## Thomas' Calculus 13th Edition

$$2$$
Directional derivative: $D_v h = \nabla h \cdot v$ Now, $$h_x= -y \sin xy +\dfrac{1}{x} \\ h_y= -x \sin xy +ze^{xy} \\ h_z=ye^{yz}+z^{-1}$$ So, $\nabla g (1,0,\dfrac{1}{2}) =\lt 1,\dfrac{1}{2},2 \gt$ Now, $v=\dfrac{u}{|u|}=\dfrac{\lt 1,2,2 \gt}{\sqrt {9}}=\dfrac{\lt 1,2,2 \gt}{3}=\lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt$ Since, $D_v \ h = \nabla h \cdot v$ or, $=\lt 1,\dfrac{1}{2},2 \gt \cdot \lt \dfrac{1}{3}, \dfrac{2}{3}, \dfrac{2}{3} \gt$ or, $=\dfrac{6}{3}$ or, $=2$