Question

НАЙТИ ПОЛНЫЕ ДИФФЕРЕНЦИАЛЫ УКАЗАННЫХ ФУНКЦИИ

Answer 1

Ответ:

$dz = Z'xdx + Z'ydy$

$Z'x = \frac{1}{2} {(3 {x}^{2} - 2 {y}^{2} + 5) }^{ - \frac{1}{2} } \times (3 {x}^{2} - 2 {y}^{2} + 5)' = \\ = \frac{1}{2 \sqrt{3 {x}^{2} - 2 {y}^{2} + 5 } } \times 6x = \frac{3x}{ \sqrt{3 {x}^{2} - 2 {y}^{2} + 5 } }$

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

$dz = \frac{3x}{ \sqrt{3 {x}^{2} - 2 {y}^{2} + 5 } } dx - \frac{2y}{ \sqrt{3 {x}^{2} - 2 {y}^{2} + 5} } dy = \\ = \frac{1}{ \sqrt{3 {x}^{2} - 2 {y}^{2} + 5} } (3xdx - 2ydy)$