To differentiate the function \( g(t) = t^9 \cos t \), we apply the **product rule**. The product rule states that:

\[
\frac{d}{dt}(u(t) \cdot v(t)) = u'(t)v(t) + u(t)v'(t)
\]

In this case, we have:

- \( u(t) = t^9 \), and
- \( v(t) = \cos t \).

Now, we'll differentiate each part:

- \( u'(t) = 9t^8 \) (power rule),
- \( v'(t) = -\sin t \) (derivative of \( \cos t \)).

Now, applying the product rule:

\[
g'(t) = u'(t)v(t) + u(t)v'(t)
\]

\[
g'(t) = 9t^8 \cos t + t^9 (-\sin t)
\]

Thus, the derivative is:

\[
g'(t) = 9t^8 \cos t - t^9 \sin t
\]

Would you like more details or have any questions?

Here are 5 related questions you can explore:
1. How does the product rule apply when differentiating multiple terms?
2. What happens if \( g(t) \) has a different trigonometric function, like \( \sin t \)?
3. How would the result change if \( g(t) \) were a sum instead of a product?
4. Can you differentiate functions involving higher powers of trigonometric functions?
5. What is the importance of trigonometric derivatives in calculus?

**Tip:** Always check for opportunities to use the product, chain, or quotient rule when dealing with combinations of functions.

Differentiate g(t) = t^9 cos(t). Find g'(t).

Learn how to differentiate g(t) = t^9 cos(t) using the product rule. This problem covers key calculus concepts, including the power rule and the derivative of trigonometric functions. Suitable for students in Grades 11-12.

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