Problem: Verify that $$\frac{d}{dt}\left[F\times(G\times H)\right]'=[(H\cdot F)G]'-[(G\cdot F)H]'.$$
I am a bit confused about the prime symbol in the derivative. According to me $$\frac{d}{dt}\left[F\times(G\times H)\right]'=\frac{d^2}{dt^2}\left[F\times(G\times H)\right].$$ But when I use the following rule $$\frac{d}{dt}[A\times B]=A\times\frac{dB}{dt}+\frac{dA}{dt}\times B$$ I get a different answer. Please explain where am I going wrong.
Edit: After some looking around, I found the following useful identity on Wikipedia $$\mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=\mathbf {b} (\mathbf {a} \cdot \mathbf {c} )-\mathbf {c} (\mathbf {a} \cdot \mathbf {b} ).$$
Thus $$\frac{d}{dt}\left[F\times(G\times H)\right]'= \frac{d}{dt}\left[{G} ( {F} \cdot {H} )-{H} ( {F} \cdot {G})\right]'$$ $$=\frac{d}{dt}\left[{G'} ( {F} \cdot {H} )-{H'} ( {F} \cdot {G})\right]$$ $$=(F\cdot H)\frac{dG'}{dt}-(F\cdot G)\frac{dH'}{dt}.$$ Still I am not able to derive the right identity.
If we interpret the prime in the formula as the derivative w/r $t$, then the right-hand side is an expression that involves first derivatives of the vector functions while the left-hand side has both first and second derivatives, so it seems unlikely that we’re going to have universal equality there.
On the other hand, $$\begin{align} [F\times(G\times H)]' &= F'\times(G\times H)+F\times(G'\times H)+F\times(G\times H') \\ &= (F'\cdot H)G - (F'\cdot G)H + (F\cdot H)G' - (F\cdot G')H + (F\cdot H')G - (F\cdot G)H' \\ &= [(F'\cdot H)G + (F\cdot H')G + (F\cdot H)G'] - [(F'\cdot G)H + (F\cdot G')H + (F\cdot G)H'] \\ &= [(F\cdot H)G]' - [(F\cdot G)H]' \\ &= [(H\cdot F)G]' - [(G\cdot F)H]'. \end{align}$$ So it would appear that the $\frac d{dt}$ on the left-hand side of the identity in your question is a typo in the problem.