Differentiation of Vector Product

Now There are two type of vector Products

a) Cross product

b) Dot product

Let A and B are two vectors.The cross product will be reprensented by AXB and the dot product is represented by A.B

The differentiation for these are given as

$\frac{d}{\mathrm{dt}}$(AXB)=$\frac{dA}{\mathrm{dt}}$XB + AX $\frac{dB}{\mathrm{dt}}$

$\frac{d}{\mathrm{dt}}$(A.B)=$\frac{dA}{\mathrm{dt}}$.B + A. $\frac{dB}{\mathrm{dt}}$

Let X(t) denote A(t) X B(t)

Now

X(t+Δt) -X(t) =A(t+Δt) X B(t+Δt) - A(t) X B(t)

≊[A(t) + $\frac{\mathrm{dA}}{\mathrm{dt}}\mathrm{\Delta t}$] X [B(t) + $\frac{\mathrm{dB}}{\mathrm{dt}}\mathrm{\Delta t}$] - A(t) X B(t)

$=\mathrm{\Delta t}[\frac{\mathrm{dA}}{\mathrm{dt}}\mathrm{XB}+\mathrm{AX}\frac{\mathrm{dB}}{\mathrm{dt}}]+{\left(\mathrm{\Delta t}\right)}^2\left[\frac{\mathrm{dA}}{\mathrm{dt}}X\frac{\mathrm{dB}}{\mathrm{dt}}\right]$

So we have

$\frac{\mathrm{dX}}{\mathrm{dt}}=\underset{\mathrm{\Delta t}->0}{lim}\frac{X(t+\mathrm{\Delta t})-X\left(t\right)}{\mathrm{\Delta t}}=\frac{\mathrm{dA}}{\mathrm{dt}}XB+\mathrm{AX}\frac{\mathrm{dB}}{\mathrm{dt}}$

Similarly other can be proved