Question

Помогите решить примеры ​

Answer 1

Ответ:

1.

а

$f'(x) = 0$

б

$f'(x) = 6 {x}^{5}$

в

$f'(x) = 15 {x}^{2} + 4$

г

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

д

$f'(x) = ((2x - 6)2 {x}^{3} )' = (4 {x}^{4} - 12 {x}^{3} ) '= \\ = 16 {x}^{3} - 36 {x}^{2}$

е

$f'(x) = \frac{(5x - 3)' {x}^{2} - ( {x}^{2} )' \times (5x - 3) }{ {x}^{4} } = \\ = \frac{5 {x}^{2} - 2x(5x - 3)}{ {x}^{4} } = \frac{5 {x}^{2} - 10 {x}^{2} + 6x }{ {x}^{4} } = \\ = \frac{ - 5 {x}^{2} + 6x}{ {x}^{4} } = \frac{6 - 5x}{ {x}^{3} }$

ж

$f'(x) = 4 \cos(x) - 2 \sin(x)$

з

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

и

$f'(x) = 2(3x + 6) \times 3 = 6(3x + 6) = \\ = 18x + 36$

к

$f'(x) = \frac{1}{2} {(2x - 9)}^{ - \frac{1}{2} } \times (2x - 9) '= \frac{1}{2 \sqrt{2x - 9} } \times 2 = \frac{1}{ \sqrt{2x - 9} }$

л

$f'(x) = \frac{1}{ \cos {}^{2} (7x - 3) } \times7$

м

$f'(x) = 5 \times ( - 4) {x}^{ - 5} + 3 \times ( - 3) {x}^{ - 4} = \\ = - \frac{20}{ {x}^{5} } - \frac{9}{ {x}^{4} }$

2.

а

$F(x) = \int\limits( - 7)dx = - 7x + C$

б

$F(x) = \int\limits {x}^{7}dx = \frac{ {x}^{8} }{8} + C$

в

$F(x) = \int\limits( \cos(x) - \sin(x)) dx = \sin(x) + \cos(x) + C$

г

$F(x) = \int\limits4 {x}^{ - 3} dx = \frac{4 {x}^{ - 2} }{( - 2)} + C = - \frac{2}{ {x}^{2} } + C$

д

$F(x) = \int\limits(3x - 2 {x}^{ - \frac{1}{2} } )dx = \frac{3 {x}^{2} }{2} - \frac{2 {x}^{ \frac{1}{2} } }{ \frac{1}{2} } + C = \\ = \frac{3 {x}^{2} }{2} - 4 \sqrt{x} + C$

е

$F(x) = \int\limits \sin(2x)dx = \frac{1}{2} \int\limits \sin(2x) d(2x) = - \frac{1}{2} \cos(2x) + C$

ж

$F(x) = \int\limits {(2x + 3)}^{4} dx = \frac{1}{2} \int\limits {(2x + 3)}^{4} d(2x + 4) = \\ = \frac{ {(2x + 3)}^{5} }{10} + C$

з

$F(x) = \int\limits \frac{dx}{ \sqrt{2x - 5} } = \frac{1}{2} \int\limits {(2x - 5)}^{ - \frac{1}{2} } d(2x - 5) = \\ = \sqrt{2x - 5} + C$

и

$F(x) = \int\limits \frac{dx}{ \cos {}^{2} (3x + 1) } = \frac{1}{3} \int\limits \frac{d(3x + 1)}{ \cos {}^{2} (3x + 1) } = tg(3x + 1) + C$