Chapter 13 - Section 13.5 - Directional Derivatives and Gradient Vectors - Exercises - Page 720: 16

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$f_x=2x$ 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 $ The directional derivative can be found as: $D_v h = \nabla h \cdot v$ Therefore, $D_v h = \nabla h \cdot v=\lt 2,4,-6 \gt \cdot \lt \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3}, \dfrac{1}{\sqrt 3} \gt=0$