Question

Вычислить производную первого порядка.

P.S: Помогите плиз, идёт кр​

Answer 1

Ответ:

$arctg( \frac{y}{x} ) = ln( \sqrt{ {x}^{2} + {y}^{2} } )$

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