Question

Хотелось бы получить решение понятное.

Answer 1

Ответ:

$z = ln(x + {y}^{10} )$

$z'_x = \frac{1}{x + {y}^{10} } \times (x + {y}^{10} )'_x = \\ = \frac{1}{x + {y}^{10} } \times 1 = \frac{1}{x + {y}^{10} }$

$z'_y = \frac{1}{ x+ {y}^{10} } \times (x + {y}^{10} )'_y = \frac{10 {y}^{9} }{x + {y}^{10} }$

$\\z''_{xx} = (z'_x)'_x = ( \frac{1}{x + {y}^{10} } )'_x = ( {(x + {y}^{10}) }^{ - 1} )'_x = \\ = - {(x + {y}^{10}) }^{ - 2} \times (x + {y}^{10} )'_x = - \frac{1}{ {(x + {y}^{10}) }^{2} }$

$\\z''_{yy} = (z'_y)'_y = \frac{(10 {y}^{9})'(x + {y}^{10} ) - (x + {y}^{10})'_y \times 10 {y}^{9} }{ {(x + {y}^{10} )}^{2} } = \\ = \frac{90 {y}^{8} (x + {y}^{10}) - 10 {y}^{9} \times 10 {y}^{9} }{ {(x + {y}^{10} )}^{2} } = \frac{10 {y}^{8}(9x + 9 {y}^{10} - 10 {y}^{10} ) }{ {(x + {y}^{10} )}^{2} } = \\ = \frac{10 {y}^{8} (9x - {y}^{10}) }{ {(x + {y}^{10} )}^{2} }$

$\\z''_{xy} = z''_{yx} = ( \frac{1}{x + {y}^{10} } )'_y = ( {(x + {y}^{10}) }^{ - 1} )'_y = \\ = - {(x + {y}^{10}) }^{ - 2} \times 10 {y}^{9} = - \frac{10 {y}^{9} }{ {(x + {y}^{10}) }^{2} }$