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
$\nabla$.($\phi$A)=A.$\nabla$$\phi$+$\phi$.$\nabla$A

Homework Equations

The Attempt at a Solution

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

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

I'm struggling to work out the last part:
$\phi$.$\nabla$A

I tried working out $\phi$.grad(A)? but the answer sheet has
div(A) = z
$\phi$div(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
$\nabla$.($\phi$A)=A.$\nabla$$\phi$+$\phi$.$\nabla$A

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.