📝 题目
8．设 $z=f\left(\mathrm{e}^{x y}, x^{2}-y^{2}\right)$ ，其中 $f(\xi, \eta)$ 有连续的二阶偏导数，求 $\frac{\partial z}{\partial y} 、 \frac{\partial^{2} z}{\partial y^{2}}$ ．
💡 答案与解析
[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：★★★☆☆