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

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To find the derivative of ( h(t) = t^\frac{1}{3}(t^2 + 4) ) using the product rule, you can follow these steps:

- Identify the two functions being multiplied: ( f(t) = t^\frac{1}{3} ) and ( g(t) = t^2 + 4 ).
- Apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.
- Calculate the derivatives of ( f(t) ) and ( g(t) ) separately.
- Apply the product rule formula to find the derivative of ( h(t) ).
- Combine the results to get the final derivative expression.

Here's the breakdown:

( f(t) = t^\frac{1}{3} ) ( f'(t) = \frac{1}{3}t^{\frac{1}{3} - 1} = \frac{1}{3}t^{-\frac{2}{3}} )

( g(t) = t^2 + 4 ) ( g'(t) = 2t )

Now, apply the product rule:

( h'(t) = f'(t)g(t) + f(t)g'(t) ) ( h'(t) = \frac{1}{3}t^{-\frac{2}{3}}(t^2 + 4) + t^\frac{1}{3}(2t) )

Simplify the expression:

( h'(t) = \frac{1}{3}t^{-\frac{2}{3}}(t^2 + 4) + 2t^\frac{4}{3} )

So, the derivative of ( h(t) ) is ( \frac{1}{3}t^{-\frac{2}{3}}(t^2 + 4) + 2t^\frac{4}{3} ).

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