Answer :
[tex]f(x) = \sin^2{3x} - \cos^2{3x} \\ \\
\text{Differentiate each separate part of the function using the chain rule} \\ \\
\frac{d}{dx}\sin^2{3x} \\
= \frac{d}{dx}(\sin{3x})^2 \\
= 2\sin{3x}\cos{3x} \times 3 \\
= 6\sin{3x}\cos{3x} \\ \\
\frac{d}{dx} \cos^2{3x} \\
= \frac{d}{dx}(\cos{3x})^2 \\
= 2\cos{3x}(-\sin{3x} \times 3) \\
= -6\sin{3x}\cos{3x} \\ \\
\text{So } f'(x) = 6\sin{3x}\cos{3x} - (-6\sin{3x}\cos{3x}) \\
= 12\sin{3x}\cos{3x} \\ \\
\text{Substitute } x \text{ for } \frac{\pi}{18} \\ \\[/tex][tex]f'(\frac{\pi}{18}) = 12\sin{\frac{3\pi}{18}\cos{\frac{3\pi}{18} \\
= 12\sin{\frac{3\pi}{18}\cos{\frac{3\pi}{18} \\
= 12(\frac{1}{2})(\frac{\sqrt{3}}{2}) \\
= 3\sqrt{3}[/tex]
[tex]f'(x)=2\sin 3x\cdot\cos 3x\cdot 3-2\cos 3x\cdot(-\sin 3x)\cdot3\\
f'(x)=6\sin 3x\cos 3x+6\sin 3x\cos 3x\\
f'(x)=12 \sin 3x \cos 3x\\\\
f'\left(\dfrac{\pi }{18}\right)=12\sin\left(3\cdot\dfrac{\pi }{18}\right)\cos\left(3\cdot\dfrac{\pi }{18}\right)\\
f'\left(\dfrac{\pi }{18}\right)=12\sin\left(\dfrac{\pi }{6}\right)\cos\left(\dfrac{\pi }{6}\right)\\
f'\left(\dfrac{\pi }{18}\right)=12\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt 3}{2}\\
f'\left(\dfrac{\pi }{18}\right)=3\sqrt3
[/tex]
