Given Function:

f(x,y,z) = xy^2yz^3

Point:

(2,-1,1)

Direction:

Outward normal to the surface x log z -y^2 = 4 at (-1,2,1)

To find the directional derivative of the given function in the given direction, we need to follow the following steps:

Step 1: Find the gradient of the given function at the given point.

Step 2: Find the equation of the plane tangent to the given surface at the given point.

Step 3: Find the equation of the line passing through the given point and perpendicular to the plane found in step 2.

Step 4: Find the dot product of the gradient found in step 1 and the unit vector in the direction of the line found in step 3.

Step 1: Gradient of the function:

∇f(x,y,z) = (y^2z^3, 2xyz^3, 3xy^2z^2)

∇f(2,-1,1) = (1, -4, -6)

Step 2: Equation of the tangent plane:

The given surface can be written as:

x log z -y^2 = 4

Differentiating w.r.t x, y, and z, we get:

log z = 4x / (z log e)

-2y = 0

x / z = 0

At the point (-1,2,1), we have:

log 1 = 4(-1) / (1 log e) = -4

-2(2) = -4

(-1) / 1 = -1

Thus, the equation of the tangent plane at point (-1,2,1) is:

(x + 1) + 2(y - 2) - (z - 1) = 0

x + 2y - z - 1 = 0

Step 3: Equation of the line perpendicular to the plane:

The line perpendicular to the plane is the normal to the plane. The normal to the plane is the same as the direction of the outward normal to the surface at the given point.

The equation of the surface is:

x log z -y^2 = 4

Differentiating w.r.t x, y, and z, we get:

log z = 4x / (z log e)

-2y = 0

x / z = 0

At the point (-1,2,1), we have:

log 1 = 4(-1) / (1 log e) = -4

-2(2) = -4

(-1) / 1 = -1

Thus, the gradient of the surface at point (-1,2,1) is:

∇f(-1,2,1) = (log 1, -4, 0) = (0, -4, 0)

The unit vector in the direction of the outward normal is:

n = (0, -4, 0) / 4 = (0, -1, 0)

The equation of the line passing through the point (2,-1,1) and in the direction of the outward normal is:

x = 2 + 0t

y = -1 - t

z = 1 + 0t

Ok