rule
$(z^n)' = nz^{n - 1}z', \tag 8$
and obtain
$qy^{q - 1}y' = px^{p - 1}, \tag 9$
whence
$y' = \dfrac{p}{q} \dfrac{x^{p - 1}}{y^{q - 1}}; \tag{10}$
now,
$y^{q - 1} = \dfrac{x^p}{y} = \dfrac{x^p}{x^{p/q}} = x^{p - p/q}, \tag{11}$
via which (10) yields
$y' = \dfrac{p}{q} \dfrac{x^{p - 1}}{x^{p - p/q}} = \dfrac{p}{q} x^{p/q - 1}. \tag{12}$
$OE\Delta$.
