The partial derivative of the Cross Product of Two Vectors?

Question

As far as I know, the partial derivative of the dot product of two vectors can be given by: $\frac{\partial(\vec A\cdot\vec B)}{\partial\vec A}=\vec B$. What if The Derivative of the Cross Product of Two Vector Valued Functions $\frac{\partial(\vec A\times\vec B)}{\partial\vec A}=?$

e.g. How to calculate $\frac{\partial|\vec \omega\times\vec r|^2}{\partial\vec r}$, where $\vec\omega$ is independent of $\vec r$.

Answer 1

Let $f_1(x) = x^T b = b^T x$, then $f_1$ is linear and so ${\partial f_1(x) \over \partial x} = b^T$.

If $f_2(x) = x \times b$, then $f_2$ is linear and so ${\partial f_2(x) \over \partial x}h = h \times b = \begin{bmatrix} e_1 \times b & e_2 \times b & e_3 \times b \end{bmatrix} h = \begin{bmatrix} 0 & -b_3 & b_2 \\ b_3 & 0 & -b_1 \\ -b_2 & b_1 & 0 \end{bmatrix} h$, or, ${\partial f_2(x) \over \partial x} = \begin{bmatrix} 0 & -b_3 & b_2 \\ b_3 & 0 & -b_1 \\ -b_2 & b_1 & 0 \end{bmatrix}$.

Answer 2

there is a missing minus sign in the other answer, if x is a vector, the general form is given by:

$$\frac{\partial \left(\vec{a}\times\vec{b}\right)}{\partial x} =\displaystyle -\left[b^{\times}\right]\frac{\partial \vec{a}}{\partial x} + \left[a^{\times}\right]\frac{\partial \vec{b}}{\partial x}$$

and $$\left[b^{\times}\right]$$ is known as the cross product matrix is given by

$$\left[b^{\times}\right] = \begin{bmatrix} 0 & -b_3 & b_2 \\ b_3 & 0 & -b_1 \\ -b_2 & b_1 & 0 \end{bmatrix}$$