Question

Найти производные функций, неопределенный интеграл

Answer 1

Ответ:

1.

a

$y' = - 2 \times 3 {x}^{2} + 3 \times 2x - 0 = - 6 {x}^{2} + 6x$

b

$y' = ( {x}^{5} )'tgx + (tgx)' \times {x}^{5} = \\ = 5 {x}^{4} tgx + \frac{ {x}^{5} }{ \cos {}^{2} (x) } = {x}^{4} (5tgx + \frac{x}{ \cos {}^{2} (x) })$

c

$y' = \frac{(4x - 1)'(7x + 1) - (7x + 1)'(4x - 1)}{ {(7x + 1)}^{2} } = \\ = \frac{4(7x + 1) - 7(4x - 1)}{ {(7x + 1)}^{2} } = \frac{28x + 4 - 28x + 7}{ {(7x + 1)}^{2} } = \\ = \frac{11}{ {(7x + 1)}^{2} }$

d

$y' = 10 {(2 + {x}^{3} )}^{9} \times (2 + {x}^{3} )' = 10 {(2 + {x}^{3} )}^{9} \times 3 {x}^{2} = \\ = 30 {x}^{2} {(2 + {x}^{3} )}^{9}$

e

$y '= - \sin(5x) \times (5x) '= - 5 \sin(5x)$

2.

$\int\limits( \frac{8}{x} - \frac{5}{1 + {x}^{2} } + 3 \sin(x) )dx = \\ = \int\limits \frac{8}{x} dx - \int\limits \frac{5}{1 + {x}^{2} } dx + \int\limits3 \sin(x) ) dx = \\ = 8 ln( |x| ) - 5arctgx - 3 \cos(x) + C$

3.

$\int\limits^{1}_{-2}(4x+5-9x^2)dx=(\frac{4x^2}{2}+5x-\frac{9x^2}{3}) |^{1}_{-2}=\\=(2x^2+5x-3x^3) |^{1}_{-2}=2+5-3-(8-10+24)=-18$