Question

Найти производные данных функций.

Answer 1

Ответ:

а)

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

б)

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

в)

$y' = ln(3) \times {3}^{arctg {x}^{3} } \times \frac{1}{1 + {x}^{6} } \times 3 {x}^{2}$

г)

$y = {(arctgx)}^{x}$

$y' = ( ln(y)) ' \times y$

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

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

д)

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