Example Find the directional derivative of $f$ at the point $P$ in the direction of ${\bf a}$．

$$f = x^2 + 3y^2 + 4z^2, P:(1,0,1), \boldsymbol{a} = -\boldsymbol{i} - \boldsymbol{j}+\boldsymbol{k}$$

Answer A directional derivative of $f$ at $P(1,0,1)$ in the direction of unit vector u is give by the following:

$$\frac{\partial f(1,0,1)}{\partial u} = \nabla f(1,0,1)\cdot {\bf u}$$

Here，${\bf u}$ is unit directional vector．That is ${\bf u} = \frac{\boldsymbol{a}}{\vert\boldsymbol{a}\vert}$

In this problem, we have

$$\frac{\partial f(1,0,1)}{\partial u} = \nabla \phi(1,0,-2) \cdot{\bf u}$$

$\boldsymbol{a} = -\boldsymbol{i} - \boldsymbol{j}+\boldsymbol{k}$. Then

$$\boldsymbol{u} = \frac{-\boldsymbol{i} - \boldsymbol{j}+\boldsymbol{k}}{\sqrt{3}}$$

Also, $\nabla f = 2x\boldsymbol{i} + 6y\boldsymbol{j} + 8z\boldsymbol{k}$. Then the directional derivative is

$$\nabla f(1,0,1)\cdot u = (2\boldsymbol{i} + 8\boldsymbol{k}) \cdot \frac{-\boldsymbol{i} - \boldsymbol{j}+\boldsymbol{k}}{\sqrt{3}}$$

$$= \frac{2 + 8}{\sqrt{3}} = \frac{10}{\sqrt{3}}$$