Question

помогите сроооочнооо!! 100 баллов!​

Answer 1

$y = \sin( \frac{\pi}{4} - 3x) , \: x _{0} = \frac{\pi}{12} \\ y' = ( \frac{\pi}{4} - 3x)'( \sin(g) )' = - 3 \cos( \frac{\pi}{4} - 3x ) \\ y'(x_{0}) = - 3 \cos( \frac{\pi}{4} - \frac{3\pi}{12} ) = - 3 \cos(0) = - 3$

$y = \tan(6x) , \: x_{0} = \frac{\pi}{24} \\ y' = (6x)'( \tan(h) )' = \frac{6}{ \cos {}^{2} (6x) } \\ y'(x_{0}) = \frac{6}{ \cos {}^{2} ( \frac{6\pi}{24} ) } = 12$

$y = \cos(2x - \frac{\pi}{3} ) , \: x _{0} = -\frac{\pi}{3} \\ y' = (2x - \frac{\pi}{3} )'( \cos(k) )' = - 2 \sin(2x - \frac{\pi}{3} ) \\ y'(x_{0}) = - 2 \sin( -\frac{2\pi}{3} - \frac{\pi}{3} ) = - 2sin(-\pi) = 0$