Answer
$F'(t)=-6(3t-1)^{4}(2t+1)^{-4} / +12(2t+1)^{-3}(3t-1)^{3}$
Work Step by Step
$F(t)=(3t-1)^{4}(2t+1)^{-3}$
Differentiate using the product rule:
$F'(t)=[(3t-1)^{4}][(2t+1)^{-3}]'+[(2t+1)^{-3}][(3t-1)^{4}]'=...$
Use the chain rule to find $[(2t+1)^{-3}]'$ and $[(3t-1)^{4}]'$:
$...=[(3t-1)^{4}][-3(2t+1)^{-4}(2t+1)']+[(2t+1)^{-3}][4(3t-1)^{3}(3t-1)']=...$
$...=[(3t-1)^{4}][-3(2t+1)^{-4}(2)]+[(2t+1)^{-3}][4(3t-1)^{3}(3)]=...$
$...=-6(3t-1)^{4}(2t+1)^{-4} / +12(2t+1)^{-3}(3t-1)^{3}$
