# How do you use the product rule to differentiate h(t)=root3(t)(t^2+4)?

Oct 30, 2017

$h ' \left(t\right) = \frac{7 {t}^{2} + 4}{3 {t}^{\frac{2}{3}}}$

#### Explanation:

$h \left(t\right) = \sqrt[3]{t} \left({t}^{2} + 4\right)$

$= {t}^{\frac{1}{3}} \cdot \left({t}^{2} + 4\right)$

Apply the product rule.

$h ' \left(t\right) = {t}^{\frac{1}{3}} \cdot \frac{d}{\mathrm{dt}} \left({t}^{2} + 4\right) + \frac{d}{\mathrm{dt}} \left({t}^{\frac{1}{3}}\right) \cdot \left({t}^{2} + 4\right)$

Apply the power rule

$h ' \left(t\right) = {t}^{\frac{1}{3}} \cdot \left(2 t + 0\right) + \frac{1}{3} {t}^{- \frac{2}{3}} \cdot \left({t}^{2} + 4\right)$

$= 2 {t}^{\frac{4}{3}} + \frac{{t}^{2} + 4}{3 {t}^{\frac{2}{3}}}$

=(6t^2+t^2+4)/(3t^(2/3)

$= \frac{7 {t}^{2} + 4}{3 {t}^{\frac{2}{3}}}$