What's the differential of the cross product wrt a vector? $$ w= u(x) \times v(x), \qquad u(x), v(x), x \in R^3$$
How to calculate the $ \frac{\partial w}{\partial x}$? Thanks very much!
 A: 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 $\mathrm{d}w(x)h = B\left(\mathrm{d}u(x)h,v(x) \right) + B\left(u(x),\mathrm{d}v(x)h \right)$.
A: 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 $:
$$
dx^i\partial_i \omega_j =   \epsilon_{jab} (  u_a dx^i\partial_i v_b + v_b dx^i\partial_i u_a ) =  \epsilon_{jab} (  u_a dv_b + v_b du_a ) =  \epsilon_{jab} (  du_a v_b + u_a dv_b   )
$$
