What is the Derivative of a Cross Product?

[plmokn2]

[SOLVED] derivative of a cross product

Homework Statement

In some lecture notes I'm reading they jump straight from $$\frac{d}{d\mathbf{r}}( \frac{m}{2} |\mathbf{\omega}\times\mathbf{r}|^2)$$
to
$$\mathbf{r}\omega^2-\mathbf{\omega}(\mathbf{\omega}.\mathbf{r})$$

Homework Equations

The Attempt at a Solution

It's easy to check this by writing out each component but this is messy so there's probably an easier way to do it. Any help appreciated.
Thanks

 

[HallsofIvy]

Homework Statement

In some lecture notes I'm reading they jump straight from $$\frac{d}{d\mathbf{r}}( \frac{m}{2} |\mathbf{\omega}\times\mathbf{r}|^2)$$
to
$$\mathbf{r}\omega^2-\mathbf{\omega}(\mathbf{\omega}.\mathbf{r})$$

Surely not! There must be an m/2 in there!
Ignoring the constant m/2, $|\omega\times r|^2= (\omega\times r)\cdot(\omega\times\r)$. Take the derivative of that, apply the product rule- which is true, for vectors, for both dot product and cross product: $(\vec{u}\cdot\vec{v})'= \vec{u} '\cdot\vec{v}+ \vec{u}\cdot\vec{v} '$ and $(\vec{u}\times\vec{v})'= \vec{u} '\times\vec{v}+ \vec{u}\times\vec{v} '$.

 

[plmokn2]

opps I forgot an m in the answer: sorry.

Thanks, I've solved it now.