What's the differential of the cross product wrt a vector?

Question

$$ w= u(x) \times v(x), \qquad u(x), v(x), x \in R^3$$ How to calculate the $ \frac{\partial w}{\partial x}$? Thanks very much!

Answer 1

Seems like you are considering $u$ and $v$ as functions over a variable $x$. Write $u(x + h) = u(x) + \mathrm{d}u(x)h + o(||h||)$ and $v(x+h) = v(x) + \mathrm{d}v(x)h + o(||h||)$. Then : \begin{align} w(x+h) &= u(x+h)\times v(w+h) \\ &= \left(u(x) + u'(x)h + o(h) \right) \times \left(v(x) + v'(x)h + o(h) \right)\\ &= u(x)\times v(x) + \left(\mathrm{d}u(x)h\right)\times v(x) + u(x)\times \left( \mathrm{d}v(x)h\right) + \left(\mathrm{d}u(x)h\right) \times \left(\mathrm{d}v(x)h\right) + o(||h||^2)\\ &= w(x) + \left(\left(\mathrm{d}u(x)h\right)\times v(x) + u(x)\times \left( \mathrm{d}v(x)h\right) \right) + o(||h||) \end{align} Thus the differential of $w$ at $x$ is the linear function $h \mapsto \left(\mathrm{d}u(x)h\right)\times v(x) + u(x)\times \left( \mathrm{d}v(x)h\right)$

In fact, the same computations show that if $B$ is bilinear and $u$, $v$ are differentiable, then $w=B(u,v)$ is differentiable and $\mathrm{d}w(x)h = B\left(\mathrm{d}u(x)h,v(x) \right) + B\left(u(x),\mathrm{d}v(x)h \right)$.

Answer 2

You already have the answer of @Dldier_ for the differential. If you are actually asking for the "Jacobian matrix", then:

$$ \partial_i \omega_j = \epsilon_{jab} \partial_i(u_a v_b ) = \epsilon_{jab} ( u_a \partial_i v_b + v_b \partial_i u_a ) = \epsilon_{jab} ( u_a \partial_i v_b - v_a \partial_i u_b ) $$

You should be able to see that is in accordance with the answer of @Dldier_ by considering $d\omega_j = dx^i\partial_i \omega_j $:

$$ dx^i\partial_i \omega_j = \epsilon_{jab} ( u_a dx^i\partial_i v_b + v_b dx^i\partial_i u_a ) = \epsilon_{jab} ( u_a dv_b + v_b du_a ) = \epsilon_{jab} ( du_a v_b + u_a dv_b ) $$