Why does $\nabla$ behave like a member of $\mathbb{R}^3$ (Euclidean vector in 3d space) in many cases? Is there a reason why $\vec{\nabla} = \left[\; \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\; \right]$, behaves like a member of $\mathbb{R}^3$ (Euclidean vector in 3d space) in so many cases:


*

*Dot products and cross products of $\vec{\nabla}$ with multivariable vector function like $\vec{\nabla} \cdot \mathbf{\vec{f}}$ and $\vec{\nabla} \times \mathbf{\vec{f}}$ or with scalar functions (gradient) are vectors (i.e. independent of particular choice of co-ordinates).





*Properties like the Lagrange's formula that are valid for vectors have direct analogues involving the $\vec{
\nabla}$ operator.





*We can write pseudo-determinants for curls just like we could with components of vectors.





*Most of its vector properties hold if $\vec{\nabla}$ is replaced by any other vector.


There will be separate proofs for all of these properties. But is there any underlying reason that works for all such properties? Is there a fundamental reason why $\vec{\nabla}$ is $\mathbb{R}^3$ vector-ish?
 A: The gradient operator acts on a scalar differentiable function $f(\vec x)$, where $\vec x \in \mathbb R^n$, and returns a vector:
$$\text{grad} \ f(\vec x) = \nabla f(\vec x) \equiv \sum_{i=1}^n \frac{\partial f (\vec x)}{\partial x_i} \vec e_i $$
where $\{\vec e_i \dots\vec e_n\}$ is an orthogonal basis of $\mathbb R^n$.
The divergence operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^n$, and returns a scalar function:
$$\text{div} \ \vec F(\vec x) = \nabla \cdot \vec F(\vec x) \equiv \sum_{i=1}^n \frac{\partial F_i (\vec x)}{\partial x_i} $$
The curl operator acts on a vector field $\vec F(\vec x)$, where $\vec x,\vec F \in \mathbb R^3$, and returns a vector field:
$$\text{curl} \ \vec F(\vec x) = \nabla \times \vec F(\vec x) = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \hat i+\left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \hat j+ \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \hat k$$
where $\hat i, \hat j, \hat k$ are the unit vectors of the three Cartesian axes. 
Notice that, unlike the gradient and divergence, the curl operator does not generalize simply in $n$ dimensions. Also, the notation $\nabla \times \vec F$ is only a mnemonic device useful when we work in cartesian coordinates: in other coordinate systems, applying $\nabla \times \vec F$ will hold the wrong result.
We should probably also mention the laplacian operator, which is the divergence of the gradient:
$$\nabla^2 f(\vec x) \equiv \text{div} \ (\text{grad} \ f(\vec x)) = \nabla \cdot (\nabla f(\vec x))$$

So, to sum up, $\nabla$ is just a useful notation that is used in the representation of three different vector operators. It turns out that we can often formally manipulate $\nabla$ as if it were a vector, but it is not a vector in the usual sense: $\nabla$ alone is meaningless. 
To see this, just consider one of the fundamental properties of vector spaces: if $v,w$ are elements of the vector space $V$, then $v+w$ is also an element of $V$.
Let's consider the vector space $\mathbb R^n$: what meaning should we give to an expression such as
$$\nabla + \vec x \ ?$$
the answer is: no meaning at all, because $\nabla$ is not a vector.
A: $$\vec{\nabla} = \left[\; \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\; \right]$$ behaves like a vector because $$\vec{\nabla}f $$  is a vector.
The vector space in which it is defined is not the usual $ \mathbb R^n$
It is a linear operator which operates on functions and the outcome is a vector whose components are partial derivatives of the function. 
