1. **State the problem:** We need to find the derivative of the function $f(x) = \cos(4x)$ and then evaluate it at $x = \frac{\pi}{12}$. 2. **Recall the formula:** The derivative of $\cos(u)$ with respect to $x$ is $-\sin(u) \cdot \frac{du}{dx}$ by the chain rule. 3. **Apply the chain rule:** Here, $u = 4x$, so $\frac{du}{dx} = 4$. 4. **Find $f'(x)$:** $$ f'(x) = -\sin(4x) \cdot 4 = -4\sin(4x) $$ 5. **Evaluate at $x = \frac{\pi}{12}$:** $$ f'\left(\frac{\pi}{12}\right) = -4\sin\left(4 \cdot \frac{\pi}{12}\right) = -4\sin\left(\frac{4\pi}{12}\right) = -4\sin\left(\frac{\pi}{3}\right) $$ 6. **Simplify $\sin\left(\frac{\pi}{3}\right)$:** $$ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} $$ 7. **Calculate the final value:** $$ f'\left(\frac{\pi}{12}\right) = -4 \times \frac{\sqrt{3}}{2} = -\cancel{4}2 \times \frac{\sqrt{3}}{\cancel{2}} = -2\sqrt{3} $$ **Final answer:** $f'\left(\frac{\pi}{12}\right) = -2\sqrt{3}$