Product Rule for Curl

Theorem
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 $\mathbf V$.

Let $\nabla \times \mathbf f$ denote the curl of $f$.

Then:

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:

where $\nabla$ denotes the del operator.

Hence we are to demonstrate that:

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:

Also presented as
This result can also be presented as:


 * $\nabla \times \paren {U \mathbf A} = \map U {\nabla \times \mathbf A} + \paren {\nabla U} \times \mathbf A$

or:


 * $\nabla \times \paren {U \mathbf A} = \map U {\nabla \times \mathbf A} - \mathbf A \times \paren {\nabla U}$

presupposing the implementations of $\curl$ and $\grad$ as operations using the del operator.