人邮高数 第6章 第6-2-5题
📝 题目
5．设 $\displaystyle z=\mathrm{e}^{-\left(\frac{1}{x}+\frac{1}{y}\right)}$ ，求证 $\displaystyle x^{2} \frac{\partial z}{\partial x}+y^{2} \frac{\partial z}{\partial y}=2 z$ ．
💡 答案解析
[AI解答]
[AI解答]
已知
$$
z = e^{-\left( \frac{1}{x} + \frac{1}{y} \right)}.
$$
先对 $x$ 求偏导：
$$
\frac{\partial z}{\partial x} = e^{-\left( \frac{1}{x} + \frac{1}{y} \right)} \cdot \frac{\partial}{\partial x} \left( -\frac{1}{x} - \frac{1}{y} \right)
= z \cdot \left( \frac{1}{x^{2}} \right).
$$
因此
$$
x^{2} \frac{\partial z}{\partial x} = x^{2} \cdot z \cdot \frac{1}{x^{2}} = z.
$$
再对 $y$ 求偏导：
$$
\frac{\partial z}{\partial y} = e^{-\left( \frac{1}{x} + \frac{1}{y} \right)} \cdot \frac{\partial}{\partial y} \left( -\frac{1}{x} - \frac{1}{y} \right)
= z \cdot \left( \frac{1}{y^{2}} \right).
$$
因此
$$
y^{2} \frac{\partial z}{\partial y} = y^{2} \cdot z \cdot \frac{1}{y^{2}} = z.
$$
将两式相加得
$$
x^{2} \frac{\partial z}{\partial x} + y^{2} \frac{\partial z}{\partial y} = z + z = 2z.
$$
证毕。
难度：★☆☆☆☆