If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k. solution (a) the gradient of f is ∇f(x, y, z) = fx(x, y, z), fy(x, y, z), fz(x, y, z) = . (b) at (2, 4, 0) we have ∇f(2, 4, 0) = < 0,0,8> . the unit vector in the direction of v = i + 3j − k is u = < 1 √11, 3 √11,− 1 √11> . therefore this equation gives duf(2, 4, 0) = ∇f(2, 4, 0) · u = 8k · 1 √11, 3 √11,− 1 √11 = − 8 √11 .

a) .

b)

Step-by-step explanation:

Given a function , this function has the following gradient:

.

(a) find the gradient of f

We have that . So

.

.

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

.

The vector is

To use v as an unitary vector, we divide each component of v by the norm of v.

So

Now, we can calculate the scalar product that is the directional derivative.

answer: 4/6 =5/7.5

step-by-step explanation: it is based on scale and the scale between the two is 1.25. the answer is correct but i dont know if i explained it right