Question

помогите решить производную функции

Answer 1

Ответ:

1.

$f(x) = 7 {x}^{3} - 5x - 1.5 \\ f'(x) = 21 {x}^{2} - 5$

$\\ f(x) = \frac{3}{ {x}^{5} } - \frac{ {x}^{9} }{9} + 10 \sqrt{x} = 3 {x}^{ - 5} - \frac{ {x}^{9} }{9} + 10 {x}^{ \frac{1}{2} } \\ f'(x) = 3 \times ( - 5) {x}^{ - 6} - {x}^{8} + 10 \times \frac{1}{2} {x}^{ - \frac{1}{2} } = - \frac{15}{ {x}^{6} } - {x}^{8} + \frac{5}{ \sqrt{x} }$

$\\ f(x) = (2 {x}^{3} - 3)(x + 8) = 2 {x}^{4} - 3x + 16 {x}^{3} - 24 \\ f'(x) = 8 {x}^{3} - 3 + 48 {x}^{2}$

$\\ f(x) = {(3 {x}^{7} - 3) }^{7} \\ f'(x) = 7 {(3 {x}^{7} - 3)}^{6} \times (3 {x}^{7} - 3)' = \\ = 7 {(3 {x}^{7} - 3) }^{6} \times 21 {x}^{6} = 147 {x}^{6} {(3 {x}^{7} - 3) }^{6}$

$\\ f(x) = tgx \times (x + 8) \\ f'(x) = (tgx)' \times (x + 8) + (x + 8) '\times tgx = \\ = \frac{x + 8}{ \cos {}^{2} (x) } + tgx$

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

$\\ f(x) = \frac{5 - {x}^{2} }{ \sin(x) } \\ f'(x) = \frac{(5 - {x}^{2} ) '\times \sin(x) - (\sin(x)) '\times (5 - {x}^{2} ) }{ \sin {}^{2} (x) } = \\ = \frac{ - 2x \sin(x) - \cos(x) \times (5 - {x}^{2} ) }{ \sin {}^{2} (x) } = \\ = - \frac{2x \sin(x) + (5 - {x}^{2} ) \cos(x) }{ \sin {}^{2} (x) }$

2.

$f(x) = 24 {x}^{2} - 5x - 4 \\ f'(x) = 48x - 5 \\ dy = (48x - 5)dx$

$\\ f(x) = \frac{8x - 5}{ \sin(x) } \\ f'(x) = \frac{(8x - 5)' \sin(x) - ( \sin(x) ) '\times (8x - 5)}{ \sin {}^{2} (x) } = \\ = \frac{8 \sin(x) - (8x - 5) \cos(x) }{ \sin {}^{2} (x) } \\ dy = \frac{8 \sin(x) - (8x - 5) \cos(x) }{ \sin {}^{2} (x) } dx$

$\\ f(x) = 9tgx \\ f'(x) = \frac{9}{ \cos {}^{2} (x) } \\ dy = \frac{9}{ \cos {}^{2} (x) } dx$