Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

If,

[tex]f(x)\quad =\quad \sin ^{ 2 }{ 3x-\cos ^{ 2 }{ 3x } } \\ \\ [/tex]

What is the value of,

[tex]f^{ ' }\left( \frac { \pi }{ 18 } \right) [/tex] ?

Sagot :

superkoopasirf
[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]
konrad509
[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]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.