I am trying to understand the expression $$ \frac{\partial(x-y)}{\partial x} \times a, $$ where $x$, $y$, and $a$ are vectors. I am unsure of how to compute this expression and would appreciate some guidance.
Specifically, I would like to know: How do I perform the cross product with vector $a$ after calculating the partial derivative $$ \frac{\partial(x-y)}{\partial x}? $$ Which is "Identity"??
I would be grateful for an explanation of the steps involved and any relevant concepts or properties that apply in this situation.
Thank you in advance for your help!
$ \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\L[\begin{array}{r}#1\end{array}\R]} $Let's use a lowercase letter for a vector in ${\mathbb R}^3$ and an uppercase letter for its cross-product matrix. This allows us to write the cross product as a matrix-vector product, i.e. $$\eqalign{ a=\m{a_1\\a_2\\a_3},\quad A=\m{0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0}\qiq a\times b=Ab,\quad b\times a=b^TA }$$ Replacing the $b$ vector by the identity matrix yields $$\eqalign{ a\times I &= AI &= A \\ I\times a &= IA &= A \\ }$$ The only other bit of information that you need is that the gradient of a constant vector $(c)$ is zero, and the gradient of a vector variable $(x)$ with respect to itself is the identity matrix $$\eqalign{ &\grad{c}{x}=0,\quad \grad{x}{x}=I \qiq &\grad{\LR{x-c}}{x} = \LR{I-0} = I \\ }$$