Question

Составить дифференциал второго порядка функции.

Answer 1

Ответ:

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

$z = {e}^{ \frac{x}{ {y}^{2} } }$

$Z'x = {e}^{ \frac{x}{ {y}^{2} } } \times \frac{1}{ {y}^{2} }$

$Z'y = {e}^{ \frac{x}{ {y}^{2} } } \times ( -2 x {y}^{ - 3} ) = \\ = - \frac{2x}{ {y}^{3} } {e}^{ \frac{x}{ {y}^{2} } }$

$Z''xx = {e}^{ \frac{x}{ {y}^{2} } } \times \frac{1}{ {y}^{4} }$

$Z''yy = 6x {y}^{ - 4} {e}^{ \frac{x}{ {y}^{2} } } - \frac{2x}{ {y}^ {3} } \times {e}^{ \frac{x}{ {y}^{2} } } \times ( - \frac{2x}{ {y}^{3} } ) = \\ = {e}^{ \frac{x}{ {y}^{2} } } ( \frac{6x}{ {y}^{4} } + \frac{4 {x}^{2} }{ {y}^{4} } ) = \\ = {e}^{ \frac{x}{ {y}^{2} } } \frac{2x}{ {y}^{4} } (3 + 2x)$

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

${d}^{2} z = {e}^{ \frac{x}{ {y}^{2} } } ( \frac{d {x}^{2} }{ {y}^{4} } - \frac{2}{ {y}^{3} }( \frac{x}{ {y}^{2} } + 1 ) dxdy + \frac{2x}{ {y}^{4} } (2x + 3) {dy}^{2} )$