Source: Chapter 2 of Classical Dynamics by David Tong (p14)
In the text the author attempts to derive the equations of motion for the system of a free particle in a frame rotating with angular velocity $\,\boldsymbol{\omega} = (0,0,\omega)$ about the $z$ axis. In the new frame our new/primed coordinates are related to the original coordinates by the following relations:
\begin{align}x' &= x \cos(\omega t) + y \sin( \omega t) \\ y' &= y \cos(\omega t) - x \sin( \omega t) \\ z' &= z \end{align}
By manipulating the above expression, we obtain the following form for the Lagrangian of the system: $$ \mathcal{L} = \cfrac{1}{2} m (\dot{\boldsymbol{r'}} + \boldsymbol{\omega} \times \boldsymbol{r'})^2$$
To obtain the first half of the equations of motion the following is done:
$$\frac{\partial\mathcal{L}}{\partial \boldsymbol{r'}} = m(\dot{\boldsymbol{r'}} \times \boldsymbol{\omega} - \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{r'})$$
Is there a way to see this without having to expand the Lagrangian out of vector form and performing the tedious calculation that way?