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To solve for \( f'(2\pi) \), we need to apply the product rule since \( f(x) = h(x) \cos(x) \) is a product of two functions.

The product rule states:

\[
(fg)'(x) = f'(x)g(x) + f(x)g'(x)
\]

In this case, let:

- \( h(x) = h(x) \) (first function)
- \( \cos(x) = g(x) \) (second function)

Thus, by applying the product rule to \( f(x) = h(x) \cos(x) \):

\[
f'(x) = h'(x) \cos(x) + h(x) (-\sin(x))
\]

Now, substitute \( x = 2\pi \):

\[
f'(2\pi) = h'(2\pi) \cos(2\pi) + h(2\pi)(-\sin(2\pi))
\]

We know:

- \( h(2\pi) = -25 \)
- \( h'(2\pi) = 19 \)
- \( \cos(2\pi) = 1 \)
- \( \sin(2\pi) = 0 \)

Substitute these values into the equation:

\[
f'(2\pi) = (19)(1) + (-25)(0)
\]

Simplify:

\[
f'(2\pi) = 19
\]

Thus, \( f'(2\pi) = 19 \).

Would you like more details on this? Here are 5 related questions you could explore:

1. What is the product rule and how is it generally applied?
2. How do you find derivatives of trigonometric functions like \( \cos(x) \) and \( \sin(x) \)?
3. What is the significance of \( \sin(2\pi) \) and \( \cos(2\pi) \) being 0 and 1 respectively?
4. How would the result change if \( h(2\pi) \) or \( h'(2\pi) \) had different values?
5. Can you solve similar problems using the chain rule instead of the product rule?

**Tip**: Always ensure the proper application of differentiation rules when handling products of functions or composite functions.

Given that f(x) = h(x) cos(x), h(2π) = -25, and h'(2π) = 19, calculate f'(2π).

Learn how to apply the product rule to find the derivative of f(x) = h(x) cos(x) at x = 2π, with h(2π) = -25 and h'(2π) = 19. This problem uses core concepts in calculus and trigonometric functions, suitable for students in Grades 11-12 and early college.

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