Final answer:
The vector field F(x,y) is not a conservative field as the cross-partials are not equal, therefore, there is no potential function for F(x,y).
Step-by-step explanation:
To determine if the vector field F(x,y)=(sin(xy)+xycos(xy), x²cos(xy)+2y) is a conservative field, one must check if the cross-partials are equal, that is, if ∂F₁/∂y = ∂F₂/∂x. Calculating these derivatives:
∂F₁/∂y = cos(xy) + xy(-sin(xy)) + xcos(xy) = cos(xy) - x²sin(xy) + xcos(xy)
∂F₂/∂x = 2xcos(xy) - x²sin(xy)
These are not equal, hence F(x,y) is not a conservative vector field, and there is no potential function.