A Compact Formula for the Derivative of a 3-D Rotation in Exponential Coordinates

Abstract

We present a compact formula for the derivative of a 3-D rotation matrix with respect to its exponential coordinates. A geometric interpretation of the resulting expression is provided, as well as its agreement with other less-compact but better-known formulas. To the best of our knowledge, this simpler formula does not appear anywhere in the literature. We hope by providing this more compact expression to alleviate the common pressure to reluctantly resort to alternative representations in various computational applications simply as a means to avoid the complexity of differential analysis in exponential coordinates.

Appendices

Appendix 1: Some Cross Product Relations

Let us use the dot notation for the Euclidean inner product $\mathbf {a}\cdot \mathbf {b}=\mathbf {a}^{\top }\mathbf {b}$. Also, let $\mathtt {G}$ be a $3\times 3$ matrix, invertible when required so that it represents a change of coordinates in $\mathbb {R}^{3}$.

$$\begin{aligned} \left[ \mathbf {a}\right] _{\times }\mathbf {a}&= \mathbf {0}\end{aligned}$$

(10)

$$\begin{aligned} \left[ \mathbf {a}\right] _{\times }\mathbf {b}&= -\left[ \mathbf {b}\right] _{\times }\mathbf {a}\end{aligned}$$

(11)

$$\begin{aligned} \left[ \mathbf {a}\right] _{\times }\left[ \mathbf {b}\right] _{\times }&= \mathbf {b}\mathbf {a}^{\top }-(\mathbf {a}\cdot \mathbf {b})\text{ Id } \end{aligned}$$

(12)

$$\begin{aligned} \mathbf {a}\times (\mathbf {b}\times \mathbf {c})&\mathop {=}\limits ^{(12)}(\mathbf {a}\cdot \mathbf {c})\mathbf {b}-(\mathbf {a}\cdot \mathbf {b})\mathbf {c}\end{aligned}$$

(13)

$$\begin{aligned} \left[ \mathbf {a}\times \mathbf {b}\right] _{\times }&= \mathbf {b}\mathbf {a}^{\top }-\mathbf {a}\mathbf {b}^{\top } \end{aligned}$$

(14)

$$\begin{aligned} \left[ \mathbf {a}\times \mathbf {b}\right] _{\times }&= \left[ \mathbf {a}\right] _{\times }\left[ \mathbf {b}\right] _{\times }-\left[ \mathbf {b}\right] _{\times } \left[ \mathbf {a}\right] _{\times }\end{aligned}$$

(15)

$$\begin{aligned} \left[ (\mathtt {G}\mathbf {a})\times (\mathtt {G}\mathbf {b})\right] _{\times }&= \mathtt {G}\left[ \mathbf {a}\times \mathbf {b}\right] _{\times } \mathtt {G}^{\top }\nonumber \\ (\mathtt {G}\mathbf {a})\times (\mathtt {G}\mathbf {b})&= \det (\mathtt {G})\mathtt {G}^{-\top }(\mathbf {a}\times \mathbf {b})\end{aligned}$$

(16)

$$\begin{aligned} \left[ \mathbf {a}\right] _{\times }\mathtt {G}+\mathtt {G}^{\top }\left[ \mathbf {a}\right] _{\times }&= {{\mathrm{trace}}}(\mathtt {G})\left[ \mathbf {a}\right] _{\times }- \left[ \mathtt {G}\mathbf {a}\right] _{\times }\\ \left[ \mathtt {G}\mathbf {a}\right] _{\times }&= \det (\mathtt {G})\mathtt {G}^{-\top }\left[ \mathbf {a}\right] _{\times }\mathtt {G}^{-1}\nonumber \end{aligned}$$

(17)

Appendix 2: Proof of Result 1

Proof

Four terms result from applying the chain rule to (1) acting on vector $\mathbf {u}$. Let us use $\theta =\Vert \mathbf {v}\Vert$ and $\bar{\mathbf {v}}=\mathbf {v}/\Vert \mathbf {v}\Vert$, then

$$\begin{aligned} \frac{\partial \mathtt {R}\mathbf {u}}{\partial \mathbf {v}}&= \sin \theta \,\frac{\partial \left[ \bar{\mathbf {v}}\right] _{\times }\mathbf {u}}{\partial \mathbf {v}} +\left[ \bar{\mathbf {v}}\right] _{\times }\mathbf {u}\,\frac{\partial \sin \theta }{\partial \mathbf {v}}\\&+(1-\cos \theta )\frac{\partial \left[ \bar{\mathbf {v}}\right] _{\times }^{2}\mathbf {u}}{\partial \mathbf {v}}+\left[ \bar{\mathbf {v}}\right] _{\times }^{2} \mathbf {u}\,\frac{\partial (1-\cos \theta )}{\partial \mathbf {v}}. \end{aligned}$$

The previous derivatives are computed next, using some of the cross product properties listed in Appendix 1:

$$\begin{aligned} \frac{\partial \left[ \bar{\mathbf {v}}\right] _{\times }\mathbf {u}}{\partial \mathbf {v}}\mathop {=}\limits ^{(11)} \frac{\partial (-\left[ \mathbf {u}\right] _{\times }\bar{\mathbf {v}})}{\partial \bar{\mathbf {v}}}\frac{\partial \bar{\mathbf {v}}}{\partial \mathbf {v}}=-\left[ \mathbf {u}\right] _{\times } \frac{\partial \bar{\mathbf {v}}}{\partial \mathbf {v}}, \end{aligned}$$

with derivative of the unit rotation axis vector

$$\begin{aligned} \frac{\partial \bar{\mathbf {v}}}{\partial \mathbf {v}}=\frac{\partial }{\partial \mathbf {v}}\left( \frac{\mathbf {v}}{\Vert \mathbf {v}\Vert }\right) = \frac{1}{\theta }(\text{ Id }-\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top }) \mathop {=}\limits ^{(4)}-\frac{1}{\theta } \left[ \bar{\mathbf {v}}\right] _{\times }^{2}. \end{aligned}$$

(18)

Also by the chain rule,

$$\begin{aligned} \frac{\partial \sin \theta }{\partial \mathbf {v}}&= \frac{\partial \sin \theta }{\partial \theta }\frac{\partial \theta }{\partial \mathbf {v}}=\cos \theta \,\bar{\mathbf {v}}^{\top },\\ \frac{\partial (1-\cos \theta )}{\partial \mathbf {v}}&= -\frac{\partial \cos \theta }{\partial \theta }\frac{\partial \theta }{\partial \mathbf {v}}=\sin \theta \,\bar{\mathbf {v}}^{\top }, \end{aligned}$$

and, applying the product rule twice,

$$\begin{aligned} \begin{aligned} \frac{\partial \left[ \bar{\mathbf {v}}\right] _{\times }^{2}\mathbf {u}}{\partial \mathbf {v}}&\mathop {=}\limits ^{(4)} \frac{\partial \bar{\mathbf {v}}(\bar{\mathbf {v}}^{\top }\mathbf {u})}{\partial \mathbf {v}}=\frac{\partial \bar{\mathbf {v}}}{\partial \mathbf {v}}(\bar{\mathbf {v}}^{\top }\mathbf {u})+\bar{\mathbf {v}}\frac{\partial (\bar{\mathbf {v}}^{\top }\mathbf {u})}{\partial \mathbf {v}}\\&= \bigl ((\bar{\mathbf {v}}^{\top }\mathbf {u})\text{ Id }+\bar{\mathbf {v}}\mathbf {u}^{\top }\bigr )\frac{\partial \bar{\mathbf {v}}}{\partial \mathbf {v}}\\&\mathop {=}\limits ^{(18)} -\frac{1}{\theta } \bigl ((\bar{\mathbf {v}}^{\top }\mathbf {u})\text{ Id }+\bar{\mathbf {v}}\mathbf {u}^{\top }\bigr )\left[ \bar{\mathbf {v}}\right] _{\times }^{2}, \end{aligned} \end{aligned}$$

which can be rewritten as a sum of cross product matrix multiplications since

$$\begin{aligned} \begin{aligned}&\bigl ((\bar{\mathbf {v}}^{\top }\mathbf {u})\text{ Id }+\bar{\mathbf {v}}\mathbf {u}^{\top }\bigr )\left[ \bar{\mathbf {v}}\right] _{\times }^{2}\\&\quad \mathop {=}\limits ^{(10)}\bigl ((\bar{\mathbf {v}}^{\top }\mathbf {u})\text{ Id }-\mathbf {u}\bar{\mathbf {v}}^{\top }+\bar{\mathbf {v}}\mathbf {u}^{\top }-\mathbf {u}\bar{\mathbf {v}}^{\top }\bigr )\left[ \bar{\mathbf {v}}\right] _{\times }^{2}\\&\quad \mathop {=}\limits ^{(12)\,(14)}-\left[ \bar{\mathbf {v}}\right] _{\times }\left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }^{2}+\left[ \mathbf {u}\times \bar{\mathbf {v}}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }^{2}\\&\quad \mathop {=}\limits ^{(15)\,(4)}-2\left[ \bar{\mathbf {v}}\right] _{\times }\left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }^{2}-\left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }. \end{aligned} \end{aligned}$$

Hence, so far the derivative of the rotated vector is

$$\begin{aligned} \begin{aligned} \frac{\partial \mathtt {R}\mathbf {u}}{\partial \mathbf {v}}&=(\cos \theta \left[ \bar{\mathbf {v}}\right] _{\times }+\sin \theta \left[ \bar{\mathbf {v}}\right] _{\times }^{2})\mathbf {u}\bar{\mathbf {v}}^{\top }+\frac{\sin \theta }{\theta } \left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }^{2}\\&\quad +\frac{1-\cos \theta }{\theta }(2\left[ \bar{\mathbf {v}}\right] _{\times }\left[ \mathbf {u}\right] _{\times } \left[ \bar{\mathbf {v}}\right] _{\times }^{2}+\left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }). \end{aligned} \end{aligned}$$

Next, multiply on the left by $\mathtt {R}^{\top }$ and use

$$\begin{aligned} \mathtt {R}^{\top }\left[ \bar{\mathbf {v}}\right] _{\times }\mathop {=}\limits ^{(3)\,(10)}\cos \theta \, \left[ \bar{\mathbf {v}}\right] _{\times }-\sin \theta \, \left[ \bar{\mathbf {v}}\right] _{\times }^{2} \end{aligned}$$

(19)

to simplify the first term of $\mathtt {R}^{\top }\partial (\mathtt {R}\mathbf {u})/\partial \mathbf {v}$,

$$\begin{aligned} \begin{aligned}&\mathtt {R}^{\top }(\cos \theta \,\left[ \bar{\mathbf {v}}\right] _{\times }+\sin \theta \,\left[ \bar{\mathbf {v}}\right] _{\times }^{2})\mathbf {u}\bar{\mathbf {v}}^{\top }\\&\quad \mathop {=}\limits ^{(19)}(\cos ^{2}\theta \,\left[ \bar{\mathbf {v}}\right] _{\times }-\sin ^{2}\theta \,\left[ \bar{\mathbf {v}}\right] _{\times }^{3})\mathbf {u}\bar{\mathbf {v}}^{\top }\\&\quad \mathop {=}\limits ^{(4)}(\cos ^{2}\theta +\sin ^{2}\theta )\,\left[ \bar{\mathbf {v}}\right] _{\times }\mathbf {u}\bar{\mathbf {v}}^{\top }\\&\quad \mathop {=}\limits ^{(11)}-\left[ \mathbf {u}\right] _{\times }\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top }. \end{aligned} \end{aligned}$$

(20)

For the remaining term of $\mathtt {R}^{\top }\partial (\mathtt {R}\mathbf {u})/\partial \mathbf {v}$, we use the transpose of (1) and apply $\left[ \bar{\mathbf {v}}\right] _{\times }\left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }\mathop {=}\limits ^{(12)}-(\mathbf {u}^{\top }\bar{\mathbf {v}})\left[ \bar{\mathbf {v}}\right] _{\times }$ to simplify

$$\begin{aligned} \begin{aligned}&\bigl (\text{ Id }-\sin \theta \,\left[ \bar{\mathbf {v}}\right] _{\times }+(1-\cos \theta )\left[ \bar{\mathbf {v}}\right] _{\times }^{2} \bigr )\cdot \bigl (\sin \theta \left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }^{2}\\&\qquad +(1-\cos \theta )(2\left[ \bar{\mathbf {v}}\right] _{\times }\left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }^{2}+ \left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times })\bigr )\\&\quad =\sin \theta \,\left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }^{2}+(1-\cos \theta ) \left[ \mathbf {u}\right] _{\times }\left[ \bar{\mathbf {v}}\right] _{\times }\\&\qquad +(\mathbf {u}^{\top }\bar{\mathbf {v}})\bigl (-2(1-\cos \theta )+\sin ^{2}\theta \, +(1-\cos \theta )^{2}\bigr )\left[ \bar{\mathbf {v}}\right] _{\times }\\&\quad =\left[ \mathbf {u}\right] _{\times }\bigl (\sin \theta \,\left[ \bar{\mathbf {v}}\right] _{\times }^{2}-(1-\cos \theta )\left[ \bar{\mathbf {v}}\right] _{\times }^{3}\bigr )\\&\quad =-\left[ \mathbf {u}\right] _{\times }(\mathtt {R}^{\top }-\text{ Id })\left[ \bar{\mathbf {v}}\right] _{\times }, \end{aligned} \end{aligned}$$

where the term in $(\mathbf {u}^{\top }\bar{\mathbf {v}})$ vanished since $\sin ^{2}\theta -2(1-\cos \theta )+(1-\cos \theta )^{2}=0$. Collecting terms,

$$\begin{aligned} \mathtt {R}^{\top }\frac{\partial \mathtt {R}\mathbf {u}}{\partial \mathbf {v}}=-\left[ \mathbf {u}\right] _{\times }\bigl (\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top } +\frac{1}{\theta }(\mathtt {R}^{\top }-\text{ Id })\left[ \bar{\mathbf {v}}\right] _{\times }\bigr ). \end{aligned}$$

(21)

Finally, multiply (21) on the left by $\mathtt {R}$ and use $\mathtt {R}\mathtt {R}^{\top }=\text{ Id }$, $\theta =\Vert \mathbf {v}\Vert$, $\bar{\mathbf {v}}=\mathbf {v}/\Vert \mathbf {v}\Vert$ to obtain (8). $\square$

Appendix 3: Proof of Result 2

Proof

Stemming from (8), we show that it is possible to write

$$\begin{aligned} \frac{\partial \mathtt {R}}{\partial v_{i}}\mathbf {u}=\mathtt {A}\mathbf {u}\end{aligned}$$

(22)

for some matrix $\mathtt {A}$ and for all vector $\mathbf {u}$ independent of $\mathbf {v}$. Thus in this operator form, $\mathtt {A}$ is indeed the representation of $\partial \mathtt {R}/\partial v_{i}$. First, observe that

$$\begin{aligned} \frac{\partial \mathtt {R}}{\partial v_{i}}\mathbf {u}=\frac{\partial }{\partial v_{i}}(\mathtt {R}\mathbf {u})=\frac{\partial }{\partial \mathbf {v}} (\mathtt {R}\mathbf {u})\,\mathbf {e}_{i}, \end{aligned}$$

then substitute (8) and simplify using the cross-product properties to arrive at (22):

(23)

After some manipulations,

$$\begin{aligned} \mathtt {R}\left[ v_{i}\mathbf {v}+\bigl (\mathbf {v}\times (\mathtt {R}^{\top }-\text{ Id })\mathbf {e}_{i}\bigr )\right] _{\times } =\left[ v_{i}\mathbf {v}+\bigl (\mathbf {v}\times (\text{ Id }-\mathtt {R})\mathbf {e}_{i}\bigr )\right] _{\times }\mathtt {R}, \end{aligned}$$

and so, substituting in (23) and using the linearity of the cross-product matrix (2), the desired formula (9) is obtained. $\square$

Appendix 4: Agreement Between Derivative Formulas

Here we show the agreement between (7) and (9). First, use $\theta =\Vert \mathbf {v}\Vert$ and the definition of the unit vector $\bar{\mathbf {v}}=\mathbf {v}/\theta$, to write (9) as

$$\begin{aligned} \frac{\partial \mathtt {R}}{\partial v_{i}}=\bar{v}_{i}\left[ \bar{\mathbf {v}}\right] _{\times }\mathtt {R}+\frac{1}{\theta } \left[ \bar{\mathbf {v}}\times (\text{ Id }-\mathtt {R})\mathbf {e}_{i}\right] _{\times }\mathtt {R}. \end{aligned}$$

(24)

Using (3) and $\left[ \bar{\mathbf {v}}\right] _{\times }\bar{\mathbf {v}}=\mathbf {0}$, it follows that

$$\begin{aligned} \left[ \bar{\mathbf {v}}\right] _{\times }\mathtt {R}=\cos \theta \,\left[ \bar{\mathbf {v}}\right] _{\times }+\sin \theta \,\left[ \bar{\mathbf {v}}\right] _{\times }^{2}. \end{aligned}$$

Also, since $\left[ \bar{\mathbf {v}}\times \mathbf {b}\right] _{\times }=\mathbf {b}\bar{\mathbf {v}}^{\top }-\bar{\mathbf {v}}\mathbf {b}^{\top }$ and $\mathtt {R}^{\top }\bar{\mathbf {v}}=\bar{\mathbf {v}}$, it yields

$$\begin{aligned} \left[ \bar{\mathbf {v}}\times (\text{ Id }-\mathtt {R})\mathbf {e}_{i}\right] _{\times }\mathtt {R}&= (\text{ Id }-\mathtt {R}) \mathbf {e}_{i}\bar{\mathbf {v}}^{\top }\mathtt {R}-\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }(\text{ Id }-\mathtt {R}^{\top })\mathtt {R}\\&= (\text{ Id }-\mathtt {R})\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }-\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }(\mathtt {R}-\text{ Id })\\&= \mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }-(\mathtt {R}\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+ \bar{\mathbf {v}}\mathbf {e}_{i}^{\top }\mathtt {R}), \end{aligned}$$

and expanding $\mathtt {R}\mathbf {e}_{i}$ and $\mathbf {e}_{i}^{\top }\mathtt {R}$ in the previous formula by means of (3), we obtain

$$\begin{aligned} \begin{aligned}&\left[ \bar{\mathbf {v}}\times (\text{ Id }-\mathtt {R})\mathbf {e}_{i}\right] _{\times }\mathtt {R}\\&\quad =\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }\\&\qquad -\bigl (\cos \theta \,\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+\sin \theta \,\left[ \bar{\mathbf {v}}\right] _{\times } \mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+(1-\cos \theta )\bar{v}_{i}\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top }\bigr )\\&\qquad -\bigl (\cos \theta \,\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }+\sin \theta \,\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }\left[ \bar{\mathbf {v}}\right] _{\times }+(1-\cos \theta )\bar{v}_{i}\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top }\bigr )\\&\quad =(1-\cos \theta )(\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }-2\bar{v}_{i} \bar{\mathbf {v}}\bar{\mathbf {v}}^{\top })\\&\qquad -\sin \theta \,(\left[ \bar{\mathbf {v}}\right] _{\times }\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+\bar{\mathbf {v}}\mathbf {e}_{i}^{\top } \left[ \bar{\mathbf {v}}\right] _{\times }). \end{aligned} \end{aligned}$$

Using property (17) with $\mathbf {a}=\bar{\mathbf {v}}$, $\mathtt {G}=\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }$ we have that

$$\begin{aligned} \left[ \bar{\mathbf {v}}\right] _{\times }\mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }\left[ \bar{\mathbf {v}}\right] _{\times }&= {{\mathrm{trace}}}(\mathbf {e}_{i}\bar{\mathbf {v}}^{\top })\left[ \bar{\mathbf {v}}\right] _{\times }-\left[ \mathbf {e}_{i}\bar{\mathbf {v}}^{\top }\bar{\mathbf {v}}\right] _{\times },\\&= {{\mathrm{trace}}}(\bar{\mathbf {v}}^{\top }\mathbf {e}_{i})\left[ \bar{\mathbf {v}}\right] _{\times }-\left[ \mathbf {e}_{i}\Vert \bar{\mathbf {v}}\Vert ^{2}\right] _{\times }\\&= \bar{v}_{i}\left[ \bar{\mathbf {v}}\right] _{\times }-\left[ \mathbf {e}_{i}\right] _{\times }\\&= \left[ \bar{v}_{i}\bar{\mathbf {v}}-\mathbf {e}_{i}\right] _{\times }. \end{aligned}$$

Finally, substituting previous results in (24), the desired result (7) is obtained.

Appendix 5: Derivative Formula with Sines and Cosines

Here, we show how to obtain formula (7). First, differentiate the Euler–Rodrigues rotation formula (3) with respect to the $i$-th component of the parametrizing vector $\mathbf {v}=\theta \bar{\mathbf {v}}$ and take into account that

$$\begin{aligned} \theta ^{2}=\Vert \mathbf {v}\Vert ^{2}\implies \frac{\partial \theta }{\partial v_{i}}=\frac{v_{i}}{\theta } =:\bar{v}_{i}. \end{aligned}$$

Applying the chain rule to (3), gives

$$\begin{aligned} \frac{\partial \mathtt {R}}{\partial v_{i}}&= -\sin \theta \,\bar{v}_{i}\text{ Id }+\cos \theta \, \bar{v}_{i}\left[ \bar{\mathbf {v}}\right] _{\times }+\sin \theta \,\bar{v}_{i}\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top }\nonumber \\&+\sin \theta \,\frac{\partial \left[ \bar{\mathbf {v}}\right] _{\times }}{\partial v_{i}}+(1-\cos \theta )\frac{\partial (\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top })}{\partial v_{i}}. \end{aligned}$$

(25)

Next, we use

$$\begin{aligned} \frac{\partial }{\partial v_{i}}\left( \frac{v_{j}}{\Vert \mathbf {v}\Vert }\right) = {\left\{ \begin{array}{ll} -\frac{1}{\theta }\bar{v}_{i}\bar{v}_{j} &{} i\ne j\\ \frac{1}{\theta }(1-\bar{v}_{i}^{2}) &{} i=j \end{array}\right. }, \end{aligned}$$

(26)

to simplify one of the terms in (25),

$$\begin{aligned} \frac{\partial \left[ \bar{\mathbf {v}}\right] _{\times }}{\partial v_{i}}=\frac{\partial }{\partial v_{i}}\left[ \frac{\mathbf {v}}{\Vert \mathbf {v}\Vert }\right] _{\times }\mathop {=}\limits ^{(26)}\frac{1}{\theta } \left[ \mathbf {e}_{i}- \bar{v}_{i}\bar{\mathbf {v}}\right] _{\times }. \end{aligned}$$

Applying the product rule to the last term in (25) and using

$$\begin{aligned} \frac{\partial \bar{\mathbf {v}}}{\partial v_{i}}=\frac{\partial }{\partial v_{i}}\left( \frac{\mathbf {v}}{\Vert \mathbf {v}\Vert }\right) =\frac{1}{\theta }(\text{ Id }-\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top })\mathbf {e}_{i}, \end{aligned}$$

(27)

gives

$$\begin{aligned} \frac{\partial (\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top })}{\partial v_{i}}\mathop {=}\limits ^{(27)} \frac{1}{\theta }\left( \mathbf {e}_{i}\bar{\mathbf {v}}^{\top }+\bar{\mathbf {v}}\mathbf {e}_{i}^{\top }-2 \bar{v}_{i}\bar{\mathbf {v}}\bar{\mathbf {v}}^{\top }\right) . \end{aligned}$$

Finally, substituting the previous expressions in (25), yields (7). A similar proof is outlined in [12] using Einstein summation index notation.