Question

If,


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

What is the value of,


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

Answer 1

`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((\pi )/(18)\right)=12\sin\left(3\cdot(\pi )/(18)\right)\cos\left(3\cdot(\pi )/(18)\right)\\ f'\left((\pi )/(18)\right)=12\sin\left((\pi )/(6)\right)\cos\left((\pi )/(6)\right)\\ f'\left((\pi )/(18)\right)=12\cdot(1)/(2)\cdot(\sqrt 3)/(2)\\ f'\left((\pi )/(18)\right)=3\sqrt3`

Answer 2

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