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.