Question

Производные

Дана функция z=√x/y

Показать что:

Answer 1

Ответ:

$z = \sqrt{ \frac{x}{y} }$

$\frac{dz}{dx} = \frac{1}{2} \times {( \frac{x}{y}) }^{ - \frac{1}{2} } \times \frac{1}{y} = \\ = \frac{1}{2y} \times \sqrt{ \frac{y}{x} } = \frac{1}{2 \sqrt{xy} }$

$\frac{dz} {dy} = \frac{ \sqrt{y} }{2 \sqrt{x} } \times ( - x {y}^{ - 2} ) = \\ = \frac{ \sqrt{y} }{2 \sqrt{x} } \times ( - \frac{x}{ {y}^{2} } ) = - \frac{ \sqrt{x} }{2y \sqrt{y} } = - \frac{ \sqrt{x} }{2 (\sqrt {y )}^{3} }$

$\frac{ {d}^{2}z }{dx ^{2} } = \frac{1}{2} \times ( - \frac{1}{2} ) {(xy)}^{ - \frac{3}{2} } \times y = \\ = - \frac{y}{4xy \sqrt{xy} } = - \frac{1}{4x \sqrt{xy} }$

${x}^{2} \times \frac{ {d}^{2} z}{dx ^{2} } - \frac{d}{dy} \times ( {y}^{2} \times \frac{dz}{dy} ) \\ {x}^{2} \times ( - \frac{1}{4x \sqrt{xy} } ) - \frac{d}{dy} ( {y}^{2} \times ( - \frac{ \sqrt{x} }{2 \sqrt{ {y}^{3} } } )) = \\ = - \frac{ \sqrt{x} }{4 \sqrt{y} } - \frac{d}{dy} ( - \frac{1}{2} \sqrt{xy} ) = \\ = - \frac{ \sqrt{x} }{4 \sqrt{y} } + \frac{1}{2} \times \frac{1}{2} \frac{1}{ \sqrt{xy} } \times x = \\ = - \frac{ \sqrt{x} }{4 \sqrt{y} } + \frac{ \sqrt{x} }{4 \sqrt{y} } = 0$