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 βf,
βf = fxi + fyj + fzk.
Comparing the field F to βf since they at equal, P = fx, Q = fy and R = fz
Since P is fx therefore;
βP βy = β βy( βf βx) = β2f βyβx
Similarly,
Since Q is fy therefore;
βQ βx = β βx( βf βy) = β2f βxβy
Which shows that βP βy = βQ βx
The same is also true for the remaining conditions given