Alternative expression of acceleration in vector form Let's imagine a one dimensional case, where a particle is moving with a velocity $v$ and an acceleration $a$. Thus
$$a=\frac{\mathrm dv}{\mathrm dt}\tag{1}$$
Applying the chain rule, equation $(1)$ can be rewritten as
$$a=\frac{\mathrm dv}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}\Longrightarrow \boxed{a=v\frac{\mathrm dv}{\mathrm dx}}\tag{2}$$
Now, if we were dealing with a 2D or a 3D case, then we would use vectors in the above expressions. Thus
\begin{alignat}{2}
a&=\frac{\mathrm dv}{\mathrm dt}&&\Longrightarrow\mathbf a=\frac{\mathrm d \mathbf v}{\mathrm dt}\tag{3}\\
a&=v\frac{\mathrm dv}{\mathrm dx}&&\Longrightarrow \mathbf a=\:\:?\tag{4}
\end{alignat}
As you can see, the vector form of equation $(1)$ (which is equation $(3)$) can be easily found, however I do not know of any way to express the equation $(2)$ in vector form.
The natural thought was to express the velocity into its components. For a 3D case, let $\mathbf v=v_x\mathbf{\hat i}+v_y\mathbf{\hat j}+v_z\mathbf{\hat k}$. Doing this, we have essentially converted the 3D case to three 1D cases. Thus using equation $(2)$:
$$\mathbf a =v_x\frac{\mathrm d v_x}{\mathrm dx}\mathbf{\hat i}+v_y\frac{\mathrm d v_y}{\mathrm dy}\mathbf{\hat j}+v_z\frac{\mathrm d v_z}{\mathrm dz}\mathbf{\hat k}\tag{5}$$
However, this expanded version doesn't seem particularly useful to me. Is there any way to express equation $(5)$ in a "closed form" (without explicitly writing out the components)? I feel that writing it in closed form might involve some common vector calculus operators (along with dot and cross products), though I am not exactly sure how to express it in a "closed form".

Justification of equation $(5)$: We kow that $\mathbf a=a_x\mathbf{\hat i}+a_y\mathbf{\hat j}+a_z\mathbf{\hat k}$
Now since
$$a_x=\frac{\mathrm dv_x}{\mathrm dt}=v_x\frac{\mathrm d v_x}{\mathrm d x}$$
Thus subsitituting this for every component, we re-obtain equation $(5)$.
 A: I am a bit confused with your notation so I chose my notation.
Assume the components of the position vector $\vec{R}=[x_1,x_2,\ldots,x_{n_R}]^T $are function of the generalized coordinates $q_1,q_2,\ldots,q_{n_Q}$   thus: $x_j=x_j(q_i)$ where $j=1,(1),n_R$ and $i=1,(1),n_Q\quad ,n_Q \le n_R$
We want to obtain the velocity vector $\vec{v}=\frac{d\vec{R}}{dt}$ 
$$\dot{x}_1=\frac{\partial x_1}{\partial q_1}\,\dot{q}_1+\frac{\partial x_1}{\partial q_2}\,\dot{q}_2+\ldots$$
$$\dot{x}_2=\frac{\partial x_2}{\partial q_1}\,\dot{q}_1+\frac{\partial x_2}{\partial q_2}\,\dot{q}_2+\ldots$$
or
$$\dot{x}_j=\sum_i^{nQ}\frac{\partial x_j}{\partial q_i}\,\dot{q}_i$$
or with vector notation (Engineer notation ) :
$$\vec{v}=\vec{\dot R}=\underbrace{\frac{\partial \vec{R}}{\partial \vec{q}}}_{n_R\times n_Q}\,\vec{\dot{q}}$$
Example:
$$\vec{R}=\left[ \begin {array}{c} x_{{1}} \left( q_{{1}},q_{{2}} \right) 
\\ x_{{2}} \left( q_{{1}},q_{{2}} \right) 
\\ x_{{3}} \left( q_{{1}},q_{{2}} \right) 
\end {array} \right] 
=\left[ \begin {array}{c} r\sin \left( q_{{1}} \right) \cos \left( q_{
{2}} \right) \\r\sin \left( q_{{1}} \right) \sin
 \left( q_{{2}} \right) \\ r\cos \left( q_{{1}}
 \right) \end {array} \right] 
$$
$$\vec{q}=\left[ \begin {array}{c} q_{{1}}\\ q_{{2}}
\end {array} \right] 
$$
$$\underbrace{\frac{\partial \vec{R}}{\partial \vec{q}}}_{3\times 2}=
 \left[ \begin {array}{cc} {\frac {\partial }{\partial q_{{1}}}}x_{{1}
} \left( q_{{1}},q_{{2}} \right) &{\frac {\partial }{\partial q_{{2}}}
}x_{{1}} \left( q_{{1}},q_{{2}} \right) \\ {\frac {
\partial }{\partial q_{{1}}}}x_{{2}} \left( q_{{1}},q_{{2}} \right) &{
\frac {\partial }{\partial q_{{2}}}}x_{{2}} \left( q_{{1}},q_{{2}}
 \right) \\{\frac {\partial }{\partial q_{{1}}}}x_{
{3}} \left( q_{{1}},q_{{2}} \right) &{\frac {\partial }{\partial q_{{2
}}}}x_{{3}} \left( q_{{1}},q_{{2}} \right) \end {array} \right] 
$$
Thus:
$$\vec{v}=\left[ \begin {array}{cc} r\cos \left( q_{{1}} \right) \cos \left( q_
{{2}} \right) &-r\sin \left( q_{{1}} \right) \sin \left( q_{{2}}
 \right) \\ r\cos \left( q_{{1}} \right) \sin
 \left( q_{{2}} \right) &r\sin \left( q_{{1}} \right) \cos \left( q_{{
2}} \right) \\ -r\sin \left( q_{{1}} \right) &0
\end {array} \right] 
\,\left[ \begin {array}{c} \dot{q}_{{1}}\\ \dot{q}_{{2}}
\end {array} \right] 
$$
Remark:
The velocity vector $\vec{v}$ is a function of $\vec{q}$ and $\vec{\dot{q}}$. Your vector v is only a function of $\vec{q}$ this is not the general case 
A: $$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{dx}{dt}\frac{\partial\mathbf{v}}{\partial x}+\frac{dy}{dt}\frac{\partial\mathbf{v}}{\partial y}+\frac{dz}{dt}\frac{\partial\mathbf{v}}{\partial z}=(\mathbf{v}\cdot\nabla)\mathbf{v}$$
