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

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Answer 1 · 2017-10-30T02:39:39

#h'(t) = (7t^2+4)/(3t^(2/3))#

#h(t) = root3t(t^2+4)#

#= t^(1/3) * (t^2+4)#

#h'(t) = t^(1/3) * d/dt(t^2+4) + d/dt(t^(1/3)) * (t^2+4)#

#h'(t) = t^(1/3) * (2t +0) + 1/3t^(-2/3) * (t^2+4)#

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

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

#= (7t^2+4)/(3t^(2/3))#