Question

с математикой. Нужно найти производную. Задание на фото

Answer 1

$arctg( \frac{y}{x} ) - 2 \cos(2x) + 3xy = 0 \\ \frac{1}{1 + \frac{ {y}^{2} }{ {x}^{2} } } \times \frac{y'x - y}{ {x}^{2} } + 4 \sin(2x) + 3y + 3xy' = 0 \\ \frac{1}{1 + \frac{ {y}^{2} }{ {x}^{2} } } \times ( \frac{y'}{x} - \frac{y}{ {x}^{2} } ) + 3xy' = - 3y - 4 \sin(2x) \\ \frac{y'}{x(1 + \frac{ {y}^{2} }{ {x}^{2} }) } - \frac{y}{ {x}^{2} (1 + \frac{ {y}^{2} }{ {x}^{2} } )} + 3xy'= - 3y - 4 \sin(2x) \\ y'( \frac{1}{x(1 + \frac{ {y}^{2} }{ {x}^{2} } )} + 3x) = \frac{y}{ {x}^{2} + {y}^{2} } - 3y - 4 \sin(2x) \\ y' = \frac{1}{3x + \frac{1}{x(1 + \frac{ {y}^{2} }{ {x}^{2} } )} } ( \frac{y}{ {x}^{2} + {y}^{2} } - 3y - 4 \sin(2x)) \\ y'= \frac{1}{3x + \frac{x}{ {x}^{2} + {y}^{2} } } ( \frac{y}{ {x}^{2} + {y}^{2} } - 3y - 4 \sin(2x)) \\ y' = \frac{ {x}^{2} + {y}^{2} }{3x( {x}^{2} + {y}^{2} ) + x} \times \frac{y - 3y( {x}^{2} + {y}^{2} ) - 4( {x}^{2} + {y}^{2}) \sin(2x) }{ {x}^{2} + {y}^{2} } \\ y' = \frac{y - ( {x}^{2} + {y}^{2} )( 3y + 4 \sin(2x)) }{x(3 {x}^{2} + 3 {y}^{2} + 1)}$