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We are given the expression for \( y \) and need to find \(\frac{dy}{dt}\). The function is:

\[
y = \left( \frac{t^4}{t^5 - 6t} \right)^3
\]

### Steps:

1. **Rewrite the function for clarity:**

   Let:
   \[
   z = \frac{t^4}{t^5 - 6t}
   \]
   So, \( y = z^3 \).

2. **Use the chain rule to differentiate:**
   Using the chain rule:
   \[
   \frac{dy}{dt} = 3z^2 \cdot \frac{dz}{dt}
   \]
   
3. **Find \(\frac{dz}{dt}\) using the quotient rule:**
   The quotient rule for \( z = \frac{t^4}{t^5 - 6t} \) is:
   \[
   \frac{dz}{dt} = \frac{(t^5 - 6t) \cdot \frac{d}{dt}(t^4) - t^4 \cdot \frac{d}{dt}(t^5 - 6t)}{(t^5 - 6t)^2}
   \]

   Simplifying:
   \[
   \frac{dz}{dt} = \frac{(t^5 - 6t) \cdot 4t^3 - t^4 \cdot (5t^4 - 6)}{(t^5 - 6t)^2}
   \]

4. **Substitute \(\frac{dz}{dt}\) into the chain rule expression for \( y \):**
   Finally, substitute back into the chain rule to find \(\frac{dy}{dt}\):
   \[
   \frac{dy}{dt} = 3z^2 \cdot \frac{dz}{dt}
   \]

Would you like to proceed with further simplification, or do you need clarification on any part of this? 

### Related Questions:

1. How do you apply the chain rule in calculus?
2. What is the quotient rule, and how is it used in differentiation?
3. Can you find the critical points of the function \( y \)?
4. How does the power rule relate to higher-order derivatives?
5. Can you analyze the behavior of this function as \( t \to \infty \)?

**Tip:** Always simplify the expression first before differentiating, as it reduces the chances of errors in computation!

Find dy/dt where y = (t^4 / (t^5 - 6t))^3.

Learn how to differentiate the function y = (t^4 / (t^5 - 6t))^3 step by step using the chain rule and quotient rule. This problem covers core calculus concepts and is suitable for undergraduate students.

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