Question

20баловПроизводные кто шарит срочно помогите

Answer 1

Ответ:

1.

$y' = 2 \cos(3x) \times ( \cos(3x))' \times (3x) '= \\ = 2 \cos(3x) \times ( - \sin(3x)) \times 3 = \\ = - 3 \sin(6x)$

2.

$y' = (x)'arctgx + (arctgx)'x = \\ = arctgx + \frac{x}{1 + {x}^{2} }$

3.

$y' = \frac{(x - 1)'lnx - (lnx)'(x - 1)}{ {ln}^{2} x} = \\ = \frac{lnx - \frac{1}{x} (x - 1)}{ {ln}^{2}x } = \\ = \frac{lnx - 1 + \frac{1}{x} }{ {ln}^{2} x} = \\ = \frac{1}{lnx} - \frac{1}{ {ln}^{2} x} + \frac{1}{x {ln}^{2}x }$

4.

$y' = \frac{( \sqrt[9]{4 {x}^{5} + 2} ), \times 3 {x}^{4} - (3 {x}^{4})' \sqrt[9]{4 {x}^{5} + 2 } }{9 {x}^{8} } = \\ = \frac{ \frac{1}{9} {(4 {x}^{5} + 2)}^{ - \frac{8}{9} } \times 20 {x}^{4} \times 3 {x}^{4} - 12 {x}^{3} \sqrt[9]{4 {x}^{5} + 2 } }{9 {x}^{8} } = \\ = \frac{ \frac{60 {x}^{8} }{9 \sqrt[9]{ {(4 {x}^{5} + 2) }^{8} } } - 12 {x}^{3} \sqrt[9]{4 {x}^{5} + 2} }{9 {x}^{8} } = \\ = \frac{20}{27 \sqrt[9]{ {(4 {x}^{5} + 2)}^{8} } } - \frac{4 \sqrt[9]{4 {x}^{5} + 2 } }{3 {x}^{5} }$