Gradient of Divergence

Theorem
Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions.

Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.

Let $\mathbf V: \R^3 \to \R^3$ be a vector field on $\R^3$:


 * $\mathbf V := \tuple {\map {V_x} {x, y, z}, \map {V_y} {x, y, z}, \map {V_z} {x, y, z} }$

Then:
 * $\map \grad {\operatorname {div} \mathbf V} = \paren {\dfrac {\partial^2 V_x} {\partial x^2} + \dfrac {\partial^2 V_y} {\partial x \partial y} + \dfrac {\partial^2 V_z} {\partial x \partial z} } \mathbf i + \paren {\dfrac {\partial^2 V_x} {\partial x \partial y} + \dfrac {\partial^2 V_y} {\partial y^2} + \dfrac {\partial^2 V_z} {\partial y \partial z} } \mathbf j + \paren {\dfrac {\partial^2 V_x} {\partial x \partial z} + \dfrac {\partial^2 V_y} {\partial y \partial z} + \dfrac {\partial^2 V_z} {\partial z^2} } \mathbf k$

where:
 * $\grad$ denotes the gradient operator
 * $\operatorname {div}$ denotes the divergence operator.

Proof
From Divergence Operator on Vector Space is Dot Product of Del Operator and definition of the gradient operator:

where $\nabla$ denotes the del operator.

Hence: