Question

Составить полный дифференциал функции.
$z=\frac{x-y}{x+y}$

Answer 1

Ответ:

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

$z = \frac{x - y}{x + y}$

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

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

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