Question

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

Answer 1

`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)`