📝 题目
25．设函数 $z=f\left(x^{2}-y^{2}, x^{y}\right)$ ，其中 $f$ 具有二阶连续偏导数，求 $\frac{\partial^{2} z}{\partial x \partial y}$ ．
💡 答案与解析
[AI解答]
[AI解答]
已知 $ z = f(u, v) $，其中 $$ u = x^2 - y^2, \quad v = x^y. $$ 函数 $ f $ 具有二阶连续偏导数，记 $$ f_1 = \frac{\partial f}{\partial u}, \quad f_2 = \frac{\partial f}{\partial v}, $$ 且 $ f_{11}, f_{12}, f_{21}, f_{22} $ 表示相应的二阶偏导，由连续性知 $ f_{12} = f_{21} $。
**第一步：求一阶偏导 $\frac{\partial z}{\partial x}$** 由链式法则： $$ \frac{\partial z}{\partial x} = f_1 \cdot \frac{\partial u}{\partial x} + f_2 \cdot \frac{\partial v}{\partial x}. $$ 计算： $$ \frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial x} = y x^{y-1}. $$ 因此： $$ \frac{\partial z}{\partial x} = 2x f_1 + y x^{y-1} f_2. $$
**第二步：求混合偏导 $\frac{\partial^2 z}{\partial x \partial y}$** 对 $\frac{\partial z}{\partial x}$ 关于 $y$ 求偏导： $$ \frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} \left( 2x f_1 + y x^{y-1} f_2 \right). $$ 分别对两项求导。
第一项 $2x f_1$ 对 $y$ 求导： $$ \frac{\partial}{\partial y}(2x f_1) = 2x \left( f_{11} \frac{\partial u}{\partial y} + f_{12} \frac{\partial v}{\partial y} \right). $$ 其中： $$ \frac{\partial u}{\partial y} = -2y, \quad \frac{\partial v}{\partial y} = x^y \ln x. $$ 所以： $$ \frac{\partial}{\partial y}(2x f_1) = 2x \left( -2y f_{11} + x^y \ln x \, f_{12} \right) = -4xy f_{11} + 2x^{y+1} \ln x \, f_{12}. $$
第二项 $y x^{y-1} f_2$ 对 $y$ 求导，使用乘积法则： $$ \frac{\partial}{\partial y} \left( y x^{y-1} f_2 \right) = \frac{\partial}{\partial y} \left( y x^{y-1} \right) f_2 + y x^{y-1} \frac{\partial f_2}{\partial y}. $$ 先计算： $$ \frac{\partial}{\partial y} \left( y x^{y-1} \right) = x^{y-1} + y \cdot x^{y-1} \ln x = x^{y-1} (1 + y \ln x). $$ 再计算： $$ \frac{\partial f_2}{\partial y} = f_{21} \frac{\partial u}{\partial y} + f_{22} \frac{\partial v}{\partial y} = -2y f_{21} + x^y \ln x \, f_{22}. $$ 由于 $f_{12}=f_{21}$，所以第二项整体为： $$ x^{y-1} (1 + y \ln x) f_2 + y x^{y-1} \left( -2y f_{12} + x^y \ln x \, f_{22} \right). $$ 整理为： $$ x^{y-1} (1 + y \ln x) f_2 - 2y^2 x^{y-1} f_{12} + y x^{2y-1} \ln x \, f_{22}. $$
**第三步：合并两项结果** 将第一项和第二项相加： $$ \frac{\partial^2 z}{\partial x \partial y} = \left[ -4xy f_{11} + 2x^{y+1} \ln x \, f_{12} \right] + \left[ x^{y-1} (1 + y \ln x) f_2 - 2y^2 x^{y-1} f_{12} + y x^{2y-1} \ln x \, f_{22} \right]. $$ 合并 $f_{12}$ 的系数： $$ 2x^{y+1} \ln x - 2y^2 x^{y-1} = 2x^{y-1} (x^2 \ln x - y^2). $$ 因此最终结果为： $$ \boxed{\frac{\partial^{2} z}{\partial x \partial y} = -4xy f_{11} + 2x^{y-1}(x^2 \ln x - y^2) f_{12} + x^{y-1}(1 + y \ln x) f_2 + y x^{2y-1} \ln x \, f_{22}}. $$
难度：★★★★☆