📝 题目
11．设 $z=f(u, x, y)$ ，而 $u=x \mathrm{e}^{y}$ ，其中函数 $f$ 有二阶连续偏导数，求 $\frac{\partial^{2} z}{\partial x^{2}} 、 \frac{\partial^{2} z}{\partial y^{2}}$ 、 $\frac{\partial^{2} z}{\partial x \partial y}$.
💡 答案与解析
[AI解答]
[AI解答]
已知 $ z = f(u, x, y) $，其中 $ u = x e^y $，且 $ f $ 具有二阶连续偏导数。 记 $ f $ 对第一个变量（即 $ u $）的偏导数为 $ f_1' $，对第二个变量（即 $ x $）的偏导数为 $ f_2' $，对第三个变量（即 $ y $）的偏导数为 $ f_3' $。类似地，二阶偏导数记作 $ f_{11}'' $、$ f_{12}'' $ 等。
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**第一步：求一阶偏导数**
由链式法则： $$ \frac{\partial z}{\partial x} = f_1' \cdot \frac{\partial u}{\partial x} + f_2' \cdot 1 + f_3' \cdot 0 $$ 而 $\displaystyle{\frac{\partial u}{\partial x} = e^y}$，所以 $$ \frac{\partial z}{\partial x} = e^y f_1' + f_2'. $$
再对 $ y $ 求偏导： $$ \frac{\partial z}{\partial y} = f_1' \cdot \frac{\partial u}{\partial y} + f_2' \cdot 0 + f_3' \cdot 1 $$ 而 $\displaystyle{\frac{\partial u}{\partial y} = x e^y}$，所以 $$ \frac{\partial z}{\partial y} = x e^y f_1' + f_3'. $$
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**第二步：求二阶偏导数 $\displaystyle{\frac{\partial^2 z}{\partial x^2}}$**
对 $\displaystyle{\frac{\partial z}{\partial x} = e^y f_1' + f_2'}$ 再对 $ x $ 求偏导： $$ \frac{\partial^2 z}{\partial x^2} = e^y \frac{\partial}{\partial x}(f_1') + \frac{\partial}{\partial x}(f_2'). $$
注意 $ f_1' $ 仍是 $ (u, x, y) $ 的函数，所以 $$ \frac{\partial}{\partial x}(f_1') = f_{11}'' \cdot e^y + f_{12}'' \cdot 1 + f_{13}'' \cdot 0 = e^y f_{11}'' + f_{12}''. $$ 而 $$ \frac{\partial}{\partial x}(f_2') = f_{21}'' \cdot e^y + f_{22}'' \cdot 1 + f_{23}'' \cdot 0 = e^y f_{21}'' + f_{22}''. $$
由于二阶连续偏导数，有 $ f_{12}'' = f_{21}'' $，于是： $$ \frac{\partial^2 z}{\partial x^2} = e^y (e^y f_{11}'' + f_{12}'') + (e^y f_{21}'' + f_{22}'') = e^{2y} f_{11}'' + 2 e^y f_{12}'' + f_{22}''. $$
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**第三步：求二阶偏导数 $\displaystyle{\frac{\partial^2 z}{\partial y^2}}$**
对 $\displaystyle{\frac{\partial z}{\partial y} = x e^y f_1' + f_3'}$ 再对 $ y $ 求偏导： $$ \frac{\partial^2 z}{\partial y^2} = x e^y f_1' + x e^y \frac{\partial}{\partial y}(f_1') + \frac{\partial}{\partial y}(f_3'). $$
其中 $$ \frac{\partial}{\partial y}(f_1') = f_{11}'' \cdot x e^y + f_{12}'' \cdot 0 + f_{13}'' \cdot 1 = x e^y f_{11}'' + f_{13}'', $$ $$ \frac{\partial}{\partial y}(f_3') = f_{31}'' \cdot x e^y + f_{32}'' \cdot 0 + f_{33}'' \cdot 1 = x e^y f_{31}'' + f_{33}''. $$
由连续性，$ f_{13}'' = f_{31}'' $，代入得： $$ \frac{\partial^2 z}{\partial y^2} = x e^y f_1' + x e^y (x e^y f_{11}'' + f_{13}'') + (x e^y f_{13}'' + f_{33}'') $$ $$ = x e^y f_1' + x^2 e^{2y} f_{11}'' + 2 x e^y f_{13}'' + f_{33}''. $$
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**第四步：求混合偏导数 $\displaystyle{\frac{\partial^2 z}{\partial x \partial y}}$**
方法：对 $\displaystyle{\frac{\partial z}{\partial x} = e^y f_1' + f_2'}$ 再对 $ y $ 求偏导： $$ \frac{\partial^2 z}{\partial x \partial y} = e^y f_1' + e^y \frac{\partial}{\partial y}(f_1') + \frac{\partial}{\partial y}(f_2'). $$
其中 $$ \frac{\partial}{\partial y}(f_1') = x e^y f_{11}'' + f_{13}'', $$ $$ \frac{\partial}{\partial y}(f_2') = x e^y f_{21}'' + f_{23}''. $$
由连续性，$ f_{12}'' = f_{21}'' $，代入得： $$ \frac{\partial^2 z}{\partial x \partial y} = e^y f_1' + e^y (x e^y f_{11}'' + f_{13}'') + (x e^y f_{12}'' + f_{23}'') $$ $$ = e^y f_1' + x e^{2y} f_{11}'' + e^y f_{13}'' + x e^y f_{12}'' + f_{23}''. $$
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**最终结果：** $$ \boxed{\frac{\partial^{2} z}{\partial x^{2}} = e^{2y} f_{11}'' + 2 e^{y} f_{12}'' + f_{22}''} $$ $$ \boxed{\frac{\partial^{2} z}{\partial y^{2}} = x e^{y} f_1' + x^{2} e^{2y} f_{11}'' + 2 x e^{y} f_{13}'' + f_{33}''} $$ $$ \boxed{\frac{\partial^{2} z}{\partial x \partial y} = e^{y} f_1' + x e^{2y} f_{11}'' + e^{y} f_{13}'' + x e^{y} f_{12}'' + f_{23}''} $$
难度：★★★☆☆