The function to differentiate is:

$h(t) = \left( \frac{t^3}{t^6 + 9} \right)^2$

To find $h'(t)$, we will use the chain rule along with the quotient rule. Here are the steps:

1. Chain Rule: First, recognize that the outer function is a square, so we will differentiate $(\text{something})^2$ first and then multiply by the derivative of the inside.

   Using the chain rule: $\frac{d}{dt} \left( u^2 \right) = 2u \cdot \frac{du}{dt}$ where $u = \frac{t^3}{t^6 + 9}$.

2. Quotient Rule: Now, to differentiate the inside part $\frac{t^3}{t^6 + 9}$, we use the quotient rule: $\frac{d}{dt} \left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2}$ where $f(t) = t^3$ and $g(t) = t^6 + 9$.

   Let's calculate the necessary derivatives:

   • $f'(t) = 3t^2$
   • $g'(t) = 6t^5$

   Substituting these into the quotient rule: $\frac{d}{dt} \left( \frac{t^3}{t^6 + 9} \right) = \frac{3t^2(t^6 + 9) - t^3(6t^5)}{(t^6 + 9)^2}$ Simplifying the numerator: $3t^2(t^6 + 9) - t^3(6t^5) = 3t^8 + 27t^2 - 6t^8 = -3t^8 + 27t^2$ So the derivative becomes: $\frac{-3t^8 + 27t^2}{(t^6 + 9)^2}$

3. Final derivative: Now applying the chain rule to get $h'(t)$: $h'(t) = 2 \cdot \left( \frac{t^3}{t^6 + 9} \right) \cdot \frac{-3t^8 + 27t^2}{(t^6 + 9)^2}$ Simplifying the final expression: $h'(t) = \frac{2t^3(-3t^8 + 27t^2)}{(t^6 + 9)^3}$

So, the derivative is:

$h'(t) = \frac{2t^3(-3t^8 + 27t^2)}{(t^6 + 9)^3}$

Would you like a more detailed explanation of any part of this solution?

Related questions:

1. How does the chain rule work in this context?
2. Can you explain the quotient rule in more depth?
3. What would change if the power was different, such as $\left( \frac{t^3}{t^6 + 9} \right)^3$?
4. How can the quotient rule be generalized for more complex functions?
5. What are some common mistakes to avoid when applying the quotient rule?

Tip: Always simplify the numerator and denominator before differentiating to make the process easier.