Question

Найти полный дифференциал указанной функции​

Answer 1

Ответ:

$dz = Z'x \times dx + Z'y \times dy$

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

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

$dz = - \frac{y}{x \sqrt{2 {x}^{2} - 2xy - {y}^{2} } } dx + \frac{1}{ \sqrt{2 {x}^{2} - 2xy - {y}^{2} } } dy = \\ = \frac{1}{ \sqrt{2 {x}^{2} - 2xy - {y}^{2} } } ( - \frac{y}{x} dx + dy)$