Question

Сделайте пожалуйста, заранее спасибо

Answer 1

Ответ:

1.

$dz = z'_xdx + z'_ydy$

a)

$z = (x + 2y) {y}^{2} = x {y}^{2} + 2 {y}^{3}$

$z'_x = {y}^{2}$

$z'_y = 2xy + 6 {y}^{2}$

$dz = {y}^{2} dx + (2xy + 6 {y}^{2}) dy$

б)

$z = \cos(2x + {e}^{y} )$

$z'_x = - \sin(2x + {e}^{y} ) \times 2 = - 2 \sin(2x + {e}^{y} )$

$z'_y = - \sin(2x + {e}^{y} ) \times {e}^{y}$

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

2.

а)

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

$z'_x = 3 {x}^{2} + 2 {y}^{2} \\ z'_y = 4xy + 9 {y}^{2}$

$z''_{xx } = 6x\\ z''_{yy} = 4x + 18y \\ z''_{xy} = 4y$

б)

$z = x ln(y) + \sqrt{ \sin(x) }$

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

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

$z''_{xx} = \frac{1}{2} \times \frac{( \cos(x)) \times \sqrt{ \sin(x) } - ( \sin(x) {}^{ \frac{1}{2} } ) \times \cos(x) }{ \sin(x) } = \\ = \frac{1}{2} \times \frac{ - \sin(x) \times \sqrt{ \sin(x) } - \frac{1}{2 \sqrt{ \sin(x) } } \times \cos(x) \times \cos(x) }{ \sin(x) } = \\ = \frac{1}{2 \sin(x) } \times ( - \sqrt{ \sin {}^{3} (x) } - \frac{ \cos {}^{2} (x) }{2 \sqrt{ \sin( x )} } ) = \\ = - \frac{ \sqrt{ \sin(x) } }{2} - \frac{ \cos {}^{2} (x) }{4 \sqrt{ \sin {}^{3} (x) } }$

$z''_{yy} = - x {y}^{ - 2} = - \frac{x}{ {y}^{2} }$

$z''_{xy }= \frac{1}{y}$