I am working on deriving equations for mechanics problem including vector calculus, more specifically, derivatives of cross product of vectors with respect to a vector.
Let $\boldsymbol{a},\boldsymbol{b}$ be vectors.
Then, I know that
$\frac{\partial}{\partial \boldsymbol{a}}\left(\boldsymbol{a} \times \boldsymbol{b}\right) = -\left[\boldsymbol{B}\right],\quad \frac{\partial}{\partial \boldsymbol{b}}\left(\boldsymbol{a} \times \boldsymbol{b}\right) = \left[\boldsymbol{A}\right],\qquad$where $\left[\boldsymbol{A}\right] = skew \left(\boldsymbol{a}\right), \quad \left[\boldsymbol{B}\right] = skew \left(\boldsymbol{b}\right)$.
$\frac{\partial}{\partial \boldsymbol{a}} \left|\left|\boldsymbol{a}\right|\right| = \frac{\boldsymbol{a}}{\left|\left|\boldsymbol{a}\right|\right|}$
Based on this,
I have been struggling with how to take the derivative of the following vector product
$ \left(1\right) \frac{\partial}{\partial \boldsymbol{a}}\left\lbrace\boldsymbol{a} \times \left( \boldsymbol{a} \times \boldsymbol{b}\right)\right\rbrace $
$ \left(2\right) \frac{\partial}{\partial \boldsymbol{a}}\left\lbrace\boldsymbol{b} \times \left( \boldsymbol{b} \times \boldsymbol{a}\right)\right\rbrace $
$ \left(3\right) \frac{\partial}{\partial \boldsymbol{a}}\left[ \left|\left| \boldsymbol{a} \right|\right| \left\lbrace \boldsymbol{b} \times \left( \boldsymbol{b} \times \boldsymbol{a}\right)\right\rbrace \right] $
Anyone who has experience of those problems, could you help me, please? And also, it would be deeply appreciated if you could show me any references such as book or website I can refer to.
Thank you.
Best Answer
Use the "BAC-CAB" rule: $$ a\times(b\times c) = b(a^Tc) - c(a^Tb) $$ to expand the cross products before taking derivatives.
For example, here is a solution for case (1).
Find the differential and then the gradient.
$$\eqalign{ y &= a\times(a\times b) \cr &= (b^Ta)a - (a^Ta)b \cr dy &= (b^Ta)\,da + (b^Tda)a - 2(a^Tda)b \cr &= \Big((b^Ta)I + ab^T - 2ba^T\Big)\,da \cr \frac{\partial y}{\partial a} &= (b^Ta)I + ab^T - 2ba^T \cr }$$ Note that I've used matrix notation for the dot products, i.e. $\,\,a^Tb=a\cdot b$
The solution to case (3) can be put in an intriguing form $$\|a\|^{-1}\big(bb^T-b^TbI\big)\big(aa^T + a^TaI\big)$$