Answer:
-2
Step-by-step explanation:
The directional derivative of the function f(x, y) = x²y³ − 3y at the point (2, −1) in the direction of the vector v = 5i + 2j is expressed using the formula
∇f(x,y)•v where ∇f(x,y) is the gradient of the function f(x,y)
∇f(x,y) = δ/δx(x²y³ − 3y)i + δ/δy(x²y³ − 3y)j
∇f(x,y) = 2xy³i+(3x²y²-3)j
∇f(x,y) at (2,-1) = 2(2)(-1)³i + {3(2)²(-1)²-3}j
∇f(x,y) at (2,-1) = -4i+(12-3)j
∇f(x,y) at (2,-1) = -4i+9j
∇f(x,y)•v = (-4i+9j)(5i+2j)
Note that i•i = j•j = 1 and i•j = 0
∇f(x,y)•v = -4(5)+9(2)
∇f(x,y)•v = -20+18
∇f(x,y)•v = -2
