设二元函数 $z=z(x, y)$ 有二阶连续偏导数, 且满足
$$
6 \frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial x \partial y}-\frac{\partial^2 z}{\partial y^2}=1,
$$
令变量 $\left\{\begin{array}{l}u=x-2 y \\ v=x+3 y\end{array}\right.$, 那么 $\frac{\partial^2 z}{\partial u \partial v}=$
【答案】 $\frac{1}{25}$.

【解析】 由复合函数的链导法则, 有
$$
\begin{aligned}
&\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}=\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}, \\
&\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}=-2 \frac{\partial z}{\partial u}+3 \frac{\partial z}{\partial v},
\end{aligned}
$$
继续求偏导函数, 得
$$
\begin{aligned}
&\frac{\partial^2 z}{\partial x^2}=\frac{\partial^2 z}{\partial u^2}+2 \frac{\partial^2 z}{\partial u \partial v}+\frac{\partial^2 z}{\partial v^2}, \frac{\partial^2 z}{\partial x \partial y}=-2 \frac{\partial^2 z}{\partial u^2}+\frac{\partial^2 z}{\partial u \partial v}+3 \frac{\partial^2 z}{\partial v^2}, \\
&\frac{\partial^2 z}{\partial y^2}=4 \frac{\partial^2 z}{\partial u^2}-12 \frac{\partial^2 z}{\partial u \partial v}+9 \frac{\partial^2 z}{\partial v^2},
\end{aligned}
$$
代人已知等式, 得

所以 $\frac{\partial^2 z}{\partial u \partial v}=\frac{1}{25}$.

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