To solve this problem, we need to find $f''(\pi)$ for the function $f(x) = \cos(4x)$.

Step 1: Find $f'(x)$

Given: $f(x) = \cos(4x)$

Using the chain rule, we find the first derivative: $f'(x) = -\sin(4x) \cdot 4 = -4 \sin(4x)$

Step 2: Find $f''(x)$

Now, we differentiate $f'(x)$ to get the second derivative: $f''(x) = -4 \cdot \cos(4x) \cdot 4 = -16 \cos(4x)$

Step 3: Substitute $x = \pi$ into $f''(x)$

Now we need to evaluate $f''(\pi)$: $f''(\pi) = -16 \cos(4\pi)$

Since $\cos(4\pi) = 1$ (as $4\pi$ is a multiple of $2\pi$), we get: $f''(\pi) = -16 \cdot 1 = -16$

Answer

The correct answer is: $\boxed{-16}$

Would you like more details, or do you have further questions?

Here are 5 related questions for practice:

1. If $f(x) = \sin(3x)$, what is $f''(x)$?
2. What is the second derivative of $f(x) = e^{5x}$?
3. For $g(x) = \cos(x^2)$, what is $g'(x)$?
4. How would you find the third derivative of $h(x) = \cos(2x)$?
5. If $f(x) = \cos(5x + 3)$, what is $f''(0)$?

Tip: When differentiating trigonometric functions with a coefficient inside the argument, remember to apply the chain rule carefully.