Предмет: Математика, автор: Zhibektoleyzhan

помогите пожалуйста решить производные.

Приложения:

Ответы

Автор ответа: Miroslava227

Ответ:

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

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

$y' = 4 {( {x}^{5} + 1) }^{3} \times ( {x}^{5} + 1) '= \\ = 4 {( {x}^{5} + 1)}^{3} \times 5 {x}^{4} = 20 {x}^{4} {( {x}^{5} + 1) }^{2}$

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

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

$y' = \cos(3 {x}^{2} + 5 ) \times (3 {x}^{2} + 5) '= 6x \cos(3 {x}^{2} + 5) \\$

$y '= 3 \sin {}^{2} (x) \times (\sin(x)) ' = 3 \sin {}^{2} (x) \cos {}^{2} (x) \\$

$y' = 4 {tg}^{3} x \times (tgx)' = \frac{4 {tg}^{3} x}{ \cos {}^{2} (x) } \\$

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

$y' = \frac{1}{2} {(ctgx)}^{ - \frac{1}{2} } \times (ctgx)' = \\ = - \frac{1}{2 \sin {}^{2} (x) \sqrt{ctgx} }$

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

$y' = 2tg(3 {x}^{2} - 1) \times (tg(3 {x}^{2} - 1))' \times (3 {x}^{2} - 1) '= \\ = 2tg(3 {x}^{2} - 1) \times \frac{1}{ \cos {}^{2} (3 {x}^{2} - 1) } \times 6x = \\ = \frac{12xtg(3 {x}^{2} - 1) }{ \cos {}^{2} (3 {x}^{2} - 1 ) }$