二元二重复合函数求导法则（B012）

问题

设函数

z

=

f (u, v)

u

=

φ (x, y)

v

=

ψ (x, y)

, 则

\frac{\partial z}{\partial x}

=

?

\frac{\partial z}{\partial y}

=

?

选项

[A].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{d z}{d u} \cdot \frac{d u}{d x} + \frac{d z}{d v} \cdot \frac{d v}{d x}, \\ \frac{\partial z}{\partial y} = \frac{d z}{d u} \cdot \frac{d u}{d y} + \frac{d z}{d v} \cdot \frac{d v}{d y} . \end{cases}

[B].

{\begin{cases} \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{cases}

[C].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{d u}{d x} + \frac{\partial z}{\partial v} \cdot \frac{d v}{d x}, \\ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{d u}{d y} + \frac{\partial z}{\partial v} \cdot \frac{d v}{d y} . \end{cases}

[D].

{\begin{cases} \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{cases}

答案

${\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{cases}$

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