Vector calculus identities

Statement

Let $U \subseteq \mathbb{R}^3$ be open, let $\phi, \psi \colon U \to \mathbb{R}$ be scalar fields, and let $A \colon U \to \mathbb{R}^3$ be a vector field. If these fields are of class $C^1$, then $$\nabla (\phi \psi) = \phi \nabla \psi + \psi \nabla \phi, \qquad \nabla \cdot (\psi A) = \psi \, \nabla \cdot A + \nabla \psi \cdot A, \qquad \nabla \times (\psi A) = \psi \, \nabla \times A + \nabla \psi \times A.$$ If these fields are of class $C^2$, then $$\nabla \times (\nabla \phi) = 0, \qquad \nabla \cdot (\nabla \times A) = 0, \qquad \nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \Delta A,$$ where $\Delta A$ applies the Laplacian to each component of $A$.

Remarks

Here $\nabla \phi$ is the gradient, $\nabla \cdot A$ the divergence, $\nabla \times A$ the curl, and $\Delta$ the Laplacian. The product rules follow from the product rule for partial derivatives, applied to each component. The identities curl of a gradient is zero and divergence of a curl is zero follow from the symmetry of second derivatives. Also, $\nabla \cdot (\nabla \phi) = \Delta \phi$ and $\Delta (\phi \psi) = \phi \Delta \psi + 2 \nabla \phi \cdot \nabla \psi + \psi \Delta \phi$.

Explanations

L1 human

These identities are the algebra rules of vector calculus. They let you expand derivatives of products and simplify repeated derivatives, the way the ordinary product rule does in one variable.