Direction vector Using differential I want to find the value of the function $g(x,y,z) = \ln \sqrt{x^2+y^2+z^2}$ if we move from (3,4,12) to a distance of 0.1 in the direction of vector $u = 3i+6j-2k$.
I first found $$|u| = \sqrt{9+36+4} = 7$$ 
so $$u_o = \frac 3 7 i+ \frac 6 7j - \frac 2 7k$$
and $$\nabla g(3,4,12) = \frac 6 {29}i + \frac 8{29}j+ \frac {24}{29}k$$
Then I don't know what to do. Any ideas?
 A: You can compute the value of $g$ at the  the point ${\bf x}_0=(3,4,12)$. It's
$$
g(3,4,12)=\ln(13)
$$ 
You want to estimate the value of $g$ at another point.
From ${\bf x}_0$, you move to the   the point ${\bf x}_1=(3,4,12)+(.1)u_0$ and you wish to estimate the value of $g$ there.
To do this, you can first consider the change in the value of $g$ between the two points. You can find an approximation to  the  change in the value of $g$, denoted $\Delta g$, between those two points  by using use the formula $$\tag{1} \Delta g\approx  \nabla g({\bf x}_0)\cdot \Delta {\bf x},$$
where $\Delta {\bf x}= {\bf x}_1-{\bf x_0} = (.1)u_0$. 
So, to approximate the value of $g$ at the point ${\bf x}_1$, add the change in the value of $g$ from ${\bf x}_0$ to ${\bf x}_1$ to the value of $g$ at ${\bf x}_0$:
$$\eqalign{
g({\bf x}_1)&= \strut g({\bf x}_0)+ \Delta g\cr &\tag{2}\approx \strut  g({\bf x}_0)+\nabla g({\bf x}_0)\cdot \Delta {\bf x}
}
$$
So you're almost there: compute $(1)$ with ${\bf x}_0=(3,4,12)$ and $\Delta{\bf x}=(.1)u_0 $ and then  add to $g(3,4,12)=\ln (13)$. (You may want to check your computation of the gradient, it seems off (you need $z=12$)) 

Note this is similar to the one variable case with the gradient playing the role of the derivative
$$
f(x_1)\approx f(x_0)+(x_1-x_0)f'(x_0).
$$ 
