Compute the directional derivative of the function g(x, y)= sin(π(x−5y)).

Step-by-step explanation:

The directional derivative of a function in a particular direction u is given as the dot product of the unit vector in the direction of u and the gradient of the function

g(x,y) = sin(π(x−5y)

∇g = [(∂/∂x)î + (∂/∂y)j + (∂/∂z)ķ] [sin(π(x−5y))

(∂/∂x) g = (∂/∂x) sin (πx−5πy) = π [cos(π(x−5y))]

(∂/∂y) g = (∂/∂y) sin (πx−5πy) = - 5π [cos (π(x−5y))]

∇g = π [cos(π(x−5y))] î - 5π [cos (π(x−5y))] j

∇g = π [cos (π(x−5y))] [î - 5j]

So, the question requires a direction vector and a point to fully evaluate this directional derivative now.

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x