Term or identity for sum of the grad of each cartesian component If we have a vector field $\vec{A} = A_x \hat{x} + A_y \hat{y} + A_z \hat{z}$, is there a special term, symbolic representation or identity that can let me represent the vector $\nabla A_x + \nabla A_y + \nabla A_z$ in terms of only $\vec{A}$, independent of any coordinate system? 
Thanks in advance... 
Updated 
As advised by @joriki, I'll summarize the answer:
$$\nabla \left(\vec{A} \cdot \left(\hat{x} + \hat{y} + \hat{z}\right)\right) = \nabla \left( A_x  + A_y  + A_z  \right) = \nabla A_x  + \nabla A_y + \nabla A_z $$
$\left(\hat{x} + \hat{y} + \hat{z}\right)$ is a constant vector of magnitude $\sqrt{3}$.
However it appears that we can't do this in a manner independent of the coordinate system--the choice of the vector $\left(\hat{x} + \hat{y} + \hat{z}\right)$ depends on the reference directions chosen to decompose $\vec{A}$.
 A: We are given a vector field ${\bf x}\mapsto{\bf A}({\bf x})$ in $\Bbb R^3$, i.e., a law that "attaches" to each point ${\bf x}$ of some domain $\Omega\subset\Bbb R^3$ a vector ${\bf A}({\bf x})\in T_{\bf x}$. Here $T_{\bf x}$ denotes the tangent space at ${\bf x}$. If one writes out everything in coordinates one has
$${\bf A}({\bf x})=\bigl[A_1(x_1,x_2,x_3),A_2(x_1,x_2,x_3),A_3(x_1,x_2,x_3)\bigr]^\top\ ,$$
so that formally ${\bf A}$ has the same appearance as a map ${\bf f}:\ \Bbb R^3\to\Bbb R^3$.
Now we want to investigate the "infinitesimal change" of ${\bf A}$ when we move from a given point ${\bf p}\in\Omega$ to a nearby point ${\bf p}+{\bf X}$. When the increment ${\bf A}({\bf p}+{\bf X})-{\bf A}({\bf p})$ satisfies a relation of the form
$${\bf A}({\bf p}+{\bf X})-{\bf A}({\bf p})=L.{\bf X} + o\bigl(|{\bf X}|\bigr)\qquad ({\bf X}\to{\bf 0})$$
for some linear map $L:\ \Bbb R^3\to\Bbb R^3$ then ${\bf A}$ is called differentiable at ${\bf p}$, and $L=:d{\bf A}({\bf p})$ is the differential or derivative of ${\bf A}$ at ${\bf p}$. 
Since $d{\bf A}({\bf p})$ is, similarly to the Jacobian of an ${\bf f}:\ \Bbb R^3\to\Bbb R^3$, a linear map $\Bbb R^3\to\Bbb R^3$ it is not possible to represent it by a vector. This contrasts the differential $df$ of a scalar function, where $df({\bf p})$ can be represented by the gradient vector $\nabla f({\bf p})$.
In terms of coordinates and matrices $d{\bf A}({\bf p})$ is described by
$$d{\bf A}({\bf p}).{\bf X}=\left[\matrix{A_{1.1}& A_{1.2}& A_{1.3} \cr
A_{2.1}& A_{2.2}& A_{2.3}\cr
A_{3.1}& A_{3.2}& A_{3.3}\cr}\right]_{\bf p}\ \left[\matrix{X_1\cr X_2\cr X_3\cr}\right]\ .$$
In the rows of the matrix $\bigl[d{\bf A}({\bf p})\bigr]$ we see the gradients $\nabla A_i$ $\ (1\leq i\leq 3)$, evaluated at ${\bf p}$, of the coordinate functions describing ${\bf A}$.
A final remark: Your vector $\hat x+\hat y+\hat z$, in my notation: ${\bf e}_1+{\bf e}_2+{\bf e}_3$, does not play any rôle here.
