Question

Найти производные функции
а) у=arcsin*√(1-3x)
б) (у/х) =arctg(x/y)
в) y=x(в степени) (-tgx)
Срочно, 15 минут

Answer 1

Ответ:

$y = arcsin \sqrt{1 - 3x}$

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

б)

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

в)

формула:

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

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