Question

Буду благодарен если поможете

Answer 1

Ответ:

$z = \frac{(x - 2y) {}^{2} }{x + 2y}$

$z'_x = \frac{( {(x - 2y)}^{2}) '\times (x - 2y)'_x \times (x + 2y) - (x + 2y)'_x \times {(x - 2y)}^{2} }{ {(x + 2y)}^{2} } = \\ = \frac{2(x - 2y) \times 1 \times (x + 2y) - 1 \times {(x - 2y)}^{2} }{ {(x + 2y)}^{2} } = \\ = \frac{(x - 2y)(2(x + 2y) - (x - 2y))}{ {(x + 2y)}^{2} } = \\ = \frac{(x - 2y)(2x + 4y - x + 2y)}{ {(x + 2y)}^{2} } = \frac{(x - 2y)(x + 6y)}{ {(x + 2y)}^{2} }$

$z'_y = \frac{( {(x - 2y)}^{2} )' \times (x - 2y)'_y \times (x + 2y) - (x + 2y)'_y \times {(x - 2y)}^{2} }{ {(x + 2y)}^{2} } = \\ = \frac{2(x - 2y) \times ( - 2) \times (x + 2y) - 2(x - 2y) {}^{2} }{ {(x + 2y)}^{2} } = \\ = \frac{(x - 2y)( - 4(x + 2y) - 2(x - 2y))}{ {(x + 2y)}^{2} } = \\ = \frac{(x - 2y)( - 4x - 8y - 2x + 4y)}{ {(x + 2y)}^{2} } = \frac{(x - 2y)( - 6x - 4y)}{ {(x + 2y)}^{2} } = \\ = - \frac{(x - 2y)(6x + 4y)}{ {(x + 2y)}^{2} }$