Find the directional derivative of the given function at the specified point $P$ in the direction $\vec { u }$ $f ( x , y ) = \cos ( x y ) , P = \left( 2 , \frac { \pi } { 6 } \right) , \vec { u } = \left\langle \frac { 3 } { \sqrt { 10 } } , \frac { 1 } { \sqrt { 10 } } \right\rangle$
A) $\frac { - 4 \sqrt { 10 } - \pi \sqrt { 30 } } { 40 }$
B) $- 3 \pi \sqrt { 3 } - 4 \sqrt { 3 }$
C) $\frac { \pi \sqrt { 3 } } { 12 } - \sqrt { 3 }$
D) $\frac { 3 \pi \sqrt { 3 } + 4 \sqrt { 3 } } { \sqrt { 10 } }$

Question

Quizplus · Accepted Answer

The directional derivative of $f$ at point $P$ in the direction of $\vec{u}$ is given by $\nabla f(P) \cdot \vec{u}$, where $\nabla f$ is the gradient of $f$. First, calculate the partial derivatives of $f$: $\frac{\partial f}{\partial x} = -y\sin(xy)$ and $\frac{\partial f}{\partial y} = -x\sin(xy)$. At $P(2, \frac{\pi}{6})$, these become $\frac{\partial f}{\partial x} = -\frac{\pi}{6}\sin(\frac{\pi}{3})$ and $\frac{\partial f}{\partial y} = -2\sin(\frac{\pi}{3})$. Since $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, we have $\frac{\partial f}{\partial x} = -\frac{\pi\sqrt{3}}{12}$ and $\frac{\partial f}{\partial y} = -\sqrt{3}$. The gradient of $f$ at $P$ is $\nabla f(P) = \left\langle -\frac{\pi\sqrt{3}}{12}, -\sqrt{3} \right\rangle$. The directional derivative is $\nabla f(P) \cdot \vec{u} = \left\langle -\frac{\pi\sqrt{3}}{12}, -\sqrt{3} \right\rangle \cdot \left\langle \frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}} \right\rangle = \frac{-3\pi\sqrt{3}}{12\sqrt{10}} - \frac{\sqrt{3}}{\sqrt{10}} = \frac{-\pi\sqrt{30} - 4\sqrt{10}}{40}$.

Find the Directional Derivative of the Given Function at the Specified