Алгебра: Найти производную функции :

Найти производную функции :

Ответ:

анлггрн765

14.05.2021 06:11

36.5

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

$y '= (5 {(3x - 2)}^{ - 1} ) '= - 5 {(3x - 2)}^{ - 2} \times (3x - 2)' = \\ = - \frac{5}{ {(3x - 2)}^{2} } \times 3 = - \frac{15}{ {(3x - 2)}^{2} }$

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

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

$y' = \frac{(3 - {x}^{2})'(4 + 2x) - (4 + 2x)'(3 - {x}^{2}) }{ {(4 + 2x)}^{2} } = \\ = \frac{ - 2x(4 + 2 {x}^{} ) - 2(3 - {x}^{2}) }{ {(2x + 4)}^{2} } = \\ = \frac{ - 8x - 4 {x}^{2} - 6 + 2 {x}^{2} }{ {(2x + 4)}^{2} } = \frac{ - 2 {x}^{2} - 8x - 6}{ {(2x + 4)}^{2} } = \\ = - \frac{2( {x}^{2} + 4x + 3)}{4 {(x + 2)}^{2} } = - \frac{ {x}^{2} + 4x + 3 }{2 {(x + 2)}^{2} }$

$y' = \frac{(2x - 5)(x - 7) - 1 \times ( {x}^{2} - 5x) }{ {(x - 7)}^{2} } = \\ = \frac{2 {x}^{2} - 14x - 5x + 35 - {x}^{2} + 5x }{ {(x - 7)}^{2} } = \\ = \frac{ {x}^{2} - 14x + 35}{ {(x - 7)}^{2} }$

36.6

$y' = \frac{3(x - 8) - 1 \times (3x + 5)}{(x - 8) {}^{2} } = \\ = \frac{3x - 24 - 3x - 5}{ {(x - 8)}^{2} } = - \frac{29}{ {(x - 8)}^{2} }$

$y '= (7 {(10x - 3)}^{ - 1} )' = - 7 {(10x - 3)}^{ - 2} \times (10x - 3)' = \\ = - \frac{70}{ {(10x - 3)}^{2} }$

$y '= \frac{4x(1 - 6x) - ( - 6) \times 2 {x}^{2} }{ {(1 - 6x)}^{2} } = \\ = \frac{4x - 24 {x}^{2} + 12 {x}^{2} }{ {(1 - 6x)}^{2} } = - \frac{12 {x}^{2} - 4x }{ {(1 - 6x)}^{2} }$

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

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

$y '= \frac{(2x + 6)(x + 2) - ( {x}^{2} + 6x)}{ {(x + 2)}^{2} } = \\ = \frac{2 {x}^{2} + 4x + 6x + 12 - {x}^{2} - 6x }{ {(x + 2)}^{2} } = \\ = \frac{ {x}^{2} + 4x + 12 }{ {(x + 2)}^{2} }$