Question

Помогите пожалуйста !!!!!!!​

Answer 1

Ответ:

1.

$z = 2 {x}^{3} {y}^{4} - 3 {x}^{4} + 2y - 6$

$z'_x = 6 {x}^{2} {y}^{4} - 12 {x}^{3} \\ z'_y = 8 {y}^{3} {x}^{3} + 2$

2.

$f(x,y) = \frac{4x - 3y}{2x + y}$

$f'_x = \frac{(4x - 3y)'_x(2x + y) - (2x + y)'_x(4x - 3y)}{ {(2x + y)}^{2} } = \\ = \frac{4(2x + y) - 2(4x - 3y)}{ {(2x + y)}^{2} } = \frac{8x - 4y - 8x + 6y}{ {(2x + y)}^{2} } = \\ = \frac{2y}{ {(2x + y)}^{2} }$

$f'_y = \frac{(4x - 3y)'_y(2x + y) - (2x + y)'_y(4x - 3y)}{ {(2x + y)}^{2} } = \\ = \frac{ - 3(2x + y) - 1 \times (4x - 3y)}{ {(2x + y)}^{2} } = \\ = \frac{ - 6x - 3y - 4x + 3y}{ {(2x + y)}^{2} } = \frac{ - 10x}{ {(2x + y)}^{2} }$

4.

$\psi(x,y) = {e}^{x} (2x - 3y)$

$\psi'_x = ( {e}^{x} )'(2x - 3y) + (2x - 3y)'_x \times {e}^{x} = \\ = {e}^{x} (2x - 3y) + e {}^{x} \times 2 = {e}^{x} (2x - 3y + 2)$

$\psi'_y = {e}^{x}(2x - 3y)'_y = - 3 e {}^{x}$

5.

$\phi(x,y) = ln(6 {x}^{2}y {}^{5} )$

$\phi'_x = \frac{1}{6 {x}^{2} {y}^{5} } \times 12x {y}^{5} = \frac{2}{x}$

$\phi'_y = \frac{1}{6 {x}^{2} {y}^{5} } \times 5 {y}^{4} \times 6 {x}^{2} = \frac{5}{y}$