If $f\left( \vec{v} \left( s,t,r\right) \right) =p\left( x,y,z\right) \vec{i} +q\left( x,y,z\right) \vec{j} +u\left( x,y,z\right) \vec{k}$
$x=x\left( s,t,r\right)$
$y=y\left( s,t,r\right)$
$z=z\left( s,t,r\right)$
We have a multivariable vector-valued function (takes inputs as vectors too) where:
The input space (parameters) is a 3D in $\left( s,t,r\right)$
The intermediate space is a 3D in $\left( x,y,z\right)$
The output space is a 3D in $\left( p,q,u\right)$
I wanna get the derivative of f ... I thought of the following as the Partial derivative of f() w.r.t (s)
$ \frac{\partial f}{\partial s} = \left[ \begin{matrix}\frac{\partial p}{\partial x} &\frac{\partial p}{\partial y} &\frac{\partial p}{\partial z} \\ \frac{\partial q}{\partial x} &\frac{\partial q}{\partial y} &\frac{\partial q}{\partial z} \\ \frac{\partial u}{\partial x} &\frac{\partial u}{\partial y} &\frac{\partial u}{\partial z} \end{matrix} \right]\ .\left[ \begin{matrix}\frac{\partial x}{\partial s} \\ \frac{\partial y}{\partial s} \\ \frac{\partial z}{\partial s} \end{matrix} \right] $
3x3 matrix multiplied by a vector (3x1) and the result is a 3 element vector (3x1)
$\frac{\partial f}{\partial s} =\left[ \begin{matrix}\frac{\partial p}{\partial x} .\frac{\partial x}{\partial s} +\frac{\partial p}{\partial y} .\frac{\partial y}{\partial s} +\frac{\partial p}{\partial z} .\frac{\partial z}{\partial s} \\ \frac{\partial q}{\partial x} \cdot \frac{\partial x}{\partial s} +\frac{\partial q}{\partial y} .\frac{\partial y}{\partial s} +\frac{\partial q}{\partial z} .\frac{\partial z}{\partial s} \\ \frac{\partial u}{\partial x} .\frac{\partial x}{\partial s} +\frac{\partial u}{\partial y} .\frac{\partial y}{\partial s} +\frac{\partial u}{\partial z} .\frac{\partial z}{\partial s} \end{matrix} \right] $
Is that correct? And can we similarly find $ \frac{\partial f}{\partial t}$ and $ \frac{\partial f}{\partial r}$ ?
Is there a compact form for the rule?
Can we use the following notation?
$\frac{\partial f}{\partial s} =\triangledown f.\frac{\partial }{\partial s} \left[ \begin{matrix}x\\ y\\ z\end{matrix} \right] =\left[ \begin{matrix}p\\ q\\ u\end{matrix} \right] \left[ \begin{matrix}\frac{\partial }{\partial x} &\frac{\partial }{\partial y} &\frac{\partial }{\partial z} \end{matrix} \right] .\frac{\partial }{\partial s} \left[ \begin{matrix}x\\ y\\ z\end{matrix} \right] $
As in simpler forms:
if
$f\left( \vec{v} \left( t\right) \right) =f\left( x,y\right)$
$\vec{v} \left( t\right) =\left[ \begin{matrix}x\left( t\right) \\ y\left( t\right) \end{matrix} \right] $
Then
$\frac{d}{dt} f\left( \vec{v} \left( t\right) \right) =\triangledown f.\frac{d}{dt} \vec{v} \left( t\right) = \left[ \begin{matrix}\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{matrix} \right] .\left[ \begin{matrix}\frac{dx}{dt} \\ \frac{dy}{dt} \end{matrix} \right] =\frac{\partial f}{\partial x} .\frac{dx}{dt} +\frac{\partial f}{\partial y} .\frac{dy}{dt} $
Yes, this is just the general form of the "chain rule". It can be written in a slightly more compact notation. You can put all the derivatives into one Jacobian matrix:
$$ D(f \circ \vec{v}) = Df \cdot D \vec{v} = \begin{pmatrix} \frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} & \frac{\partial p}{\partial z} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y} & \frac{\partial q}{\partial z} \\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \end{pmatrix} \begin {pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} & \frac{\partial x}{\partial r} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} & \frac{\partial y}{\partial r} \\ \frac{\partial z}{\partial s} & \frac{\partial z}{\partial t} & \frac{\partial z}{\partial r} \end {pmatrix} = \begin {pmatrix} \frac{\partial p}{\partial s} & \frac{\partial p}{\partial t} & \frac{\partial p}{\partial r} \\ \frac{\partial q}{\partial s} & \frac{\partial q}{\partial t} & \frac{\partial q}{\partial r} \\ \frac{\partial u}{\partial s} & \frac{\partial u}{\partial t} & \frac{\partial u}{\partial r} \end {pmatrix} $$
What you wrote is just looking at one column of $D \vec{v}$ at a time.