Theorem: Product Rule for Curl

Let $R$ be a region of space embedded in Cartesian $3$ space $\R^3$.
Let $\mathbf A$ be a vector field over $\mathbf V$.
Let $U$ be a scalar field over $\mathbf V$.
Then:
{{begin-eqn}}
{{eqn | l = \map \curl {U \mathbf A}
      | r = U \curl \mathbf A + \grad U \times \mathbf A
}}
{{eqn | r = U \curl \mathbf A - \mathbf A \times \grad U 
}}
{{end-eqn}}
where:
:$\curl$ denotes the curl operator
:$\grad$ denotes the gradient operator
:$\times$ denotes vector cross product

Proof:
From Curl Operator on Vector Space is Cross Product of Del Operator and definition of the gradient operator:
{{begin-eqn}}
{{eqn | l = \curl \mathbf A
      | r = \nabla \times \mathbf A
}}
{{eqn | l = \grad \mathbf U
      | r = \nabla U
}}
{{end-eqn}}
where $\nabla$ denotes the del operator.
Hence we are to demonstrate that:
{{begin-eqn}}
{{eqn | l = \nabla \times \paren {U \mathbf A}
      | r = \map U {\nabla \times \mathbf A} + \paren {\nabla U} \times \mathbf A
}}
{{eqn | r = \map U {\nabla \times \mathbf A} - \mathbf A \times \paren {\nabla U}
}}
{{end-eqn}}
Let $\mathbf A$ be expressed as a vector-valued function on $\mathbf V$:
:$\mathbf A := \tuple {\map {A_x} {\mathbf r}, \map {A_y} {\mathbf r}, \map {A_z} {\mathbf r} }$
where $\mathbf r = \tuple {x, y, z}$ is the position vector of an arbitrary point in $R$.
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
Then:
{{begin-eqn}}
{{eqn | l = \nabla \times \paren {U \mathbf A}
      | r = \paren {\dfrac {\partial U A_z} {\partial y} - \dfrac {\partial U A_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial U A_x} {\partial z} - \dfrac {\partial U A_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial U A_y} {\partial x} - \dfrac {\partial U A_x} {\partial y} } \mathbf k
      | c = {{Defof|Curl Operator}}
}}
{{eqn | r = \paren {U \dfrac {\partial A_z} {\partial y} + \dfrac {\partial U} {\partial y} A_z - U \dfrac {\partial A_y} {\partial z} - \dfrac {\partial U} {\partial z} A_y} \mathbf i
      | c = Product Rule for Derivatives
}}
{{eqn | o = 
      | ro= +
      | r = \paren {U \dfrac {\partial A_x} {\partial z} + \dfrac {\partial U} {\partial z} A_x - U \dfrac {\partial A_z} {\partial x} - \dfrac {\partial U} {\partial x} A_z} \mathbf j
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {U \dfrac {\partial A_y} {\partial x} + \dfrac {\partial U} {\partial x} A_y - U \dfrac {\partial A_x} {\partial y} - \dfrac {\partial U} {\partial y} A_x } \mathbf k
      | c = 
}}
{{eqn | r = \map U {\paren {\dfrac {\partial A_z} {\partial y} - \dfrac {\partial A_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial A_x} {\partial z} - \dfrac {\partial A_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial A_y} {\partial x} - \dfrac {\partial A_x} {\partial y} } \mathbf k}
      | c = rearrangement
}}
{{eqn | o = 
      | ro= +
      | r = \paren {\dfrac {\partial U} {\partial y} A_z - \dfrac {\partial U} {\partial z} A_y} \mathbf i + \paren {\dfrac {\partial U} {\partial z} A_x - \dfrac {\partial U} {\partial x} A_z} \mathbf j + \paren {\dfrac {\partial U} {\partial x} A_y - \dfrac {\partial U} {\partial y} A_x} \mathbf k
      | c = 
}}
{{eqn | r = \map U {\nabla \times \mathbf f} + \paren {\dfrac {\partial U} {\partial y} A_z - \dfrac {\partial U} {\partial z} A_y} \mathbf i + \paren {\dfrac {\partial U} {\partial z} A_x - \dfrac {\partial U} {\partial x} A_z} \mathbf j + \paren {\dfrac {\partial U} {\partial x} A_y - \dfrac {\partial U} {\partial y} A_x} \mathbf k
      | c = {{Defof|Curl Operator}}
}}
{{eqn | r = \map U {\nabla \times \mathbf f} + \paren {\dfrac {\partial U} {\partial x} \mathbf i + \dfrac {\partial U} {\partial y} \mathbf j + \dfrac {\partial U} {\partial z} \mathbf k} \times \paren {A_x \mathbf i + A_y \mathbf j + A_z \mathbf k}
      | c = {{Defof|Cross Product}}
}}
{{eqn | r = \map U {\nabla \times \mathbf f} + \paren {\nabla U} \times \mathbf f
      | c = {{Defof|Gradient Operator}}, {{Defof|Vector (Linear Algebra)|Vector}}
}}
{{end-eqn}}
{{qed}}