(i) To find \(f'(x)\), we first differentiate \(f(x) = 4 \cos^2 3x\) using the chain rule. Let \(u = \cos 3x\), then \(f(x) = 4u^2\). The derivative \(\frac{du}{dx} = -3 \sin 3x\).
Using the chain rule, \(\frac{df}{dx} = 8u \cdot \frac{du}{dx} = 8 \cos 3x (-3 \sin 3x) = -24 \cos 3x \sin 3x\).
Now, evaluate \(f'(\frac{1}{9}\pi)\):
\(\cos 3(\frac{1}{9}\pi) = \cos \frac{\pi}{3} = \frac{1}{2}\)
\(\sin 3(\frac{1}{9}\pi) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)
Thus, \(f'(\frac{1}{9}\pi) = -24 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -6\sqrt{3}\).
(ii) To integrate \(f(x) = 4 \cos^2 3x\), use the identity \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\).
\(f(x) = 4 \cdot \frac{1 + \cos 6x}{2} = 2 + 2 \cos 6x\).
Integrate term by term:
\(\int (2 + 2 \cos 6x) \, dx = \int 2 \, dx + \int 2 \cos 6x \, dx\)
\(= 2x + \frac{2}{6} \sin 6x + C = 2x + \frac{1}{3} \sin 6x + C\).
