What is the Last Term in the Expression for \nabla\cdot(\phi\vec{A})?

[SAMSAM12]

Homework Statement

If \phi= xy^{2}
A=xzi-z^{2}j+xy^{2}k
B=zi+xj+yk

Verify that
\nabla.(\phiA)=A.\nabla\phi+\phi.\nablaA

Homework Equations

The Attempt at a Solution

I have worked out the first two parts of the question:
\phiA = (x^{2}y^{2}z, -xy^{2}z^{2},x^{2}y^{4})
div(\phiA) = 2xy^{2}z-2xyz^{2}

A.grad(\phi) = (xy^{2}z-2xyz^{2})

I'm struggling to work out the last part:
\phi.\nablaA

I tried working out \phi.grad(A)? but the answer sheet has
div(A) = z
\phidiv(A) = xy^{2}z

why?
Any help appreciated.

Merry Christmas and Happy new year

 

[brew_guru]

Ok, so to start compute \nabla A which will just be (\frac{\partial}{\partial x}\vec{A},\frac{\partial}{\partial y}\vec{A},\frac{\partial}{\partial z}\vec{A})
You will end up with a scalar, which you can multiply by your scalar \phi and you should end up with xy^2z.

 

[vela]

Homework Statement

If \phi= xy^{2}
A=xzi-z^{2}j+xy^{2}k
B=zi+xj+yk

Verify that
\nabla.(\phiA)=A.\nabla\phi+\phi.\nablaA

The last term of the expression doesn't make sense. You can't dot a scalar into anything. It should be
$$\nabla\cdot(\phi \vec{A}) = \vec{A}\cdot\nabla \phi + \phi\nabla\cdot\vec{A}$$

 

[SAMSAM12]

The last term of the expression doesn't make sense. You can't dot a scalar into anything. It should be
$$\nabla\cdot(\phi \vec{A}) = \vec{A}\cdot\nabla \phi + \phi\nabla\cdot\vec{A}$$

Thank you.