Question

помогите на решение задачу срошно надо ???)​

Answer 1

Ответ:

2

$z = x ln( \frac{y}{x} )$

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

$z'_y = x \times \frac{x}{y} \times \frac{1}{x} = \frac{x}{y}$

Подставим в равенство:

$x \times ( ln( \frac{y}{x} ) - 1) + y \times \frac{x}{y} = \\ = x ln( \frac{y}{x} ) - x + x = x ln( \frac{y}{x} ) = z$

равенство выполняется

3

$z = \frac{ {x}^{ - 3} {y}^{6} }{ 3} - {x}^{2} + {y}^{2} + 4x - 4y$

$dz = z'_x \times dx + z'_y \times dy$

${dz}^{2} = z''_{xx}dx {}^{2} + z''_{xy}dxdy + z''_{yy} \times dy {}^{2}$

$z'_x = \frac{1}{3} \times ( - 3) {x}^{ - 4} {y}^{6} - 2x + 4 = \\ = - \frac{ {y}^{6} }{ {x}^{4} } - 2x + 4$

$z'_y = \frac{6 {y}^{5} }{ 3{x}^{3} } + 2y - 4 = \frac{2 {y}^{5} }{ {x}^{3} } + 2y - 4$

$dz = ( - \frac{ {y}^{6} }{ {x}^{4} } - 2x + 4)dx + ( \frac{2 {y}^{5} }{ {x}^{3} } + 2y - 4) dy$

_____________

$z''_{xx} = - {y}^{6} \times ( - 4 {x}^{ - 5} ) - 2 = \frac{4 {y}^{6} }{ {x}^{5} } - 2$

$z''_{yy} = \frac{10 {y}^{4} }{ {x}^{3} } + 2$

$z''_{xy} = - \frac{6 {y}^{5} }{ {x}^{4} }$

$dz {}^{2} = (\frac{4 {y}^{6} }{ {x}^{5} } - 2)dx {}^{2} - \frac{6 {y}^{5} }{ {x}^{4} } dxdy + ( \frac{10 {y}^{4} }{ {x}^{3} } + 2)dy {}^{2}$