how to find directional derivative I am trying to find the directional derivative of the following problem
$F(x,y,z) = 4x^2+ 3y−3xz+ 2z^2$ 
at the point $(2,1,2)$ in the direction $i−k$;
I worked out the derivatives of $F(x,y,z)$ as
$f_x = 8x -3z$ 
$f_y = 3$
$f_z = -3x+4z$ 
But I don't know to do next; can someone explain how to find the directional derivative here please?
Thank you
 A: Let the gradient $\nabla F$ of $F$ be
$$\nabla F = (f_x, f_y, f_z).$$
Then the gradient evaluated at $(2,1,2)$,
$$\nabla F(2, 1, 2) = (8 \cdot 2 - 3 \cdot 2, 3, -3 \cdot 2 + 4 \cdot 2) = (10, 3, 2)$$
can be dotted with the normalized direction $\frac{1}{\sqrt{2}}(1, 0, -1) = \frac{1}{\sqrt{2}}(i - k)$ to arrive at the directional derivative you are searching for
$$ \nabla F(2, 1, 2) \bullet \frac{1}{\sqrt{2}}(1, 0,-1) = (10, 3, 2)\bullet \frac{1}{\sqrt{2}} (1, 0, -1) = \frac{1}{\sqrt{2}}(10 - 2) = \frac{8}{\sqrt{2}}.
$$
Using the method described by J.W. Tanner, we notice that the same conclusion is found by computing it simply as a weighting of the directional derivatives in directions $x, y$ and $z$,
$$
\begin{align}
\frac{1}{\sqrt{2}}\cdot f_x(2, 1, 2) + 0 \cdot f_y(2, 1, 2) - \frac{1}{\sqrt{2}}\cdot f_z(2,1,2) &= \\ \frac{1}{\sqrt{2}}\cdot 10 + \frac{0}{\sqrt{2}}\cdot 3 - \frac{1}{\sqrt{2}} \cdot 2 &= \\\frac{10 - 2}{\sqrt{2}} &= \\\frac{8}{\sqrt{2}}
\end{align}
$$
A: Firstly, you need to figure out the gradient of $F$ at the point, $$ \nabla F = ( \frac {\partial F} {\partial x}, \frac {\partial F} {\partial y}, \frac {\partial F} {\partial z})$$.
Say that the direction vector of interest is $\vec l$. Take the dot product between $\nabla F$ and the unit vector in the direction of $\vec l$, which should give $$\frac {\partial F} {\partial \vec l} = \nabla F \cdot \frac {\vec l} {\Vert \vec l \Vert}$$
