Answer: The field F has a continuous partial derivative on R.
Step-by-step explanation:
For the field F has a continuous partial derivative on R, fxy must be equal to fyx and since our field F is ∇f,
∇f = fxi + fyj + fzk.
Comparing the field F to ∇f since they at equal, P = fx, Q = fy and R = fz
Since P is fx therefore;
∂P ∂y = ∂ ∂y( ∂f ∂x) = ∂2f ∂y∂x
Similarly,
Since Q is fy therefore;
∂Q ∂x = ∂ ∂x( ∂f ∂y) = ∂2f ∂x∂y
Which shows that ∂P ∂y = ∂Q ∂x
The same is also true for the remaining conditions given