Directional derivative of multi-variable function I have the next function: $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ and i need to find it's directional derivative of the point $(0,0,0)$ and the vector $v=(\frac1{\sqrt{3}},-\frac1{\sqrt{3}},\frac1{\sqrt{3}})$
So I've started by finding the partial derivatives: 
${f'}_x(x,y,z)=\frac x{\sqrt{x^2+y^2+z^2}}$, ${f'}_y(x,y,z)=\frac y{\sqrt{x^2+y^2+z^2}}$, ${f'}_z(x,y,z)=\frac z{\sqrt{x^2+y^2+z^2}}$
Now i'm pretty much stuck. Since placing $(0,0,0)$ makes an undefined expression, What can i do from this point on?
 A: (Edited to address technical ambiguities in the earlier version):
If we use the usual definition of the directional derivative at $\mathbf{x}$ along $\mathbf{v}$,
$$ D_{\mathbf{v}}f(\mathbf{x}) = \lim_{t\rightarrow 0}\frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$
where there is no restriction on $t$, then the directional derivative in fact does not exist at $\mathbf{0}$ since this limit is different depending on whether $t\rightarrow 0^+$ or $t\rightarrow 0^-$.
For the former case $t>0$ so
\begin{align}\lim_{t\rightarrow 0^+} \frac{f(\mathbf{0}+t\mathbf{v})-f(\mathbf{0})}{t} &= \lim_{t\rightarrow 0^+}\frac{f\big(t\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\big)}{t} \\
&= \lim_{t\rightarrow 0^+}\frac{\sqrt{t^2}}{t} \\
&=1,
\end{align}
while for the latter $t<0$ so
$$ \lim_{t\rightarrow 0^-} \frac{f(\mathbf{0}+t\mathbf{v})-f(\mathbf{0})}{t} = \ \cdots\ =-1. $$
These are related to the direction along $\mathbf{v} = \left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$ in which we are instantaneously travelling: you can think of $t\rightarrow 0^+$ as giving the rate of change along the direction of $\mathbf{v}$ going away from $\mathbf{0}$ $[*]$, while $t\rightarrow 0^-$ gives the rate of change coming into $\mathbf{0}$ travelling "against" $\mathbf{v}$.

Intuition-builder:
Note that without doing any work at all, you can guess that the "directional derivative" (in the sense of $[*]$) is $1$.
  This because $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ is the distance of $(x,y,z)$ from the origin $\mathbf{0}$.
  The directional derivative at the origin is then the rate of change of $f$ with respect to the distance travelled away from $\mathbf{0}$ along $\mathbf{v}=\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$.
  Since $f$ is the distance from the origin, this rate is of course $1$.

To summarize, the derivative you are looking for does not exist, but if we restrict the limit to the range $t>0$ then the limit is $1$, and it makes sense to call this a "directional derivative" of some sort.
