二元二重复合函数求导法则(B012)

问题

设函数 $z$ $=$ $f(u, v)$, $u$ $=$ $\varphi(x, y)$, $v$ $=$ $\psi(x, y)$, 则 $\frac{\partial z}{\partial x}$ $=$ $?$, $\frac{\partial z}{\partial y}$ $=$ $?$

选项

[A].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} \cdot \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{array}\right.$

[B].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}+\frac{\partial z}{\partial v} \cdot \frac{\mathrm{d} v}{\mathrm{d} x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\mathrm{d} u}{\mathrm{d} y}+\frac{\partial z}{\partial v} \cdot \frac{\mathrm{d} v}{\mathrm{d} y} . \end{array}\right.$

[C].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{array}\right.$

[D].   $\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\mathrm{d} z}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}+\frac{\mathrm{d} z}{\mathrm{d} v} \cdot \frac{\mathrm{d} v}{\mathrm{d} x}, \\ \frac{\partial z}{\partial y}=\frac{\mathrm{d} z}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} y}+\frac{\mathrm{d} z}{\mathrm{d} v} \cdot \frac{\mathrm{d} v}{\mathrm{d} y} . \end{array}\right.$


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$\left\{\begin{array}{l} \frac{\partial z} {\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{array}\right.$


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