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

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Aug 18, 2015

#dy/dt# = [#t^(1/3)# (2t)] +[ (#t^2# + 4) (#1/3##t^(-2/3)#)]

#dy/dt#=[#2t^(4/3)# ] +[ (#t^2# + 4) (#1/(3t^(2/3)#)]

#dy/dt#=[#2t^(4/3)# ] +[ # (t^2 + 4)/(3t^(2/3)#]

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

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

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