Derivative of cross-product of two vectors In finding the derivative of the cross product of two vectors $\frac{d}{dt}[\vec{u(t)}\times \vec{v(t)}]$, is it possible to find the cross-product of the two vectors first before differentiating?
 A: You can evaluate this expression in two ways:


*

*You can find the cross product first, and then differentiate it.

*Or you can use the product rule, which works just fine with the cross product:


$$
\frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \frac{d\mathbf{u}}{dt} \times \mathbf{v} + \mathbf{u} \times \frac{d\mathbf{v}}{dt}
$$
Picking a method depends on the problem at hand. For example, the product rule is used to derive Frenet Serret formulas.  
A: Working from the first principles:
\begin{aligned}\vec{u}\left(t+\delta t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t\right) & =\vec{u}\left(t+\delta t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t+\delta t\right)\\
 & \quad + \vec{u}\left(t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t\right)\\
 & =\left[\vec{u}\left(t+\delta t\right)-\vec{u}\left(t\right)\right]\times\vec{v}\left(t+\delta t\right)\\
 & \quad + \vec{u}\left(t\right)\times\left[\vec{v}\left(t+\delta t\right)-\vec{v}\left(t\right)\right]
\end{aligned}
Now divide by $\delta t$ and take limit as $\delta t\to 0$, which gives
$$
\frac{d}{dt}\left( \vec{u}\times\vec{v} \right) = \frac{d\vec{u}}{dt}\times\vec{v} + \vec{u}\times\frac{d\vec{v}}{dt}
$$
On the other hand
$$\frac{d}{dt}\left|\begin{array}{ccc}
i & j & k\\
v_{x} & v_{y} & v_{z}\\
u_{x} & u_{y} & u_{z}
\end{array}\right|=\left|\begin{array}{ccc}
i & j & k\\
\frac{dv_{x}}{dt} & \frac{dv_{y}}{dt} & \frac{dv_{z}}{dt}\\
u_{x} & u_{y} & u_{z}
\end{array}\right|+\left|\begin{array}{ccc}
i & j & k\\
v_{x} & v_{y} & v_{z}\\
\frac{du_{x}}{dt} & \frac{du_{y}}{dt} & \frac{du_{z}}{dt}
\end{array}\right|$$
Using the rule of differentiation of a determinant. One useful application of it is in the proof of Abel's identity (which before Wikipedia was known to me as Ostrogradski-Liouville formula)
