Question

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xyz , (3, 3, 9), v = −1, −2, 2 duf(3, 3, 9)

Answer 1

The derivative of $f(x,y,z)$ in the direction of a vector $\mathbf u$ is

$\nabla_{\mathbf u}f=\nabla f\cdot\dfrac{\mathbf u}{\|\mathbf u\|}$

With $f(x,y,z)=xyz$, we get

$\nabla f=(yz,xz,xy)$

and $\mathbf u=(-1,-2,2)$,

$\|\mathbf u\|=\sqrt{(-1)^2+(-2)^2+2^2}=3$

Then

$\nabla_{(-1,-2,2)}f(3,3,9)=(27,27,9)\cdot\dfrac{(-1,-2,2)}3=-21$