Question

Для функции z = arctg(x-3y) найти все частные производные второго порядка и убедиться, что Z''xy = Z''yx

Answer 1

Ответ:

$z = arcctg(x - 3y)$

$Z'x = - \frac{1}{1 + {(x - 3y)}^{2} } \times 1 = \\ = - \frac{1}{1 + {x}^{2} - 6xy + 9 {y}^{2} }$

$Z'y = - \frac{1}{1 + {(x - 3y)}^{2} } \times ( - 3) = \\ = \frac{3}{1 + {(x - 3y)}^{2} } = \\ = \frac{3}{1 + {x}^{2} - 6xy + 9 {y}^{2} }$

$Z''xx = - ( - 1) {(1 + {x}^{2} - 6xy + 9 {y}^{2} )}^{ - 2} \times (2x - 6y) = \\ = \frac{2x - 6y}{ {(1 + {x}^{2} - 6xy + 9 {y}^{2}) }^{2} }$

$Z''yy = - 3 {(1 + {x}^{2} - 6xy + 9 {y}^{2} ) }^{ - 2} \times ( - 6x + 18y) = \\ = \frac{ - 3( - 6x + 18)}{ {(1 + {x}^{2} - 6xy + 9 {y}^{2}) }^{2} } = \\ = \frac{18x - 54}{ {(1 + {x}^{2} - 6xy + 9 {y}^{2} )}^{2} }$

$Z''xy = - ( - 1) {(1 + {x}^{2} - 6xy + 9 {y}^{2} )}^{ - 2} \times ( - 6x + 18y) = \\ = \frac{18 y- 6x}{ {(1 + {x}^{2} - 6xy + 9 {y}^{2}) }^{2} }$

$Z''yx = - \frac{3}{ {(1 + {x}^{2} - 6xy + 9 {y}^{2}) }^{2} } \times (2x - 6y) = \\ = \frac{ - 6x + 18y}{ {(1 + {x}^{2} - 6xy + 9 {y}^{2} )}^{2} }$

выполняется равенство

$Z''xy = Z''yx$