Question

найдите частные производные первого и второго порядка(любые три) ​

Answer 1

Ответ:

1.

$z = {x}^{2} + {y}^{2} + xy - 4x - 5y$

$z'_x = 2x + y - 4$

$z'_y = 2y + x - 5$

$z''_{xx} = 2$

$z''_{yy} = 2$

$z''_{xy} = 1$

3.

$z = {x}^{3} {y}^{2} (2 - x - y) = \\ = 2 {x}^{3} {y}^{2} - {x}^{4} {y}^{2} - {x}^{3} {y}^{3}$

$z'_x = 6 {x}^{2} {y}^{2} - 4 {x}^{3} {y}^{2} - 3 {x}^{2} {y}^{3}$

$z'_y = 4 {x}^{3} y - 2 {x}^{4} y - 3 {x}^{3} {y}^{2}$

$z''_{xx} = 12x {y}^{2} - 12 {x}^{2} {y}^{2} - 6x {y}^{3}$

$z''_{yy} = 4 {x}^{3} - 2 {x}^{4} - 6 {x}^{3} y$

$z''_{xy} = 12 {x}^{2} {y}^{} - 8 {x}^{3} y - 9 {x}^{2} {y}^{2}$

5.

$z = 2 {x}^{3} - x {y}^{2} + 5 {x}^{2} + {y}^{2}$

$z'_x = 6 {x}^{2} - {y}^{2} + 10x$

$z'_y = - 2xy + 2y$

$z''_{xx} = 12x + 10$

$z''_{yy} = - 2x + 2$

$z''_{xy} = - 2y$