Question

Требуется найти производные функций:

Answer 1

Ответ:

$a)y' = 6x \cos(5x) - 5 \sin(5x) \times 3 {x}^{2} + ln(2) \times {2}^{ - 8x} \times ( - 8) = 6x \cos(5x) - 15 {x}^{2} \sin(5x) - 8 ln(2) \times {2}^{ - 8x}$

$b)y' = - 2 {x}^{ - 2} + ln(6) \times {6}^{ \frac{1 - 3x}{2x + 7} } \times ( \frac{ - 3(2x + 7) - 2(1 - 3x)}{ {(2x + 7)}^{2} } ) = - \frac{2}{ {x}^{2} } + ln(6) \times {6}^{ \frac{1 - 3x}{2x + 7} } \times \frac{ - 6x - 21 - 2 + 6x}{ {(2x + 7)}^{2} } = - \frac{2}{ {x}^{2} } + ln(6) \times {6}^{ \frac{1 - 3x}{2x + 7} } \times ( - \frac{23}{ {(2x + 7)}^{2} } )$

$c)y' = \frac{1}{tg(8x)} \times \frac{1}{ { \cos( 8x ) }^{2} } \times 8 = \frac{ \cos(8x) }{ \sin(8x) } \times \frac{1}{ { \cos(8x) }^{2} } \times 8 = \frac{8}{ \sin(8x) \cos(8x) }$

$d)y' = 3 {arctg}^{2} (5 {x}^{4} + 2) \times \frac{1}{1 + {(5 {x}^{4} + 2)}^{2} } \times 20 {x}^{3} = 3 {arctg}^{2} (5 {x}^{4} + 2) \times \frac{20 {x}^{3} }{1 + 25 {x}^{8} + 20 {x}^{4} + 4 } = 3 {arctg}^{2} (5 {x}^{4} + 2) \times \frac{20 {x}^{3} }{ 25 {x}^{8} + 20 {x}^{4} + 5 } =$