I have to prove the following relation. I am looking for a solution beyond the obvious brute force method of considering $\mathbf{u}=u_1\mathbf{i}+u_2\mathbf{j}+u_3\mathbf{k}\;and\;\mathbf{v}=v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k}$ putting in these values and simply showing that the two sides come to the same result through a long and tedious calculation. Any thoughts?
$$∇(\mathbf{u}·\mathbf{v})=(\mathbf{v}·∇)\mathbf{u}+(\mathbf{u}·∇)\mathbf{v}+\mathbf{v}×(∇×\mathbf{u})+\mathbf{u}×(∇×\mathbf{v})\;where\;\mathbf{u},\mathbf{v}\;are\;vectors$$
P.S: I was not very sure whether I have added the correct tags or not. Feel free to add or remove tags as you deem appropriate.
A rather vanilla, if tedious, application associativity of the clifford algebra's geometric product and grade projection gets the job done.
$$\langle \nabla (uv)\rangle_1 = (\nabla \cdot u) v + (\nabla \wedge u) \cdot v + (u \cdot \nabla) v - (u \wedge \nabla) \cdot v = \nabla (u \cdot v) + \nabla \cdot (u \wedge v)$$
Apply the BAC-CAB rule to the last term:
$$\nabla \cdot (u \wedge v) = (\nabla \cdot u) v + (u \cdot \nabla) v - (\nabla \cdot v) u - (v \cdot \nabla) u$$
You'll note that the $(\nabla \cdot u) v + (u \cdot \nabla) v$ terms cancel, but some other terms do not:
$$\nabla (u \cdot v) = (\nabla \wedge u) \cdot v - (u \wedge \nabla) \cdot v + (\nabla \cdot v)u + (v \cdot \nabla) u$$
Not much to do but directly attack that $(u \wedge \nabla) \cdot v$ term:
$$(u \wedge \nabla) \cdot v = (\nabla \cdot v) u - \dot \nabla (\dot v \cdot u)$$
The dots mean that $\dot \nabla$ differentiates only $\dot v$ in $\dot v \cdot u$; effectively, $u$ is "held constant."
Great, so what is $\dot \nabla (\dot v \cdot u)$? Let's grade project again.
$$\langle (\nabla v) u \rangle_1 = (\nabla \cdot v) u + (\nabla \wedge v) \cdot u = \dot \nabla (\dot v \cdot u) + \dot \nabla \cdot (\dot v \wedge u) =\dot \nabla (\dot v \cdot u) + (\nabla \cdot v) u - (u \cdot \nabla) v$$
Putting that all together gives us
$$(u \wedge \nabla) \cdot v = (\nabla \cdot v) u - (u \cdot \nabla) v - (\nabla \wedge v) \cdot u$$
So the final result is
$$\nabla (u \cdot v) = (\nabla \wedge u) \cdot v + (u \cdot \nabla) v + (\nabla \wedge v) \cdot u + (v \cdot \nabla) u $$
Finally, note that the wedge and generalized dot products here are related to cross products or curl like so:
$$(\nabla \wedge u) \cdot v = [i(\nabla \times u)] \cdot v = i[ (\nabla \times u) \wedge v ] = -(\nabla \times u) \times v$$
That allows us to recover the more familiar result.