Answered

If,


\(f(x)\quad =\quad \sin ^{ 2 }{ 3x-\cos ^{ 2 }{ 3x } } \\ \\\)


What is the value of,


\(f^{ ' }\left( \frac { \pi }{ 18 } \right)\) ?

Verified Answer :

superkoopasirf

\(f(x) = \sin^2{3x} - \cos^2{3x} \\ \\ \text{Differentiate each separate part of the function using the chain rule} \\ \\ \frac{d}{dx}\sin^2{3x} \\ = \frac{d}{dx}(\sin{3x})^2 \\ = 2\sin{3x}\cos{3x} \times 3 \\ = 6\sin{3x}\cos{3x} \\ \\ \frac{d}{dx} \cos^2{3x} \\ = \frac{d}{dx}(\cos{3x})^2 \\ = 2\cos{3x}(-\sin{3x} \times 3) \\ = -6\sin{3x}\cos{3x} \\ \\ \text{So } f'(x) = 6\sin{3x}\cos{3x} - (-6\sin{3x}\cos{3x}) \\ = 12\sin{3x}\cos{3x} \\ \\ \text{Substitute } x \text{ for } \frac{\pi}{18} \\ \\\)\(f'(\frac{\pi}{18}) = 12\sin{\frac{3\pi}{18}\cos{\frac{3\pi}{18} \\ = 12\sin{\frac{3\pi}{18}\cos{\frac{3\pi}{18} \\ = 12(\frac{1}{2})(\frac{\sqrt{3}}{2}) \\ = 3\sqrt{3}\)

konrad509

\(f'(x)=2\sin 3x\cdot\cos 3x\cdot 3-2\cos 3x\cdot(-\sin 3x)\cdot3\\ f'(x)=6\sin 3x\cos 3x+6\sin 3x\cos 3x\\ f'(x)=12 \sin 3x \cos 3x\\\\ f'\left(\dfrac{\pi }{18}\right)=12\sin\left(3\cdot\dfrac{\pi }{18}\right)\cos\left(3\cdot\dfrac{\pi }{18}\right)\\ f'\left(\dfrac{\pi }{18}\right)=12\sin\left(\dfrac{\pi }{6}\right)\cos\left(\dfrac{\pi }{6}\right)\\ f'\left(\dfrac{\pi }{18}\right)=12\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt 3}{2}\\ f'\left(\dfrac{\pi }{18}\right)=3\sqrt3\)

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