Question

помогите решить срочно

Answer 1

Ответ:

$y' = \frac{(1 - \cos(4x)) '\times \sin(4x) - ( \sin(4x))' \times (1 - \cos(4x)) }{ \sin {}^{2} (4x) } = \\ = \frac{4 \sin(4x) \times \sin(4x) - 4 \cos(4x) (1 - \cos(4x)) }{ \sin {}^{2} (4x) } = \\ = \frac{4 \sin {}^{2} (4x) - 4 \cos(4x) + 4 \cos {}^{2} (4x) }{ \sin {}^{2} (4x) } = \\ = \frac{1 - 4 \cos(4x) }{ \sin {}^{2} (4x) }$

$y'( - \frac{\pi}{8} ) = \frac{ 1 - \cos( - \frac{\pi}{2} ) }{ \sin {}^{2} ( - \frac{\pi}{2} ) } = \frac{1}{1} = 1$