Question

Зачетная работа. II вариант. 1), 2), 3), 4)

Answer 1

Ответ:

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

$f'( \frac{\pi}{8} ) = \frac{2}{ { \sin( \frac{\pi}{4} ) }^{2} } = \frac{2}{ {( \frac{ \sqrt{2} }{2} )}^{2} } = 2 \times \frac{4}{2} = 4$

$2)f'(x) = \frac{2}{ \sqrt{ \sin(2x) } } \times \frac{1}{2} { \sin(2x) }^{ - \frac{1}{2} } \times \cos(2x) \times 2 = \frac{4 \cos(2x) }{ \sqrt{ \sin(2x) } \times \sqrt{ \sin(2x) } } = 4ctg(2x)$

$f'( \frac{\pi}{8}) = 4ctg( \frac{\pi}{4} ) = 4$

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

$f'(0) = \frac{ - {e}^{0} }{1 + {e}^{0} } = \frac{ - 1}{2} = - 0.5$

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

$f'( \frac{\pi}{4} ) = 2 + 2 \cos( \frac{\pi}{2} ) = 2$