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Find the directional derivative off at the given point in the direction indicated by the angle 0. f(x, y) = arctan(xy), (6,-5), θ= 3π/4 Duf(6, -5) =

User Asitis
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Final answer:

The directional derivative of the function f(x, y) = arctan(xy) at the point (6, -5) in the direction indicated by the angle θ = 3π/4 is -25/91 * sqrt(2).

Step-by-step explanation:

The directional derivative of a function f(x, y) in the direction of an angle θ is given by the dot product of the gradient of f with the unit vector parallel to the given direction.

In this case, the function is f(x, y) = arctan(xy), and we need to find the directional derivative at the point (6, -5) in the direction indicated by the angle θ = 3π/4.

To find the gradient of f, we take the partial derivatives with respect to x and y.

The partial derivative of f with respect to x is ∂f/∂x = y/(1 + x^2y^2), and the partial derivative of f with respect to y is ∂f/∂y = x/(1 + x^2y^2).

The gradient of f at (6, -5) is ∇f = (∂f/∂x, ∂f/∂y) = (-25/1826, 30/1826).

To find the unit vector in the direction of the angle θ, we use the trigonometric identities cos(θ) = cos(3π/4) = -sqrt(2)/2 and sin(θ) = sin(3π/4) = sqrt(2)/2.

The directional derivative Duf at (6, -5) in the direction indicated by the angle θ = 3π/4 is given by the dot product of the gradient of f and the unit vector in the direction of θ:

Duf(6, -5) = ∇f · u = (-25/1826, 30/1826) · (-sqrt(2)/2, sqrt(2)/2) = -55/1826 * sqrt(2) + 30/1826 * sqrt(2) = -25/91 * sqrt(2).

User Albeiro
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