Derivative of outer product

Question

I want to calculate the first generalized coordinate derivative $\frac{\partial}{\partial q}$ being $q=x$ or $q=y$ or $q=z$ of the outer product between two identical vectors $R=\bf{r}\bf{r}^{T}$ where $r=(x-x_0,y-y_0,z-z_0)$. In particular it would be interesting to know how the result can be expressed in terms of an arbitrary unitary matrix $\bf{U}$ that represents the rotation, i.e $\bf{r}'=\bf{U}r$, where $\bf{r}'$ represents the new coordinates and $\bf{r}$ the old coordinates.

Answer 1

You can abbreviate the coordinate derivative as $\partial_k$ where k={1,2,3} corresponds to the derivative with respect to {x,y,z}.

The your derivative can be calculated as $$\eqalign{ \partial_k(rr^T) &= (\partial_kr)r^T + r(\partial_kr^T) \cr &= (\partial_kr)r^T + r(\partial_kr^T) \cr &= e_kr^T + re_k^T \cr }$$ where the $\{e_k\}$ are the standard basis vectors.