Find the directional derivative of the scalar field Find the directional derivative of the scalar field:

$f(x,y,z)=\log(x^2+y^2+z^2)$ at $P_0(1,1,1)$ in the direction of the straight line $\ P_0P $  where $P=(3,2,1)$

What I have done: $\nabla(f)=(2/3,2/3,2/3)$ at $P_0$ and I know the eqation of $P_0P$ is 

$(x-1)/2=(y-1)/1=(z-1)/0=t(say)$ 

Now the required unit vector is $\frac{(3-1,2-1,1-1)}{\sqrt 5}$ taking dot product with $\nabla f$ my final result comes out to be $\frac{2}{\sqrt{5}}$ . But the answer does not match. 
Also my confusion arises seeing this article. According to the article, answer comes out as $\frac{9}{\sqrt{14}}$ . But the answer given in my exercisebook is $\frac{8}{3\sqrt5}$. Please help.
 A: One can just do it directly, the unit vector in your direction is $\frac{1}{\sqrt{5}}(2,1,0)$
So we have $(x,y,z)=(1+\frac{2}{\sqrt{5}}t,1+\frac{1}{\sqrt{5}}t,1)$
and $x^2+y^2+z^2=t^2+\frac{6}{\sqrt{5}}t+3$
So we must differentiate 
$$\ln \left(t^2+\frac{6}{\sqrt{5}}t+3\right)$$ at zero, 
the derivative is 
$$\frac{2t+\frac{6}{\sqrt{5}}}{t^2+\frac{6}{\sqrt{5}}t+3}$$
and when $t=0$ we get 
$$\frac{2}{\sqrt{5}}$$
So basically you are right and all those other people are wrong.
A: Your answer is correct, and the exercise book is wrong.
When I follow the steps in the article you linked (which appears not to have your exact problem statement), I arrive at your answer as well.
A: You're answer is correct:
$\frac{\partial f}{\partial x} = \frac{2x}{x^2+y^2+z^2}$
$P = (3,2,1)$, 
$P_0 = (1,1,1)$
$\nabla f$ at $P = \left(\frac{2}{3},\frac{2}{3},\frac{2}{3}\right)$
Indeed the unit vector in the direction of $P$ from $P_0$ is given by:
$\vec{n} = \frac{1}{\sqrt{5}}(P-P_0)$
So the directional derivative at $P_0$ in the direction $P$ is 
$\nabla f \cdot \vec{n} = \frac{2\cdot 2 + 1\cdot 2 }{3 \sqrt{5}} = \frac{2}{\sqrt{5}}$
