EXAMPLE 5 If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (1, 2, 0) in the direction of v = i + 4j − k. SOLUTION (a) The gradient of f is ∇f(x, y, z) = fx(x, y, z), fy(x, y, z), fz(x, y, z)

Step-by-step explanation:

Given f(x, y, z) = x sin(yz), the formula for calculating the gradient of the function is expressed as ∇f(x, y, z) = fx(x, y, z)i+ fy(x, y, z)j+fz(x, y, z)k where;

fx, fy and fz are the differential of the functions with respect to x, y and z respectively.

a) ∇f(x, y, z) = sin(yz)i + xzcos(yz)j + xycos(yz)k

The gradient of f =  sin(yz)i + xzcos(yz)j + xycos(yz)k

b) Directional derivative of f at (1,2,0) in the direction of v = i + 4j − k is expressed as ∇f(1, 2, 0)*v

∇f(1, 2, 0) = sin(2(0))i +1*0cos(2*0)j + 1*2cos(2*0)k

∇f(1, 2, 0) = sin0i +0cos(0)j + 2cos(0)k

∇f(1, 2, 0) = 0i +0j + 2k

Given v = i + 4j − k

∇f(1, 2, 0)*v (note that this is the dot product of the two vectors)

∇f(1, 2, 0)*v =  (0i +0j + 2k)*(i + 4j − k )

Given i.i = j.j = k.k =1 and i.j=j.i=j.k=k.j=i.k = 0

∇f(1, 2, 0)*v = 0(i.i)+4*0(j.j)+2(-1)k.k

∇f(1, 2, 0)*v = 0(1)+0(1)-2(1)

∇f(1, 2, 0)*v =0+0-2

∇f(1, 2, 0)*v= -2

Hence, the directional derivative of f at (1, 2, 0) in the direction of v = i + 4j − k is -2

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?