To find the greatest value of the directional derivatives of the function f = x^2yz^3 at a given point (2, 1, -1), we need to find the direction in which the derivative is maximized.



The directional derivative of a function f in the direction of a unit vector u is given by the dot product of the gradient of f and the unit vector u.

The gradient of a function f = x^2yz^3 is given by:

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z) = (2xyz^3, x^2z^3, 3x^2yz^2)

Let's find the gradient at the given point (2, 1, -1):

∇f(2, 1, -1) = (2(2)(1)(-1)^3, (2)^2(-1)^3, 3(2)^2(1)(-1)^2) = (-4, -4, -12)

The magnitude of the gradient vector is given by:

|∇f| = sqrt((-4)^2 + (-4)^2 + (-12)^2) = sqrt(16 + 16 + 144) = sqrt(176) = 4√11



To find the direction in which the derivative is maximized, we need to find a unit vector in the same direction as the gradient vector.

Unit vector u = (u1, u2, u3) = (-4/|∇f|, -4/|∇f|, -12/|∇f|) = (-4/(4√11), -4/(4√11), -12/(4√11)) = (-√11/√11, -√11/√11, -3√11/√11) = (-1/√11, -1/√11, -3/√11)



The directional derivative in the direction of the unit vector u is given by:

Duf = ∇f · u = (-4, -4, -12) · (-1/√11, -1/√11, -3/√11) = 4/√11 + 4/√11 + 36/√11 = 8/√11 + 36/√11 = 44/√11

To find the greatest value of the directional derivative, we need to rationalize the denominator:

(44/√11) * (√11/√11) = 44√11/11

Therefore, the greatest value of the directional derivative of the function f = x^2yz^3 at (2, 1, -1) is 44√11/11.

The approximate value of 44√11/11 is between 13.0 and 13.7.