Can we differentiate with respect to a vector?

[KnightTheConqueror]

We know that a = vdv/dx
But is it applicable in only one dimensional components or is this actually a vector equation? If so then how do we exactly differentiate with respect to position 'vector'?

 

[KnightTheConqueror]

Thanks for answering, but I know that I can resolve it into components and solve. My problem is not that I am unable to use that equation in 3D. Rather, I am just curious whether we can write that equation as an vector equation directly without resolving.

Let me elaborate...

v = u + at here, although we can apply this equairon to x, y, z component separately and then add them but resolving into components is just something to make our work easier. v = u + at is still a vector equation and we can solve it using other methods of vector too for example geometry.
Likewise, I can apply a = vdv/dx to x, or vdv/dy to y
But
Can we just write that equation in vector form without needing to resolve into one dimensional components? Can we say that acceleration vector = velocity vector times d of velocity vector upon x of position vector.
If so then what exactly do we mean by differentiating with respect to a vector?

 

[TSny]

We know that a = vdv/dx
But is it applicable in only one dimensional components or is this actually a vector equation? If so then how do we exactly differentiate with respect to position 'vector'?

Let $s$ denote arc length along the trajectory of the particle. Consider the velocity of the particle as a function of arc length: $\vec v(s)$. Use the chain rule to express the acceleration as $\vec a = \dfrac{d\vec v}{ds} \dfrac{ds}{dt}$ and note that $\dfrac{ds}{dt}$ is the speed $v$ of the particle. So, $\vec a = v\dfrac{d\vec v}{ds}$.
(This expression is undefined at points of the trajectory where $v = 0$, just as the 1-D expression $a = v \dfrac{dv}{dx}$ is undefined when $v = 0$.)