Curl and Divergence etc algebra manipulating quick question

[binbagsss]

$\nabla p = \rho \nabla \phi$

My textbook says that by taking the curl we get:

$0=\nabla \rho X \nabla \phi$ **

I don't follow. I understand the LHS is zero, by taking the curl of a divergence.
But I'm unsure as to how we get it into this form, from which it is clear that the gradients of $\rho$ and
$\phi$ are parallel, since I get:

$\nabla X \rho \nabla \phi$, I know that the curl acting on a scalar field doesn't make sense, I would get $\rho \nabla X \nabla \phi$, taking the scalar field $\rho$ to the left since it can not be operated on by a curl. I don't see how you would get **

Many thanks in advance

 

[Fightfish]

It is rather straightforward to prove that
<br /> \nabla \times (a \mathbf{B}) = (\nabla a) \times \mathbf{B} + a (\nabla \times \mathbf{B})<br />
where $a$ is a scalar field and $\mathbf{B}$ is a vector field. This formula is also readily available in most lists of vector calculus identities (although I would strongly recommend you try to prove it yourself)

 

[Charles Link]

One quick comment: The LHS is the curl of a "gradient". (The curl of a gradient is zero and so is the divergence of a curl).