Question

Для заданной функции многих переменных определить частные производные первого и второго порядка

Answer 1

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

$\frac{dz}{dx} = (x)' \times {e}^{ - \frac{y}{x} } + ( {e}^{ - \frac{y}{x} } )'_x \times ( - \frac{y}{x} )'_x \times x = \\ = {e}^{ - \frac{y}{x} } + {e}^{ - \frac{y}{x} } \times \frac{y}{ {x}^{2} } \times x = \\ = {e}^{ - \frac{y}{x} } (1 + \frac{y}{x} )$

$\frac{dz}{dy} = x \times ( {e}^{ - \frac{y}{x} } )'_y = x {e}^{ - \frac{y}{x} } \times ( - \frac{y}{x} ) = \\ = x {e}^{ - \frac{y}{x} } \times ( - \frac{1}{x} ) = - {e}^{ - \frac{y}{x} }$

$\frac{ {d}^{2}z }{dx {}^{2} } = {e}^{ - \frac{y}{x} } \times \frac{y}{ {x}^{2} } (1 + \frac{y}{x} ) + ( - \frac{y}{ {x}^{2} } ) {e}^{ - \frac{y}{x} } = \\ = {e}^{ - \frac{y}{x} } ( \frac{y}{ {x}^{2} } + \frac{ {y}^{2} }{ {x}^{3} } - \frac{y}{ {x}^{2} } ) = \frac{ {y}^{2} }{ {x}^{3} } {e}^{ - \frac{y}{ x} }$

$\frac{ {d}^{2}z }{dy {}^{2} } = - {e}^{ - \frac{y}{x} } \times ( - \frac{1}{x} ) = \frac{1}{x} {e}^{ - \frac{y}{x} } \\$

$\frac{ {d}^{2} z}{dxdy} = ( - {e}^{ - \frac{y}{x} } )'_x = - {e}^{ - \frac{y}{x} } \times \frac{y}{ {x}^{2} } = \\ = - \frac{y}{ {x}^{2} } {e}^{ - \frac{y}{ x} }$