The derivative of $x^r$ at zero. I am taking this directly from the Wikipedia article.
===Derivatives of elementary functions===
...
Derivatives of powers: if
$f(x) = x^r,\,$
where ''r'' is any real number, then
$f'(x) = rx^{r-1},\,$
wherever this function is defined. ...
and the derivative function is defined only for positive ''x'', not for $x=0$.
What is the problem at zero? What is the problem for negative numbers?
 A: $r$ is any real number. It can even be negative, in which case $f(0)$ is not even defined. Anyway, Ben's remark is worth noting. For $r \in (0, 1)$, $f(0) = 0$, but $f'(0)$ is not defined.
A: You are thinking about differentiation only mechanically, without any thought of what it means.  The derivative of a function at a point is the limit of the difference quotient there, and is the slope of the tangent line there. So, draw the graphs of $y=x^r$ for various $r$.
For $r=1$, one has a line, and the tangent is the line itself, with slope $1$.
For $r=2$, one has a parabola, and the tangent at $(0, 0)$ is the $x$-axis, with slope $0$.
For $r=\frac12$, one has what turns out to be half the parabola $x=y^2$, and the tangent line is vertical, with infinite slope.  Also, this is not defined for $x < 0$, so the limit does not exist anyway.
For $x=\frac13$, the function exists for all $x$, but the tangent at $(0,0)$ is still vertical, the slope is infinite, and the derivative does not exist at $0$.
A: Recall that $x^r$ is not defined for $x<0$ and $r$ irrational, and also not defined when $x<0$ and $r=p/q$, $p\in\mathbb{Z}$, $q\in\mathbb{N}$, $\gcd(p,q)=1$ and $q$ even.
At $x=0$, $(x^r)'=\lim\limits_{h\to 0}\dfrac{h^r}{h}$. We obtain
\begin{eqnarray}
\lim_{h\to 0+}\frac{h^r}{h} & = & \begin{cases}
 1 & \mbox{if } r=1 \\ 0 & \mbox{if } r>1 \\ \infty & \mbox{if } r<1. \end{cases} \\
\lim_{h\to 0-}\frac{h^r}{h} & = & \begin{cases}
 1 & \mbox{if } r=1 \\  0 & \mbox{if } r>1 \\ 
 -\infty & \mbox{if } r<1,r=p/q,q \mbox{ odd}, p \mbox{ even},\gcd(p,q)=1 \\ \infty & \mbox{if } r<1,~r=p/q,q \mbox{ odd}, p \mbox{ odd},\gcd(p,q)=1. \end{cases}
\end{eqnarray}
Therefore, at $x=0$, the derivative of $x^r$ is given by
$$\big((x^r)'\mbox{ at }x=0\big) = \begin{cases} 1 & \mbox{if } r=1 \\ 0 & \mbox{if } r>1. \end{cases}$$
In all other cases, the derivative of $x^r$ at $x=0$ does not exist.
