Математика: Знайдіть похідні наступних функцій:

Знайдіть похідні наступних функцій:

Ответ:

mulanmango

15.02.2021 03:19

$y = (x + \frac{1}{x} + \frac{1}{ {x}^{2} } )( {x}^{2} + x + 1) = \\ = {x}^{3} + {x}^{2} + x + x + 1 + \frac{1}{x} + 1 + \frac{1}{x} + \frac{1}{ {x}^{2} } = \\ = {x}^{3} + {x}^{2} + 2x + 2 + \frac{2}{x} + \frac{1}{ {x}^{2} } = \\ = {x}^{3} + {x}^{2} + 2x + 2 + 2 {x}^{ - 1} + {x}^{ - 2}$

$y' = 3 {x}^{2} + 2x + 2 + 0 - 2 {x}^{ - 2} - 2 {x}^{ - 3} = \\ = 3 {x}^{2} + 2x + 2 - \frac{2}{ {x}^{2} } - \frac{2}{ {x}^{3} }$

$y = ( {x}^{4} - \frac{1}{ {x}^{4} } )( {x}^{3} + \frac{1}{ {x}^{3} } ) = \\ = {x}^{7} + x - \frac{1}{x} - \frac{1}{ {x}^{7} } = \\ = {x}^{7} + x - {x}^{ - 1} - {x}^{ - 7}$

$y' = 7 {x}^{6} + 1 + {x}^{ - 2} + 7 {x}^{ - 8} = \\ = 7 {x}^{6} + 1 + \frac{1}{ {x}^{2} } + \frac{7}{ {x}^{8} }$

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

$y = {x}^{2} \times \sqrt{x} = {x}^{ \frac{5}{2} } \\$

$y' = \frac{5}{2} {x}^{ \frac{3}{2} } = \frac{5}{2} x \sqrt{x} \\$

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

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