Show that: $\nabla \times (\phi F) = \nabla \phi \times F + \phi \nabla \times F$. Where F is any vector field, and \phi is any scalar field.
Let F = (P,Q,R). Now by observation, the first term of the RHS of the identity is zero since the curl of a gradient field is 0.
Hence we are trying to prove that: $$\nabla \times (\phi F) = \phi \nabla \times F$$
Now if I compute the LHS I get:
$$\nabla \times (\phi F) = (\phi _y R_y -\phi _z Q_z) \hat i - (\phi _x R_x - \phi _z R_z) \hat j + (\phi_x Q_x - \phi_y P_y) \hat k$$
and the RHS:
$$\phi \nabla \times F = \phi [(R_y -Q_z) \hat i - (R_x - R_z) \hat j + ( Q_x - P_y) \hat k]$$
But this is not equivalent??
Any know what I have done wrong?