题目
(3)已知函数 =f(xy,(e)^x+y) ,且f(x,y)具有二阶连续偏导数.则-|||-dfrac ({partial )^2z}(partial xpartial y)= __
题目解答
答案
解析
步骤 1：定义中间变量
设 $u=xy$ 和 $v=e^{x+y}$，则 $z=f(u,v)$。
步骤 2：计算一阶偏导数
根据链式法则，计算 $z$ 对 $x$ 和 $y$ 的一阶偏导数。
$$
\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial f}{\partial u} y + \frac{\partial f}{\partial v} e^{x+y}
$$
$$
\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial f}{\partial u} x + \frac{\partial f}{\partial v} e^{x+y}
$$
步骤 3：计算二阶混合偏导数
根据链式法则，计算 $z$ 对 $x$ 和 $y$ 的二阶混合偏导数。
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{\partial z}{\partial y} \right) = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial u} x + \frac{\partial f}{\partial v} e^{x+y} \right)
$$
$$
= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial u} x \right) + \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial v} e^{x+y} \right)
$$
$$
= \frac{\partial^2 f}{\partial u^2} y x + \frac{\partial^2 f}{\partial u \partial v} e^{x+y} x + \frac{\partial^2 f}{\partial v \partial u} e^{x+y} + \frac{\partial^2 f}{\partial v^2} e^{2(x+y)}
$$
由于混合偏导数相等（克莱罗定理），即 $\frac{\partial^2 f}{\partial u \partial v} = \frac{\partial^2 f}{\partial v \partial u}$，因此最终表达式简化为：
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 f}{\partial u^2} y x + 2 \frac{\partial^2 f}{\partial u \partial v} e^{x+y} x + \frac{\partial^2 f}{\partial v^2} e^{2(x+y)}
$$
设 $u=xy$ 和 $v=e^{x+y}$，则 $z=f(u,v)$。
步骤 2：计算一阶偏导数
根据链式法则，计算 $z$ 对 $x$ 和 $y$ 的一阶偏导数。
$$
\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial f}{\partial u} y + \frac{\partial f}{\partial v} e^{x+y}
$$
$$
\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial f}{\partial u} x + \frac{\partial f}{\partial v} e^{x+y}
$$
步骤 3：计算二阶混合偏导数
根据链式法则，计算 $z$ 对 $x$ 和 $y$ 的二阶混合偏导数。
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{\partial z}{\partial y} \right) = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial u} x + \frac{\partial f}{\partial v} e^{x+y} \right)
$$
$$
= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial u} x \right) + \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial v} e^{x+y} \right)
$$
$$
= \frac{\partial^2 f}{\partial u^2} y x + \frac{\partial^2 f}{\partial u \partial v} e^{x+y} x + \frac{\partial^2 f}{\partial v \partial u} e^{x+y} + \frac{\partial^2 f}{\partial v^2} e^{2(x+y)}
$$
由于混合偏导数相等（克莱罗定理），即 $\frac{\partial^2 f}{\partial u \partial v} = \frac{\partial^2 f}{\partial v \partial u}$，因此最终表达式简化为：
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 f}{\partial u^2} y x + 2 \frac{\partial^2 f}{\partial u \partial v} e^{x+y} x + \frac{\partial^2 f}{\partial v^2} e^{2(x+y)}
$$