设 =f(dfrac (x)(y),dfrac (y)(z)), f可微,则偏导数计算错误的是 () .-|||-(4分)-|||-A dfrac (partial u)(partial z)=-dfrac (y{f)_(2)'}({z)^2}-|||-B 其他选项中有正确的选项-|||-C dfrac (partial u)(partial x)=dfrac ({f)_(1)'}(y)-|||-D dfrac (partial u)(partial y)=-dfrac (x{f)_(1)'}({y)^2}+dfrac (y{f)_(2)'}(z)

步骤 1：计算 $\dfrac {\partial u}{\partial x}$
根据链式法则，$\dfrac {\partial u}{\partial x} = \dfrac {\partial f}{\partial (\dfrac {x}{y})} \cdot \dfrac {\partial (\dfrac {x}{y})}{\partial x} + \dfrac {\partial f}{\partial (\dfrac {y}{z})} \cdot \dfrac {\partial (\dfrac {y}{z})}{\partial x}$
由于 $\dfrac {\partial (\dfrac {x}{y})}{\partial x} = \dfrac {1}{y}$，$\dfrac {\partial (\dfrac {y}{z})}{\partial x} = 0$，所以 $\dfrac {\partial u}{\partial x} = \dfrac {1}{y} \cdot \dfrac {\partial f}{\partial (\dfrac {x}{y})} = \dfrac {1}{y} \cdot f_1'$
步骤 2：计算 $\dfrac {\partial u}{\partial y}$
根据链式法则，$\dfrac {\partial u}{\partial y} = \dfrac {\partial f}{\partial (\dfrac {x}{y})} \cdot \dfrac {\partial (\dfrac {x}{y})}{\partial y} + \dfrac {\partial f}{\partial (\dfrac {y}{z})} \cdot \dfrac {\partial (\dfrac {y}{z})}{\partial y}$
由于 $\dfrac {\partial (\dfrac {x}{y})}{\partial y} = -\dfrac {x}{y^2}$，$\dfrac {\partial (\dfrac {y}{z})}{\partial y} = \dfrac {1}{z}$，所以 $\dfrac {\partial u}{\partial y} = -\dfrac {x}{y^2} \cdot \dfrac {\partial f}{\partial (\dfrac {x}{y})} + \dfrac {1}{z} \cdot \dfrac {\partial f}{\partial (\dfrac {y}{z})} = -\dfrac {x}{y^2} \cdot f_1' + \dfrac {1}{z} \cdot f_2'$
步骤 3：计算 $\dfrac {\partial u}{\partial z}$
根据链式法则，$\dfrac {\partial u}{\partial z} = \dfrac {\partial f}{\partial (\dfrac {x}{y})} \cdot \dfrac {\partial (\dfrac {x}{y})}{\partial z} + \dfrac {\partial f}{\partial (\dfrac {y}{z})} \cdot \dfrac {\partial (\dfrac {y}{z})}{\partial z}$
由于 $\dfrac {\partial (\dfrac {x}{y})}{\partial z} = 0$，$\dfrac {\partial (\dfrac {y}{z})}{\partial z} = -\dfrac {y}{z^2}$，所以 $\dfrac {\partial u}{\partial z} = -\dfrac {y}{z^2} \cdot \dfrac {\partial f}{\partial (\dfrac {y}{z})} = -\dfrac {y}{z^2} \cdot f_2'$