3.8. \(\frac{\partial z}{\partial x}\), \(\frac{\partial z}{\partial y}\), если \(xe^{\frac{x+y}{z}} - xyz - 10x + z^2 - 2 = 0\).

Решение:

Для решения этой задачи нам нужно найти частные производные \(\frac{\partial z}{\partial x}\) и \(\frac{\partial z}{\partial y}\) от неявной функции, заданной уравнением \(xe^{\frac{x+y}{z}} - xyz - 10x + z^2 - 2 = 0\).

Продифференцируем уравнение по \(x\), считая \(z\) функцией от \(x\) (то есть \(z=z(x)\)):

\[ \frac{\partial}{\partial x} \left( xe^{\frac{x+y}{z}} - xyz - 10x + z^2 - 2 \right) = \frac{\partial}{\partial x} (0) \]\[ 1 \cdot e^{\frac{x+y}{z}} + x \cdot e^{\frac{x+y}{z}} \cdot \frac{\partial}{\partial x} \left( \frac{x+y}{z} \right) - yz - xy \frac{\partial z}{\partial x} + 2z \frac{\partial z}{\partial x} = 0 \]\[ e^{\frac{x+y}{z}} + x e^{\frac{x+y}{z}} \cdot \frac{1 \cdot z - (x+y) \frac{\partial z}{\partial x}}{z^2} - yz - xy \frac{\partial z}{\partial x} + 2z \frac{\partial z}{\partial x} = 0 \]\[ e^{\frac{x+y}{z}} + \frac{xz e^{\frac{x+y}{z}}}{z^2} - \frac{x(x+y)e^{\frac{x+y}{z}}}{z^2} - yz - xy \frac{\partial z}{\partial x} + 2z \frac{\partial z}{\partial x} = 0 \]\[ e^{\frac{x+y}{z}} \left( 1 + \frac{x}{z} - \frac{x(x+y)}{z^2} \right) - yz = \frac{\partial z}{\partial x} \left( xy - 2z \right) \]\[ \frac{\partial z}{\partial x} = \frac{e^{\frac{x+y}{z}} \left( 1 + \frac{x}{z} - \frac{x^2+xy}{z^2} \right) - yz}{xy - 2z} \]\[ \frac{\partial z}{\partial x} = \frac{e^{\frac{x+y}{z}} \left( \frac{z^2 + xz - x^2 - xy}{z^2} \right) - yz}{xy - 2z} \]\[ \frac{\partial z}{\partial x} = \frac{e^{\frac{x+y}{z}} (z^2 + xz - x^2 - xy) - yz^3}{z^2 (xy - 2z)} \]

Продифференцируем уравнение по \(y\), считая \(z\) функцией от \(y\) (то есть \(z=z(y)\)):

\[ \frac{\partial}{\partial y} \left( xe^{\frac{x+y}{z}} - xyz - 10x + z^2 - 2 \right) = \frac{\partial}{\partial y} (0) \]\[ x e^{\frac{x+y}{z}} \cdot \frac{\partial}{\partial y} \left( \frac{x+y}{z} \right) - xz - xy \frac{\partial z}{\partial y} + 2z \frac{\partial z}{\partial y} = 0 \]\[ x e^{\frac{x+y}{z}} \cdot \frac{1 \cdot z - (x+y) \frac{\partial z}{\partial y}}{z^2} - xz - xy \frac{\partial z}{\partial y} + 2z \frac{\partial z}{\partial y} = 0 \]\[ \frac{xz e^{\frac{x+y}{z}}}{z^2} - \frac{x(x+y)e^{\frac{x+y}{z}}}{z^2} - xz - xy \frac{\partial z}{\partial y} + 2z \frac{\partial z}{\partial y} = 0 \]\[ -xz + \frac{\partial z}{\partial y} (2z - xy) = \frac{x(x+y)e^{\frac{x+y}{z}}}{z^2} - \frac{xz e^{\frac{x+y}{z}}}{z^2} \]\[ \frac{\partial z}{\partial y} (2z - xy) = xz + \frac{x(x+y)e^{\frac{x+y}{z}} - xz e^{\frac{x+y}{z}}}{z^2} \]\[ \frac{\partial z}{\partial y} = \frac{xz + \frac{x^2 e^{\frac{x+y}{z}}}{z^2} + \frac{xy e^{\frac{x+y}{z}}}{z^2} - \frac{xz e^{\frac{x+y}{z}}}{z^2}}{2z - xy} \]\[ \frac{\partial z}{\partial y} = \frac{x z^3 + x^2 e^{\frac{x+y}{z}} + xy e^{\frac{x+y}{z}} - xz e^{\frac{x+y}{z}}}{z^2 (2z - xy)} \]

Ответ:

• \(\frac{\partial z}{\partial x} = \frac{e^{\frac{x+y}{z}} (z^2 + xz - x^2 - xy) - yz^3}{z^2 (xy - 2z)}\)
• \(\frac{\partial z}{\partial y} = \frac{xz^3 + x^2 e^{\frac{x+y}{z}} + xy e^{\frac{x+y}{z}} - xz e^{\frac{x+y}{z}}}{z^2 (2z - xy)}\)