题目
设 $z = f(\ln x + \frac{1}{y})$, 其中函数 $f(u)$ 可微，则 $x \frac{\partial z}{\partial x} + y^{2} \frac{\partial z}{\partial y} = ?$
解析
设 $u = \ln x + \frac{1}{y}$, 则：
$$
z = f(u).
$$
于是：
$$
\frac{\partial z}{\partial x} =
$$
$$
\frac{\partial f(u)}{\partial u} \frac{\partial u}{\partial x} =
$$
$$
\frac{\partial f(u)}{\partial u} \frac{1}{x};
$$
$$
\frac{\partial z}{\partial y} =
$$
$$
\frac{\partial f(u)}{\partial u} \frac{\partial u}{\partial y} =
$$
$$
\frac{\partial f(u)}{\partial u} (-\frac{1}{y^{2}}).
$$
于是：
$$
x \frac{\partial z}{\partial x} + y^{2} \frac{\partial z}{\partial y} =
$$
$$
\frac{\partial f(u)}{\partial u} – \frac{\partial f(u)}{\partial u} = 0.
$$
综上可知，正确答案为 $0$.
EOF