What is the difference between $(\phi\vec{\nabla}) \cdot \vec{A}$ and $\phi(\vec{\nabla}\cdot\vec{A})$? $\vec{\nabla} \cdot (\phi\vec{A}) = (\phi\vec{\nabla}) \cdot \vec{A} + \phi(\vec{\nabla}\cdot\vec{A})$
But what is the difference between $(\phi\vec{\nabla}) \cdot \vec{A}$ and $\phi(\vec{\nabla}\cdot\vec{A})$ ?
I cannot get that. They are different, but when I try to work it out, they both becomes the same.
Please state  the difference between the above two quantities and please provide an example.
I am in my First Semester of Physics Undergrad course.
 A: I am not familiar with the notation you are using, but it is clear to me that $\phi \nabla$ is meant to be the derivative of $\phi$. If one uses einstein notation, we can see that the given equation should follow from just the 1D product rule you are familiar with,
$$ \partial_i (\phi A_i) = \partial_i\phi A_i + \phi(\partial_i A_i) $$
Since clearly $\partial_iA_i = \nabla\cdot A$, we are forced to conclude that $\partial_i \phi = \phi \nabla$.
Note that this is distinct from the term in (say) the material derivative, which in coordinates is
$$ [(u\cdot \nabla) u]_i = u_j\partial_ju_i$$

user470992 has kindly requested a computed example with the functions $$A = \begin{pmatrix} 3xyz^2 \\ + 2xy^3\\ - x^2yz \end{pmatrix}$$ and $\phi = 3x^2 - yz$ ; calculating $\nabla \cdot ( \phi \vec{A}) $ at $p=(1,-1,1)^T$ :
$$\nabla \phi|_p = \left.\begin{pmatrix} 6x \\-z\\-y \end{pmatrix}\right|_p = \begin{pmatrix} 6 \\-1\\1 \end{pmatrix} \\ \nabla \cdot A|_p = (3yz^2+6xy^2-x^2y)|_p = -3+6+1=4$$
This gives $ 6(-3) + (-1)(-2) + 1(1) + 4(4)=-15+16 = 1$.
A: It should be $${\nabla} \cdot (\phi\vec{A}) = ({\nabla_\phi \phi}) \cdot \vec{A} + \phi({\nabla}_A\cdot\vec{A})$$
where the frist $\nabla_\phi$ works on $\phi$, and the second $\nabla_A$ works on $\vec{A}$.
A: The equation
$\vec \nabla \cdot (\phi \vec A) = (\phi \vec \nabla)\cdot \vec A + \phi \vec \nabla \cdot \vec A \tag 1$
is wrong as written.  The correct expression for $\vec \nabla \cdot (\phi \vec A)$ is
$\vec \nabla \cdot (\phi \vec A) = (\vec \nabla \phi )\cdot \vec A + \phi \vec \nabla \cdot \vec A. \tag 2$
(2) may be verified by evaluating the terms in a cartesian coordinate system.
If
$A = (A_x, A_y, A_z), \tag 3$
then
$\dfrac{\partial}{\partial x}(\phi A_x) = \dfrac{\partial \phi}{\partial x} A_x + \phi \dfrac{\partial A_x}{\partial x}, \tag 4$
$\dfrac{\partial}{\partial y}(\phi A_y) = \dfrac{\partial \phi}{\partial y} A_y + \phi \dfrac{\partial A_y}{\partial y}, \tag 5$
$\dfrac{\partial}{\partial z}(\phi A_z) = \dfrac{\partial \phi}{\partial z} A_z + \phi \dfrac{\partial A_z}{\partial z}; \tag 6$
summing (4)-(6) yields
$\vec \nabla \cdot (\phi A) = \dfrac{\partial}{\partial x}(\phi A_x) + \dfrac{\partial}{\partial y}(\phi A_y) + \dfrac{\partial}{\partial z}(\phi A_z)$
$= \dfrac{\partial \phi}{\partial x} A_x + \dfrac{\partial \phi}{\partial y} A_y + \dfrac{\partial \phi}{\partial z} A_z + \phi \dfrac{\partial A_x}{\partial x} + \phi \dfrac{\partial A_y}{\partial y} + \phi \dfrac{\partial A_z}{\partial z} = \vec \nabla \phi \cdot \vec A + \phi \vec \nabla \cdot \vec A, \tag 7$
validating (2).
The usual notation is that operators act from left to right.  Therefore,
$(\phi \vec \nabla)\cdot \vec A = \phi \vec \nabla \cdot \vec A, \tag 8$
as my easily seen via operator associativity or direct evaluation in a cartesian frame as was done above in the case of (2); there is in fact no essential difference 'twixt $(\phi \vec \nabla) \cdot \vec A$ and $\phi \vec \nabla \cdot \vec A$. Thus, (1) also may read
$\vec \nabla \cdot (\phi \vec A) = 2\phi \vec \nabla \cdot \vec A, \tag 9$
which omits the derivatives of $\phi$ required by the Leibniz rule for differentiating products. 
For example, if we take
$\vec A = (x, y, z) \tag{10}$
and
$\phi = y, \tag{11}$
we obtain
$\vec \nabla \cdot (\phi \vec A) = \vec \nabla \cdot (xy, y^2, yz) = y + 2y + y = 4y, \tag{12}$
$\vec \nabla \phi \cdot \vec A = (0, 1, 0) \cdot (x, y, z) = y, \tag{13}$
$\phi \vec \nabla \cdot A = y \vec \nabla \cdot (x, y, z) = 3y; \tag{14}$
one sees that (2) holds here.  Evaluating (1) or (9) however yields
$4y = 2y(3) = 6y, \tag{15}$
which is not, in general, the case.
