人邮高数 第6章 第6-3-8题

[AI解答]

[AI解答]

已知 $ z = f(e^{xy}, x^2 - y^2) $，其中 $ f(\xi, \eta) $ 具有连续的二阶偏导数。
记 $\xi = e^{xy}$，$\eta = x^2 - y^2$，则 $ z = f(\xi, \eta) $。

**第一步：求一阶偏导 $\displaystyle{\frac{\partial z}{\partial y}}$**
由链式法则：
$$
\frac{\partial z}{\partial y}
= \frac{\partial f}{\partial \xi} \cdot \frac{\partial \xi}{\partial y}
+ \frac{\partial f}{\partial \eta} \cdot \frac{\partial \eta}{\partial y}
$$
计算：
$$
\frac{\partial \xi}{\partial y} = x e^{xy}, \quad
\frac{\partial \eta}{\partial y} = -2y
$$
记 $ f_\xi = \frac{\partial f}{\partial \xi} $，$ f_\eta = \frac{\partial f}{\partial \eta} $，则
$$
\frac{\partial z}{\partial y} = f_\xi \cdot x e^{xy} + f_\eta \cdot (-2y)
= x e^{xy} f_\xi - 2y f_\eta
$$

**第二步：求二阶偏导 $\displaystyle{\frac{\partial^2 z}{\partial y^2}}$**
对 $\displaystyle{\frac{\partial z}{\partial y}}$ 再对 $y$ 求偏导：
$$
\frac{\partial^2 z}{\partial y^2}
= \frac{\partial}{\partial y} \left( x e^{xy} f_\xi - 2y f_\eta \right)
$$
分别对两项求导：

第一项 $ x e^{xy} f_\xi $：
$$
\frac{\partial}{\partial y}(x e^{xy} f_\xi)
= x \cdot \frac{\partial}{\partial y}(e^{xy}) \cdot f_\xi
+ x e^{xy} \cdot \frac{\partial f_\xi}{\partial y}
$$
其中 $\frac{\partial}{\partial y}(e^{xy}) = x e^{xy}$，而
$$
\frac{\partial f_\xi}{\partial y}
= \frac{\partial f_\xi}{\partial \xi} \cdot \frac{\partial \xi}{\partial y}
+ \frac{\partial f_\xi}{\partial \eta} \cdot \frac{\partial \eta}{\partial y}
= f_{\xi\xi} \cdot x e^{xy} + f_{\xi\eta} \cdot (-2y)
$$
所以第一项为：
$$
x^2 e^{xy} f_\xi + x e^{xy} \left( x e^{xy} f_{\xi\xi} - 2y f_{\xi\eta} \right)
= x^2 e^{xy} f_\xi + x^2 e^{2xy} f_{\xi\xi} - 2xy e^{xy} f_{\xi\eta}
$$

第二项 $ -2y f_\eta $：
$$
\frac{\partial}{\partial y}(-2y f_\eta)
= -2 f_\eta - 2y \cdot \frac{\partial f_\eta}{\partial y}
$$
而
$$
\frac{\partial f_\eta}{\partial y}
= \frac{\partial f_\eta}{\partial \xi} \cdot \frac{\partial \xi}{\partial y}
+ \frac{\partial f_\eta}{\partial \eta} \cdot \frac{\partial \eta}{\partial y}
= f_{\eta\xi} \cdot x e^{xy} + f_{\eta\eta} \cdot (-2y)
$$
由于二阶偏导连续，有 $ f_{\xi\eta} = f_{\eta\xi} $，所以：
$$
\frac{\partial}{\partial y}(-2y f_\eta)
= -2 f_\eta - 2y \left( x e^{xy} f_{\xi\eta} - 2y f_{\eta\eta} \right)
= -2 f_\eta - 2xy e^{xy} f_{\xi\eta} + 4y^2 f_{\eta\eta}
$$

第三步：合并结果
将两项相加：
$$
\frac{\partial^2 z}{\partial y^2}
= \left[ x^2 e^{xy} f_\xi + x^2 e^{2xy} f_{\xi\xi} - 2xy e^{xy} f_{\xi\eta} \right]
+ \left[ -2 f_\eta - 2xy e^{xy} f_{\xi\eta} + 4y^2 f_{\eta\eta} \right]
$$
合并同类项：
$$
\frac{\partial^2 z}{\partial y^2}
= x^2 e^{xy} f_\xi - 2 f_\eta
+ x^2 e^{2xy} f_{\xi\xi}
+ 4y^2 f_{\eta\eta}
- 4xy e^{xy} f_{\xi\eta}
$$

最终结果：
$$
\boxed{\frac{\partial z}{\partial y} = x e^{xy} f_\xi - 2y f_\eta}
$$
$$
\boxed{\frac{\partial^2 z}{\partial y^2}
= x^2 e^{xy} f_\xi - 2 f_\eta
+ x^2 e^{2xy} f_{\xi\xi}
+ 4y^2 f_{\eta\eta}
- 4xy e^{xy} f_{\xi\eta}}
$$

难度：★★★☆☆