# 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?

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when x=0, the derivative will be zero anyway... –  DanielY May 10 '13 at 9:20
It looks like it's talking about the specific example of $f(x) = x^{1/4}$, so $f'(x) = (1/4)x^{-3/4}$ is defined for positive $x$. –  Ben May 10 '13 at 9:28

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 $0$.

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$r$ 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.

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Thank you, sir. And why is $f'(0)$ not defined for r in the unit interval? –  Student May 10 '13 at 9:51
It could be easily seen from the expression $f'(x) = rx^{r - 1}$, which is not valid at $x = 0$ when $r \in (0, 1)$. A (probably) more proper way may be to start with the case $r \in (1, \infty)$, in which $f'(x)$ is defined at $x = 0$, and is in fact $0$. Since $f'(0) = 0$, you would expect that its inverse function has slope of infinity at this point. $g(x) = x^{1/r}$ is the inverse of $f$, so $g'(0)$ should not be defined. –  Tunococ May 10 '13 at 11:18
Anyway, I'll add the following for completeness. If $g(f(x)) = x$ and $g'(f(x_0))$ and $f'(x_0)$ exist, then $g'(f(x_0))f'(x_0) = 1$ by the chain rule. That means if $f'(x_0) = 0$, $g'(f(x_0))$ cannot exist. –  Tunococ May 10 '13 at 11:22