Differentiation of a Vector with respect to a vector

Question

I am studying a paper and I am going crazy about one differentiation which it is written on it but not explained. I think I am missing something and probably something easy.

I would love if someone of you could give me an help.

My problem it is explained in the following figure:

I thank you in advance for any kind of help.

I am really stuck with it.

EDIT (Thanks to LutzL comments):

Ok to be more clear. I have the following formula. I don't understand how is it possible that when I differentiate the formula: \begin{equation} \beta_{ij}=^iR\frac{\mathbf{p}_{ij}}{\delta_{ij}} \end{equation} I end up with the following thing: \begin{equation} \dot{\beta}_{ij}= \frac{^iR}{\delta_{ij}}\left[\dot{\psi}_iS\mathbf{p}_{ij}+\left(I_{3x3}-\mathbf{\hat{p}}_{ij}\mathbf{\hat{p}}_{ij}^T\right)\dot{\mathbf{p}}_{ij}\right] \end{equation}

I understood the first part of the equation. I think it is due to the differentiation of the rotation matrix $^iR$.

1. $\mathbf{\hat{p}}_{ij}=\left(\frac{\mathbf{p}_{ij}}{\delta_{ij}}\right)$ $\in \mathbb{R}^3$

2. $\delta_{ij}=||\mathbf{p}_{ij}||$

3. $^iR\in \mathbb{R}^{3\times3}$ is a rotation matrix $=\begin{bmatrix} \cos(\psi_u) & -\sin(\psi_u) & 0\\ \sin(\psi_u) & \cos(\psi_u) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

4. $I_{3\times3}$ is the identity matrix.

5. $S$ is a skew-symmetric matrix which is appearing into the formula for the differentiation of the rotation matrix $^iR$

I hope now it is more clear.

Answer 1

Usually $$ \frac{\partial F}{\partial v}(x)=F'(x)·v $$ denotes the directional derivative of $F$ in direction $v$, while $$ \frac{\partial F}{\partial x}(x)=F'(x) $$ denotes the Jacobian matrix. Your formula is mixing this somewhat.

At its core you are differentiating, omitting the indices $$ \frac{d}{dt}_{\Big|t=0}\,\frac{p+t·\dot p}{\|p+t·\dot p\|} =\frac{\dot p}{\|p\|}-\frac{\langle p,\dot p\rangle·p}{\|p\|^3} $$ which then leads to your formula.

Answer 2

Thanks to LutzL I was searching and I found a solution. So the second term that I was looking for is due to the derivative of vector norm.

The following is used.

\begin{equation} \frac{\partial }{\partial \mathbf{x}}\frac{x-a}{||x-a||_2}= \frac{\mathbf{I}}{||x-a||_2}-\frac{\left(x-a\right)\left(x-a\right)^T}{||x-a||_2^3} \end{equation}

Thanks for the help.

Answer 3

$ \def\LR#1{\left(#1\right)} \def\BR#1{\Big(#1\Big)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\d{\delta} \def\e{{\widehat p}} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\dgrad#1#2{\frac{d #1}{d #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} $Drop the distracting $(i,j)$ subscripts and define the variables $$\eqalign{ \d &\doteq \|p\|,\qquad \e &\doteq \frac{p}{\d} \qquad\qquad\quad\qquad \\ }$$ First differentiate $\d$ using the product rule $$\eqalign{ \d^2 &= p^Tp \\ 2\d\,d\d &= 2p^Tdp \\ \grad{\d}{p} &= \fracLR{2p}{2\d} \;=\; \e \qquad\qquad\qquad\qquad\qquad \\ }$$ then differentiate $\e$ using the quotient rule $$\eqalign{ \e &= \frac{p}{\d} \\ d\e &= \fracLR{\d\,dp-p\,d\d}{\d^2} &= \fracLR{I-\e\e^T}{\d}dp \\ \grad{\e}{p} &= \fracLR{I-\e\e^T}{\d} \\ }$$ Applying this last result to the derivative in your question $$\eqalign{ \grad{(R\e)}{p} &= R\gradLR{\e}{p} &= \fracLR{R}{\d}\BR{I-\e\e^T} \ \quad\quad\; \\ }$$ yields the desired result.