Question

решите задачу пожалуйста

Answer 1

1.

а)

$f(x) = - \cos(2x) + \sin(2x)$

$f'(x) = \sin(2x) \times 2 + \cos(2x) \times 2 = \\ = 2( \sin(2x) + \cos(2x))$

б)

$f(x) = 3x + \cos(4x)$

$f'(x) = 3 - 4 \sin(4x)$

в)

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

$f'(x) = 3 {x}^{2} - 4 \cos(2x)$

г)

$f(x) = 2tg2x$

$f'(x) = \frac{4}{ { \cos }^{2}(2x) }$

2.

а)

$f( x)= \cos(x) + 1 \\ f'(x) = - \sin(x) \\ f'( \frac{\pi}{6} ) = - \sin( \frac{\pi}{6} ) = - \frac{1}{2}$

б)

$f(x) = tgx - 2 \\ f'(x) = \frac{1}{ { \cos }^{2}(x) } \\ f'( \frac{\pi}{3} ) = \frac{1}{ { \cos }^{2}( \frac{\pi}{3} ) } = \frac{1}{ {( \frac{1}{2}) }^{2} } = 4$

в)

$f(x) = \frac{2}{3} \sin(x) \\ f'(x) = \frac{2}{3} \cos(x) \\ f '(\frac{\pi}{3} ) = \frac{2}{3} \cos( \frac{\pi}{3} ) = \frac{2}{3} \times \frac{1}{2} 3 \frac{1}{3}$

г)

$f(x) = ctgx + \frac{1}{3} tgx$

$f'(x) = - \frac{1}{ { \sin }^{2} (x)} + \frac{1}{3 { \cos }^{2}(x) }$

$f'( \frac{\pi}{3} ) = - \frac{1}{ {( \frac{ \sqrt{3} }{2} )}^{2} } + \frac{1}{3 \times {( \frac{1}{2} )}^{2} } = \\ = - \frac{4}{3} + \frac{4}{3} = 0$