Vector cross product formula without a second term (Spiegel, Theoretical Mechanics) In Spiegel's Outline Of Theoretical Mechanics (more precisely in the Moving Coordinate Systems chapter, § "Derivative Operators") I find (both in the 1968 and the 1977 edition) the following formula:
$D_F \equiv D_M +\omega \times$
where $D_F$ and $D_M$ denote time derivative operators, and $\omega$ stands for the angular velocity of the moving frame $M$ (with respect to a fixed frame $F$).
Is there some explanation for this notation, or is it a misprint?

 A: The answers are already given suffice, so I wanted to offer another use of the $\boldsymbol{r}\times$ term.
Find the mass moment of inertia tensor about the origin of a point mass $m$ located at $\boldsymbol{r}$.
There are two formulas for this, and they both produce the same result

*

*Use $$\mathrm{I}_0 = -m [\boldsymbol{r} \times] [\boldsymbol{r} \times] = -m \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}$$


*Use $$\mathrm{I}_0 = m \left( \boldsymbol{r} \cdot \boldsymbol{r} - \boldsymbol{r} \odot \boldsymbol{r} \right) = m \left( \begin{bmatrix} x \\ y \\ z \end{bmatrix}^\top \begin{bmatrix} x \\ y \\ z \end{bmatrix} - \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}^\top \right) $$
If you work out the math both result in
$$  \mathrm{I}_0 = m \begin{bmatrix} y^2+z^2 & -x y & -x z\\ -x y & x^2+z^2 & - y z \\ -x z & -y z & x^2+y^2 \end{bmatrix} $$
My preference is to use $[\boldsymbol{r}\times]$ notation because it keeps the connection to cross products evident, and it codes easily in programming.
The above produces the parallel axis theorem in 3D, from the center of mass C to the origin 0 as
$$ \mathrm{I}_0 = \mathrm{I}_c - m [ \boldsymbol{c}\times][\boldsymbol{c}\times] $$
The above has a direct connection if the angular momentum about a pivot point which is
$$ \begin{aligned} \boldsymbol{L}_0 & = \boldsymbol{L}_c  
 + \boldsymbol{c} \times m \boldsymbol{v}_c \\ 
& = \mathrm{I}_c \boldsymbol{\omega} + \boldsymbol{c} \times m ( \boldsymbol{\omega} \times \boldsymbol{c} ) \\
 & = \mathrm{I}_c \boldsymbol{\omega} - m\, \boldsymbol{c} \times (\boldsymbol{c} \times \boldsymbol{\omega}) \\ & = \mathrm{I}_0 \boldsymbol{\omega} \end{aligned} $$
Additionally, constructing the 6×6 spatial inertia matrix you also use this term
$$ \boldsymbol{I} = \begin{Bmatrix} m \boldsymbol{1} & -m [ \boldsymbol{c} \times] \\ m [\boldsymbol{c} \times] & \mathrm{I}_c - m [\boldsymbol{c}\times][\boldsymbol{c}\times] \end{Bmatrix} $$
Now back to taking derivatives on rotating frames, this is done with spatial algebra with the following operator
$$ \mathbf{v} \times = \begin{Bmatrix} [\boldsymbol{\omega} \times & [\boldsymbol{v}\times] \\ [0] & [\boldsymbol{\omega}\times] \end{Bmatrix} $$
So the acceleration due to the motion of joint axis $\mathbf{s}$ is calculated with
$$ \mathbf{a} = \mathbf{v} \times \mathbf{s} \,\,\dot{q} $$
As you can see this notation has extensive usage in robotics and all levels of rigid body dynamics.
A: What you're seeing there is an operator equation.  What it means is that the two operations on each side of the equation have the same effect on any vector $\mathbf{A}$:
$$
D_F \mathbf{A} = D_M \mathbf{A} + \pmb{\omega} \times \mathbf{A}.
$$
By analogy, if we have two functions $f$ and $g$, saying that $f = g$ is a more compact way of saying $f(x) = g(x)$ for all $x$.
A: $\newcommand{\e}{\boldsymbol=}$
$\newcommand{\m}{\boldsymbol-}$
$\newcommand{\x}{\boldsymbol\times}$
Given two vectors $\:\boldsymbol\omega\:$ and $\:\mathbf x\:$ represented by their cartesian coordinates
\begin{equation}
\boldsymbol\omega\e 
\begin{bmatrix}
\omega_1 \vphantom{\dfrac{a}{b}}\\
\omega_2 \vphantom{\dfrac{a}{b}}\\
\omega_3 \vphantom{\dfrac{a}{b}}
\end{bmatrix}\qquad \texttt{and} \qquad
\mathbf x\e
\begin{bmatrix}
x_1 \vphantom{\dfrac{a}{b}}\\
x_2 \vphantom{\dfrac{a}{b}}\\
x_3 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{01}\label{01}  
\end{equation}
their outer (cross) product is
\begin{equation}
\boldsymbol\omega\x\mathbf x\e
\begin{bmatrix}
\mathbf i & \mathbf j & \mathbf k \vphantom{\dfrac{a}{b}}\\
\omega_1  & \omega_2 & \omega_3 \vphantom{\dfrac{a}{b}}\\
x_1  & x_2 & x_3 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\e
\begin{bmatrix}
\omega_2 x_3\m \omega_3 x_2 \vphantom{\dfrac{a}{b}}\\
\omega_3 x_1\m \omega_1 x_3  \vphantom{\dfrac{a}{b}}\\
\omega_1 x_2\m \omega_2 x_1 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\e
\begin{bmatrix}
\hphantom{\m} 0 & \m \omega_3 & \hphantom{\m}\omega_2 \vphantom{\dfrac{a}{b}}\\
\hphantom{\m}\omega_3  & \hphantom{\m} 0 & \m\omega_1 \vphantom{\dfrac{a}{b}}\\
\m\omega_2  & \hphantom{\m}\omega_1 & \hphantom{\m}0 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\begin{bmatrix}
x_1 \vphantom{\dfrac{a}{b}}\\
x_2 \vphantom{\dfrac{a}{b}}\\
x_3 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{02}\label{02}  
\end{equation}
So it's reasonable to use the symbol $\:\boldsymbol\omega\x\:$ for the linear operator (transformation) represented by a $\:3\times 3\:$ antisymmetric matrix
\begin{equation}
\boldsymbol\omega\x \boldsymbol\equiv
\begin{bmatrix}
\hphantom{\m} 0 & \m \omega_3 & \hphantom{\m}\omega_2 \vphantom{\dfrac{a}{b}}\\
\hphantom{\m}\omega_3  & \hphantom{\m} 0 & \m\omega_1 \vphantom{\dfrac{a}{b}}\\
\m\omega_2  & \hphantom{\m}\omega_1 & \hphantom{\m}0 \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{03}\label{03}  
\end{equation}
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
Related 1 : Velocity in a turning reference frame.
Related 2 : Vector product in a 4-dimensional Minkowski spacetime.
