Answer
The directional derivative at $P = \left( {0,0,0} \right)$ in the direction of ${\bf{v}} = {\bf{j}} + {\bf{k}}$ is $\frac{1}{{\sqrt 2 }}$.
Work Step by Step
We are given $f\left( {x,y,z} \right) = \sin \left( {xy + z} \right)$ and ${\bf{v}} = {\bf{j}} + {\bf{k}}$.
First, we normalize the direction vector:
${\bf{u}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {0,1,1} \right)}}{{\sqrt {\left( {0,1,1} \right)\cdot\left( {0,1,1} \right)} }} = \left( {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right)$
Next, we compute the gradient at $P = \left( {0,0,0} \right)$:
$\nabla f = \left( {y\cos \left( {xy + z} \right),x\cos \left( {xy + z} \right),\cos \left( {xy + z} \right)} \right)$
$\nabla {f_P} = \left( {0,0,1} \right)$
Using Theorem 3 of Section 15.5, we obtain the directional derivative in the direction of ${\bf{v}} = {\bf{j}} + {\bf{k}}$ at $P = \left( {0,0,0} \right)$:
${D_{\bf{u}}}f\left( P \right) = \nabla {f_P}\cdot{\bf{u}}$
${D_{\bf{u}}}f\left( P \right) = \left( {0,0,1} \right)\cdot\left( {0,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right) = \frac{1}{{\sqrt 2 }}$
So, the directional derivative at $P = \left( {0,0,0} \right)$ in the direction of ${\bf{v}} = {\bf{j}} + {\bf{k}}$ is $\frac{1}{{\sqrt 2 }}$.
