Suppose we want to take the partial derivative of $(\vec{B} \times \vec{A})^2$ with respect to $\vec{A}$.
We write $(\vec{B} \times \vec{A})^2 = (\vec{B} \times \vec{A}) \cdot (\vec{B} \times \vec{A}) $
1/ How we define $\frac {\partial}{\partial {\vec{A}}}$?
Suppose $\vec{A}=A_x(t)\vec{i}+A_y(t)\vec{j}+A_z(t)\vec{k}$.
Should we define $\frac {\partial}{\partial {\vec{A}}}$ first for ?
$\frac {\partial}{\partial {A_{x}}}, \frac {\partial}{\partial {A_{y}}}, \frac {\partial}{\partial {A_{z}}}$
And then somehow obtain $\frac {\partial}{\partial {\vec{A}}}(\vec{B} \times \vec{A}) \cdot (\vec{B} \times \vec{A})$?
2/ Can we show that $\frac{{\partial (\vec{B} \times \vec{A})^2}}{\partial \vec{A}}=2 \vec{B} \times (\vec{A} \times \vec{B})$
Best Answer
Define the vector $$c=(b\times a)$$ Then calculate the differential and gradient of the function using standard vector algebra. $$\eqalign{ \phi &= c\cdot c \\ d\phi &= dc\cdot c \;+\; c\cdot dc \\ &= 2c\cdot dc \\ &= 2c\cdot(b\times da) \\ &= 2(c\times b)\cdot da \\ \frac{\partial\phi}{\partial a} &= 2(c\times b) \\ &= 2(b\times a)\times b \\ &= 2b\times(a\times b) \\ }$$ The nice thing about the differential approach is that the differential of a vector has the same dimensions as the underlying vector and obeys all of the rules of vector algebra.