Hi I am currently revising for an exam and have seen a question from a past paper and I am not sure the best way to go about it, do I change the $\nabla$ $\phi$ and $\nabla$ $\psi$ to vector fields?And then use
$\nabla$ $\bullet$ ($F$ $\times$ $G$)= ($\nabla$$\times$ $F$)$\bullet$ $G$ -$F$$\bullet$($\nabla$ $\times$$G$)?
Here is the equation to prove;
$\nabla$ $\bullet$($\nabla$ $\phi$ $\times$$\nabla$ $\psi$)$f$)=($\nabla$ $\phi$ $\times$$\nabla$ $\psi$)$\bullet$$\nabla$$f$.
Where $f$,$\phi$ and $\psi$ are smooth scalar fields and $F$ & $G$ are vector fields. We may use other differential identities stated at the top of the paper( The usual 7). Any help would be appreciated as its tomorrow.
$\nabla \phi$ and $\nabla \psi$ are already vector fields. (You don't need to think about "changing them into vector fields"!)
And $\nabla \phi \times \nabla \psi$ is also a vector field!
Therefore, a good way to start is to apply the general vector identity $$\nabla . (f\mathbf A) = \nabla f.\mathbf A + f(\nabla . \mathbf A) . $$ to show that $$ \nabla . (f(\nabla \phi \times \nabla \psi)) = \nabla f. (\nabla \phi \times \nabla \psi) + f (\nabla . (\nabla \phi \times \nabla \psi)).$$
Having done this, you could then try to use the vector identity $$ \nabla . (\mathbf A \times \mathbf B) = (\nabla \times \mathbf A) . \mathbf B - \mathbf A . (\nabla \times \mathbf B)$$ to show that
$$\nabla . (\nabla \phi \times \nabla \psi) = (\nabla \times \nabla \phi) . \nabla \psi - \nabla \phi . (\nabla \times \nabla \psi),$$ which is zero, because the curl of a gradient is always zero.
You can find a full list of vector identities here.