Question

Найти производные функций
Помогите пожалуйста!!

Answer 1

Ответ:

a)

$y' = 4 {(2 {x}^{3} - 3x)}^{3} \times 6 {x}^{2} = 24 {x}^{2} {(2 {x}^{2} } - 3x)^{3}$

б)

$y = arctg \sqrt{ {x}^{5} }$

$y '= \frac{1}{1 + {x}^{5} } \times \frac{5}{2} {x}^{ \frac{3}{2} } = \\ = \frac{5x \sqrt{x} }{2(1 + {x}^{5} )}$

в)

$y = {e}^{x} ( {x}^{3} + 7 {x}^{2} )$

$y' = {e}^{x} ( {x}^{3} + 7 {x}^{2} ) + {e}^{x} (3 {x}^{2} + 14x) = \\ = {e}^{x} ( {x}^{3} + 10 {x}^{2} + 14x)$

г)

$y = \frac{ {x}^{3} }{tg( 6x - 1)}$

$y' = \frac{3 {x}^{2} tg(6x - 1) - \frac{1}{ { \cos}^{2} (6x - 1)} \times 6 \times {x}^{3} }{ {tg}^{2}(6x - 1) } = \\ = \frac{3 {x}^{2} }{tg(6x - 1)} - \frac{6 {x}^{3} }{ { \cos}^{2}(6x - 1) {tg}^{2} (6x - 1) } = \\ = \frac{3 {x}^{2} }{tg(6x - 1)} - \frac{6 {x}^{3} }{ { \sin }^{2}(6x - 1) }$

д)

$y = 11 \sqrt[11]{2 {x}^{4} - \frac{1}{x} }$

$y' = 11 \times \frac{1}{11} {(2 {x}^{4} - \frac{1}{x}) }^{ - \frac{10}{11} } \times (8 {x}^{3} + {x}^{ - 2} ) = \\ = \frac{8 {x}^{3} + \frac{1}{ {x}^{2} } }{ \sqrt[11]{ {(2 {x}^{4} - \frac{1}{x} )}^{10} } } = \frac{8 {x}^{5} + 1}{ {x}^{2} \sqrt[11]{ {(2 {x}^{4} + \frac{1}{x} )}^{10} } }$