## Calculus 8th Edition

$\dfrac{25}{6}$
Formula to calculate the directional derivative: $D_uf=\nabla f(x,y,z) \cdot u$ or, $D_uf=|\nabla f(x,y)||u| \cos \theta$ Given: $f(x,y,z)=x^2y+x\sqrt {1+z},(1,2,3)$ and in the direction towards the point $v=2i+j-2k$ Thus, $u=\dfrac{v}{|v|}$ $D_uf(x,y)=\dfrac{v}{|v|}=\sqrt{(2)^2+(1)^2+(-2)^2}=3$ This implies From the given data, we have : $(x,y,z)=$ $(1,2,3)$ $D_uf(1,2,3)=\nabla f(x,y,z) \cdot u=4+\dfrac{1}{3}-\dfrac{1}{6}$ $D_uf(1,2,3)=\dfrac{25}{6}$