# How do you use the product Rule to find the derivative of h(t)=t^(1/3)(t^2+4)?

$\frac{\mathrm{dy}}{\mathrm{dt}}$ = [${t}^{\frac{1}{3}}$ (2t)] +[ (${t}^{2}$ + 4) ($\frac{1}{3}$${t}^{- \frac{2}{3}}$)]
$\frac{\mathrm{dy}}{\mathrm{dt}}$=[$2 {t}^{\frac{4}{3}}$ ] +[ (${t}^{2}$ + 4) (1/(3t^(2/3))]
$\frac{\mathrm{dy}}{\mathrm{dt}}$=[$2 {t}^{\frac{4}{3}}$ ] +[  (t^2 + 4)/(3t^(2/3)]
$\frac{\mathrm{dy}}{\mathrm{dt}}$=  (6t^2+t^2 + 4)/(3t^(2/3)
$\frac{\mathrm{dy}}{\mathrm{dt}}$=  (7t^2 + 4)/(3t^(2/3)