Show that: $\nabla \times (\phi F) = \nabla \phi \times F + \phi \nabla \times F$ 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. 
My attempt:
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?
 A: Also 
\begin{align}
\nabla\times(\phi\mathbf{F})
&=\sum_{i,j,k=1}^3\varepsilon_{ijk}\frac{\partial}{\partial x_j}(\phi F_k)\mathbf{e}_i=\\
&=\sum_{i,j,k=1}^3\varepsilon_{ijk}\left[\left(\frac{\partial}{\partial x_j}\phi\right)F_k+\phi\frac{\partial}{\partial x_j}F_k\right]\mathbf{e} 
  _i=\\
&=\sum_{i,j,k=1}^3\varepsilon_{ijk}\left(\frac{\partial}{\partial x_j}\phi\right)F_k\mathbf{e}_i+\sum_{i,j,k=1}^3\varepsilon 
  _{ijk}\phi\frac{\partial}{\partial x_j}F_k\mathbf{e}_i=\\
&=(\nabla\phi)\times\mathbf{F}+\phi\nabla\times\mathbf{F}
\end{align}
A: The first term of RHS is not zero, it is
\begin{equation}
(\nabla \phi) \times F=\det
 \begin{bmatrix}
       \hat{i} & \hat{j} & \hat{k}           \\[0.3em]
       \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y}           & \frac{\partial \phi}{\partial z} \\[0.3em]
       F_x           & F_y & F_z
     \end{bmatrix},
\end{equation}
and the LHS is,
\begin{equation}
\nabla  \times (\phi F)=\det
 \begin{bmatrix}
       \hat{i} & \hat{j} & \hat{k}           \\[0.3em]
       \frac{\partial }{\partial x} & \frac{\partial }{\partial y}           & \frac{\partial }{\partial z} \\[0.3em]
       \phi F_x           & \phi F_y & \phi F_z
     \end{bmatrix},
\end{equation}
where you have to keep in mind the product rule for differentiation, for example,
\begin{equation}
\frac{\partial (\phi F_y)}{\partial x} = \frac{\partial \phi }{\partial x}F_y + \phi \frac{\partial F_y}{\partial x},
\end{equation} 
if you keep all this in mind I think you will be able to figure it out.
