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The function \( h(t) \) given is:

\[
h(t) = (t + 2)^{\frac{2}{3}} \cdot (2t^2 - 3)^3
\]

We need to find the derivative \( h'(t) \). This involves using the **product rule** and the **chain rule**.

### Steps:
1. **Product Rule**: If \( h(t) = u(t) \cdot v(t) \), then
   \[
   h'(t) = u'(t) v(t) + u(t) v'(t)
   \]
   Here, \( u(t) = (t + 2)^{\frac{2}{3}} \) and \( v(t) = (2t^2 - 3)^3 \).

2. **Differentiate \( u(t) \)**: 
   \[
   u'(t) = \frac{d}{dt} \left( (t + 2)^{\frac{2}{3}} \right)
   \]
   Using the chain rule:
   \[
   u'(t) = \frac{2}{3} (t + 2)^{-\frac{1}{3}} \cdot 1 = \frac{2}{3} (t + 2)^{-\frac{1}{3}}
   \]

3. **Differentiate \( v(t) \)**: 
   \[
   v'(t) = \frac{d}{dt} \left( (2t^2 - 3)^3 \right)
   \]
   Using the chain rule:
   \[
   v'(t) = 3(2t^2 - 3)^2 \cdot \frac{d}{dt}(2t^2 - 3) = 3(2t^2 - 3)^2 \cdot 4t = 12t(2t^2 - 3)^2
   \]

4. **Apply the Product Rule**: Now substitute \( u'(t) \), \( u(t) \), \( v'(t) \), and \( v(t) \):
   \[
   h'(t) = u'(t) v(t) + u(t) v'(t)
   \]
   So:
   \[
   h'(t) = \frac{2}{3} (t + 2)^{-\frac{1}{3}} \cdot (2t^2 - 3)^3 + (t + 2)^{\frac{2}{3}} \cdot 12t(2t^2 - 3)^2
   \]

This is the derivative of the function.

Do you need further details on any part of the solution, or have any questions?

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Here are 5 related questions:
1. How would you simplify the expression for \( h'(t) \)?
2. Can the quotient rule also be applied in cases like this?
3. What happens to the derivative if \( t = 0 \)?
4. What if the powers in \( h(t) \) were different? How would that affect the derivative?
5. How would you approach finding the second derivative \( h''(t) \)?

**Tip:** When applying the product rule, always differentiate each function separately before combining, and don't forget the chain rule if needed!

Find the derivative of the function: h(t) = (t + 2)^(2/3) * (2t^2 - 3)^3.

Learn how to find the derivative of h(t) = (t + 2)^(2/3) * (2t^2 - 3)^3 using both the product and chain rules. The detailed step-by-step solution covers applying the rules in differentiation, suitable for Calculus I students.

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