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