📝 题目
5．设 $z=\mathrm{e}^{-\left(\frac{1}{x}+\frac{1}{y}\right)}$ ，求证 $x^{2} \frac{\partial z}{\partial x}+y^{2} \frac{\partial z}{\partial y}=2 z$ ．
💡 答案与解析
[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。
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