# Derivative of cross-product of two vectors \frac{d}{dt} [\vec{u(t)} \times \vec{v(t)}],

Derivative of cross-product of two vectors $\frac{d}{dt}\left[\stackrel{\to }{u\left(t\right)}×\stackrel{\to }{v\left(t\right)}\right]$, is it possible to find the cross-product of the two vectors first before differentiating?
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xandir307dc
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:
$\frac{d}{dt}\left(u×v\right)=\frac{du}{dt}×v+u×\frac{dv}{dt}$
Picking a method depends on the problem at hand. For example, the product rule is used to derive Frenet Serret formulas.

Working from the first principles:
$\stackrel{\to }{u}\left(t+\delta t\right)×\stackrel{\to }{v}\left(t+\delta t\right)-\stackrel{\to }{u}\left(t\right)×\stackrel{\to }{v}\left(t\right)=\stackrel{\to }{u}\left(t+\delta t\right)×\stackrel{\to }{v}\left(t+\delta t\right)-\stackrel{\to }{u}\left(t\right)×\stackrel{\to }{v}\left(t+\delta t\right)+$
$=\stackrel{\to }{u}\left(t\right)×\stackrel{\to }{v}\left(t+\delta t\right)-\stackrel{\to }{u}\left(t\right)×\stackrel{\to }{v}\left(t\right)=$
$=\left[\stackrel{\to }{u}\left(t+\delta t\right)-\stackrel{\to }{u}\left(t\right)\right]×\stackrel{\to }{v}\left(t+\delta t\right)+$
$=\stackrel{\to }{u}\left(t\right)×\left[\stackrel{\to }{v}\left(t+\delta t\right)-\stackrel{\to }{v}\left(t\right)\right]$
Now divide by $\delta$ and take limit as $\delta t⇒0$
On the other hand
$\frac{d}{dt}\left[\begin{array}{ccc}i& j& k\\ {v}_{x}& {v}_{y}& {v}_{z}\\ {u}_{x}& {u}_{y}& {u}_{z}\end{array}\right]=\left[\begin{array}{ccc}i& j& k\\ \frac{d{v}_{x}}{dt}& \frac{d{v}_{y}}{dt}& \frac{d{v}_{z}}{dt}\\ {u}_{x}& {u}_{y}& {u}_{z}\end{array}\right]+\left[\begin{array}{ccc}i& j& k\\ {v}_{x}& {v}_{y}& {v}_{z}\\ \frac{d{u}_{z}}{dt}& \frac{d{u}_{y}}{dt}& \frac{d{u}_{z}}{dt}\end{array}\right]$
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 OstrograZSKi-Liouville formula)