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?

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!