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The problem asks us to find the derivative \( \frac{dy}{dt} \) of the given function:

\[
y = (t \cdot \tan t)^9
\]

### Steps:

1. **Apply the Chain Rule:**
   Since the expression is raised to the power 9, we'll first differentiate using the chain rule. This means we differentiate the outer function (power rule) and then multiply by the derivative of the inner function \( t \cdot \tan t \).

   Using the chain rule:
   \[
   \frac{dy}{dt} = 9 \cdot (t \cdot \tan t)^8 \cdot \frac{d}{dt}(t \cdot \tan t)
   \]

2. **Differentiate \( t \cdot \tan t \):**
   Now we need to find \( \frac{d}{dt}(t \cdot \tan t) \), which requires the product rule. Let:
   - \( u = t \)
   - \( v = \tan t \)

   Using the product rule \( \frac{d}{dt}(u \cdot v) = u'v + uv' \):
   \[
   \frac{d}{dt}(t \cdot \tan t) = 1 \cdot \tan t + t \cdot \sec^2 t
   \]
   So, \( \frac{d}{dt}(t \cdot \tan t) = \tan t + t \cdot \sec^2 t \).

3. **Substitute the derivative:**
   Now substitute this result back into the expression for \( \frac{dy}{dt} \):

   \[
   \frac{dy}{dt} = 9 \cdot (t \cdot \tan t)^8 \cdot (\tan t + t \cdot \sec^2 t)
   \]

This is the derivative of the given function.

Would you like a deeper explanation or further breakdown of any step?

### Related Questions:
1. How does the chain rule work when differentiating powers of a function?
2. Can the product rule be extended to more than two functions?
3. What are the common trigonometric derivatives for functions like \( \tan t \)?
4. How would the derivative change if the function were \( (t^2 \cdot \tan t)^9 \)?
5. Why do we multiply by 9 in the first step when differentiating \( (t \cdot \tan t)^9 \)?

**Tip:** Always break down complex expressions into smaller components before applying rules like the chain rule or product rule to avoid errors.

Find the derivative dy/dt of the function y = (t * tan(t))^9

Learn how to differentiate the function y = (t * tan(t))^9 step-by-step. This calculus problem covers key concepts like the chain rule, product rule, and trigonometric derivatives. Suitable for students in advanced high school calculus or undergraduate-level math courses.

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