Question

17.9 и 17.10

ПОМОГИТЕ ОЧЕНЬ НУЖНО ДАМ 37 БАЛЛОВ
ТОЛЬКО С РЕШЕНИЕМ ПРОСТО ОТВЕТ НЕ НУЖЕН

Answer 1

Ответ:

17.9

а)

$f'(x) = ( \cos(x)) ' \times ( \cos(x) - 1) + ( \cos(x) - 1)' \times \cos(x) = \\ = - \sin(x) \times ( \cos(x) - 1) - \sin(x) \times \cos(x) = \\ = - \sin(x) \cos(x) + \sin(x) - \sin(x) \cos(x) = \\ = \sin(x) - 2 \sin(x) \cos(x) = \sin(x) - \sin(2x)$

б)

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

в)

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

г)

$f'(x) = (4x - 1) '\times \sin(x) + ( \sin(x))' \times (4x - 1) = \\ = 4 \sin(x) + \cos(x) (4x - 1) = \\ = 4 \sin(x) + (4x - 1) \cos(x)$

17.10

а)

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

б)

$f'(x) = 6 \sin(2x) \times ( \sin(2x)) ' \times (2x) '+ 2 = \\ = 6 \sin(2x) \times \cos(2x) \times 2 + 2 = \\ = 6 \sin(4x) + 2$

в)

$f'(x) = 2( \sin(2x) + 1) \times ( \sin(2x) + 1) '= \\ =( 2 \sin(2x) + 2) \times 2 \cos(2x) = \\ = 4 \sin(2x) \cos(2x) + 4 \cos(2x) = \\ = 2 \sin(4x) + 4 \cos(2x)$

г)

$f'(x) = 3 {( \cos(2x) + \sin(2x)) }^{2} \times ( \cos(2x) + \sin(2x) ) '= \\ = 3 {( \cos(2x) + \sin(2x) )}^{2} \times ( - 2 \sin(2x) + 2 \cos(2x) ) = \\ = - 6 {( \cos(2x) + \sin(2x)) }^{2} \times ( \cos(2x) - \sin(2x))$