## Calculus: Early Transcendentals 8th Edition

$h'(t)=12t(t+1)^{2/3}(2t^{2}-1)^{2}+\dfrac{2}{3}(t+1)^{-1/3}(2t^{2}-1)^{3}$
$h(t)=(t+1)^{2/3}(2t^{2}-1)^{3}$ 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 $[(t+1)^{2/3}]'$: $...=[(t+1)^{2/3}][3(2t^{2}-1)^{2}(2t^{2}-1)']+[(2t^{2}-1)^{3}][\dfrac{2}{3}(t+1)^{-1/3}(t+1)']=...$ $...=[(t+1)^{2/3}][3(2t^{2}-1)^{2}(4t)]+[(2t^{2}-1)^{3}][\dfrac{2}{3}(t+1)^{-1/3}(1)]=...$ $...=12t(t+1)^{2/3}(2t^{2}-1)^{2}+\dfrac{2}{3}(t+1)^{-1/3}(2t^{2}-1)^{3}$