## Thomas' Calculus 13th Edition

$$0$$
Now, $f_x=2x$ and $f_y=4y$ and $f_z=-6z$ and $\nabla f (1,0,1) =\lt 2,4, -6 \gt$ Now, $v=\dfrac{u}{|u|}=\dfrac{\lt 1,1,1 \gt}{\sqrt {3}}=\lt \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3} \gt$ Since, Directional derivative: $D_v \ h = \nabla h \ \cdot v$ or, $=\lt 2,4,-6 \gt \cdot \lt \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3} \gt$ or, $=0$