题目
已知u=f(xy,x)，求u=f(xy,x)
已知
，求
题目解答
答案
本题考查偏导数的计算
，
解析
步骤 1：计算 $\dfrac {\partial u}{\partial x}$
根据链式法则，$\dfrac {\partial u}{\partial x} = \dfrac {\partial f}{\partial (xy)} \cdot \dfrac {\partial (xy)}{\partial x} + \dfrac {\partial f}{\partial x} \cdot \dfrac {\partial x}{\partial x}$
其中，$\dfrac {\partial (xy)}{\partial x} = y$，$\dfrac {\partial x}{\partial x} = 1$
因此，$\dfrac {\partial u}{\partial x} = y \cdot \dfrac {\partial f}{\partial (xy)} + \dfrac {\partial f}{\partial x}$
步骤 2：计算 $\dfrac {{\partial }^{2}u}{\partial x\partial y}$
首先，对 $\dfrac {\partial u}{\partial x}$ 关于 $y$ 求偏导数
$\dfrac {{\partial }^{2}u}{\partial x\partial y} = \dfrac {\partial }{\partial y} \left( y \cdot \dfrac {\partial f}{\partial (xy)} + \dfrac {\partial f}{\partial x} \right)$
根据乘积法则，$\dfrac {\partial }{\partial y} \left( y \cdot \dfrac {\partial f}{\partial (xy)} \right) = \dfrac {\partial f}{\partial (xy)} + y \cdot \dfrac {\partial }{\partial y} \left( \dfrac {\partial f}{\partial (xy)} \right)$
其中，$\dfrac {\partial }{\partial y} \left( \dfrac {\partial f}{\partial (xy)} \right) = \dfrac {\partial }{\partial (xy)} \left( \dfrac {\partial f}{\partial (xy)} \right) \cdot \dfrac {\partial (xy)}{\partial y} = \dfrac {\partial }{\partial (xy)} \left( \dfrac {\partial f}{\partial (xy)} \right) \cdot x$
因此，$\dfrac {{\partial }^{2}u}{\partial x\partial y} = \dfrac {\partial f}{\partial (xy)} + y \cdot \dfrac {\partial }{\partial (xy)} \left( \dfrac {\partial f}{\partial (xy)} \right) \cdot x + \dfrac {\partial }{\partial y} \left( \dfrac {\partial f}{\partial x} \right)$
根据链式法则，$\dfrac {\partial u}{\partial x} = \dfrac {\partial f}{\partial (xy)} \cdot \dfrac {\partial (xy)}{\partial x} + \dfrac {\partial f}{\partial x} \cdot \dfrac {\partial x}{\partial x}$
其中，$\dfrac {\partial (xy)}{\partial x} = y$，$\dfrac {\partial x}{\partial x} = 1$
因此，$\dfrac {\partial u}{\partial x} = y \cdot \dfrac {\partial f}{\partial (xy)} + \dfrac {\partial f}{\partial x}$
步骤 2：计算 $\dfrac {{\partial }^{2}u}{\partial x\partial y}$
首先，对 $\dfrac {\partial u}{\partial x}$ 关于 $y$ 求偏导数
$\dfrac {{\partial }^{2}u}{\partial x\partial y} = \dfrac {\partial }{\partial y} \left( y \cdot \dfrac {\partial f}{\partial (xy)} + \dfrac {\partial f}{\partial x} \right)$
根据乘积法则，$\dfrac {\partial }{\partial y} \left( y \cdot \dfrac {\partial f}{\partial (xy)} \right) = \dfrac {\partial f}{\partial (xy)} + y \cdot \dfrac {\partial }{\partial y} \left( \dfrac {\partial f}{\partial (xy)} \right)$
其中，$\dfrac {\partial }{\partial y} \left( \dfrac {\partial f}{\partial (xy)} \right) = \dfrac {\partial }{\partial (xy)} \left( \dfrac {\partial f}{\partial (xy)} \right) \cdot \dfrac {\partial (xy)}{\partial y} = \dfrac {\partial }{\partial (xy)} \left( \dfrac {\partial f}{\partial (xy)} \right) \cdot x$
因此，$\dfrac {{\partial }^{2}u}{\partial x\partial y} = \dfrac {\partial f}{\partial (xy)} + y \cdot \dfrac {\partial }{\partial (xy)} \left( \dfrac {\partial f}{\partial (xy)} \right) \cdot x + \dfrac {\partial }{\partial y} \left( \dfrac {\partial f}{\partial x} \right)$