Question

ПОЖАЛУЙСТА МАТЕМАТИКУ РЕШИТЕ ДАЮ 20 БАЛЛОВ

Answer 1

Ответ:

3)

1.

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

$\frac{dz}{dx} = {e}^{x - y} - \frac{1}{y}$

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

2.

$z = \sin(x) \sin(y) \cos(x - y)$

$\frac{dz}{dx} = \sin(y) \times ( \cos(x) \times \cos(x - y) - \sin(x - y) \times \sin(x)) = \\ = \sin(y) \times \cos(x + x - y)) = \\ = \sin(y) \times \cos(2x - y)$

$\frac{dz}{dy} = \sin(x) \times ( \cos(y) \times \cos(x - y) - \sin(x - y) \times ( - 1) \times \sin(y)) = \\ = \sin(x) ( \cos(x) \cos(x - y) + \sin(x) \sin(x - y)) = \\ = \sin(x) \times \cos(x - x + y) = \\ = \sin(x) \times \cos(y)$

4)

$z = \frac{ {x}^{4} - 8x {y}^{3} }{x - 2y}$

$\frac{dz}{dx} = \frac{(4 {x}^{3} - 8 {y}^{3})(x - 2y) - ( {x}^{4} - 8x {y}^{3}) }{ {(x - 2y)}^{2} } = \\ = \frac{4 {x}^{4} - 8y {x}^{3} - 8x {y}^{3} + 16 {y}^{4} - {x}^{4} + 8x {y}^{3} }{ {(x - 2y)}^{2} } = \\ = \frac{3 {x}^{4} - 8 {x}^{3}y + 16 {y}^{4} }{ {(x - 2y)}^{2} }$

$\frac{ {d}^{2}x }{ {dx}^{2} } = \frac{(12 {x}^{3} - 12 {x}^{2}y) {(x - 2y)}^{2} - 2(x - 2y)(3 {x}^{4} - 8 {x}^{3}y + 16 {y}^{4} )}{ {(x - 2y)}^{4} } = \\ = \frac{(x - 2y)(12 {x}^{4} - 24 {x}^{3} y - 12 {x}^{3} y + 24 {x}^{2} {y}^{2} - 6 {x}^{4} + 16 {x}^{3}y - 36 {y}^{4} ) }{ {(x - 2y)}^{4} } = \\ = \frac{6 {x}^{4} - 20 {x}^{3}y + 24 {x}^{2} {y}^{2} - 36 {y}^{4} }{ {(x - 2y)}^{3} }$

$\frac{ {d}^{3} z}{ {dx}^{2} dy} = \frac{( - 20 {x}^{3} + 48 {x}^{2}y - 144 {y}^{3}) {(x - 2y)}^{3} - 3 {(x - 2y)}^{2} \times ( - 2) \times (6 {x}^{4} -20 {x}^{3}y + 24 {x}^{2} {y}^{2} - 36 {y}^{4} ) }{ {(x - 2y)}^{6} } = \\ = \frac{ {(x - 2y)}^{2}( - 20 {x}^{4} + 40 {x}^{3}y + 48 {x}^{3}y - 96 {x}^{2} {y}^{2} - 144x {y}^{3} + 288 {y}^{4} + 36 {x}^{4} - 120 {x}^{3} y + 144 {x}^{2} {y}^{2} -216 {y}^{4} }{ {(x - 2y)}^{6} } = \\ = \frac{268 {x}^{4} - 48 {x}^{3}y + 48 {x}^{2} {y}^{2} - 144x {y}^{3} + 72 {y}^{4} }{ {(x - 2y)}^{4} }$