## Thomas' Calculus 13th Edition

$$2$$
The directional derivative can be expressed as: $D_v g = \nabla g \cdot v$ $g_x= 3e^x \cos y z \\ g_y= -3ze^x \sin yz\\ g_z=-3ye^x \sin yz$ $\nabla g (0,0,0) =\lt 3,0,0 \gt$ $\implies v=\dfrac{u}{|u|}=\dfrac{\lt 2,1,-2 \gt}{\sqrt {9}}=\lt \dfrac{2}{3}, \dfrac{1}{3}, \dfrac{-2}{3} \gt$ and $$D_v \ g = \nabla \ g \cdot \ v \\ =\lt 3,0,0 \gt \cdot \lt \dfrac{2}{3}, \dfrac{1}{3}, \dfrac{-2}{3} \gt \\ =2$$