📝 题目
7．求下列函数的 $\frac{\partial^{2} z}{\partial x^{2}}, \frac{\partial^{2} z}{\partial y^{2}}$ 和 $\frac{\partial^{2} z}{\partial x \partial y}$ ： （1）$z=x^{4}+y^{4}-4 x^{2} y^{2}$ ； （2）$z=\arctan \frac{y}{x}$ ； （3）$z=y^{x}$ ．
💡 答案与解析
[AI解答]
[AI解答]
**（1）** $ z = x^{4} + y^{4} - 4x^{2}y^{2} $
先求一阶偏导： $$ \frac{\partial z}{\partial x} = 4x^{3} - 8x y^{2}, \quad \frac{\partial z}{\partial y} = 4y^{3} - 8x^{2}y $$
再求二阶偏导： $$ \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial}{\partial x}(4x^{3} - 8x y^{2}) = 12x^{2} - 8y^{2} $$ $$ \frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial}{\partial y}(4y^{3} - 8x^{2}y) = 12y^{2} - 8x^{2} $$ $$ \frac{\partial^{2} z}{\partial x \partial y} = \frac{\partial}{\partial y}(4x^{3} - 8x y^{2}) = -16xy $$
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**（2）** $ z = \arctan \frac{y}{x} $
一阶偏导： $$ \frac{\partial z}{\partial x} = \frac{1}{1+(y/x)^{2}} \cdot \left(-\frac{y}{x^{2}}\right) = -\frac{y}{x^{2}+y^{2}} $$ $$ \frac{\partial z}{\partial y} = \frac{1}{1+(y/x)^{2}} \cdot \frac{1}{x} = \frac{x}{x^{2}+y^{2}} $$
二阶偏导： $$ \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial}{\partial x}\left(-\frac{y}{x^{2}+y^{2}}\right) = -y \cdot \frac{-2x}{(x^{2}+y^{2})^{2}} = \frac{2xy}{(x^{2}+y^{2})^{2}} $$ $$ \frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial}{\partial y}\left(\frac{x}{x^{2}+y^{2}}\right) = x \cdot \frac{-2y}{(x^{2}+y^{2})^{2}} = -\frac{2xy}{(x^{2}+y^{2})^{2}} $$ $$ \frac{\partial^{2} z}{\partial x \partial y} = \frac{\partial}{\partial y}\left(-\frac{y}{x^{2}+y^{2}}\right) = -\frac{(x^{2}+y^{2}) - y(2y)}{(x^{2}+y^{2})^{2}} = -\frac{x^{2} - y^{2}}{(x^{2}+y^{2})^{2}} $$
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**（3）** $ z = y^{x} $
改写为指数形式：$ z = e^{x \ln y} $
一阶偏导： $$ \frac{\partial z}{\partial x} = e^{x\ln y} \cdot \ln y = y^{x} \ln y $$ $$ \frac{\partial z}{\partial y} = e^{x\ln y} \cdot \frac{x}{y} = x y^{x-1} $$
二阶偏导： $$ \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial}{\partial x}(y^{x} \ln y) = y^{x} (\ln y)^{2} $$ $$ \frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial}{\partial y}(x y^{x-1}) = x (x-1) y^{x-2} $$ $$ \frac{\partial^{2} z}{\partial x \partial y} = \frac{\partial}{\partial y}(y^{x} \ln y) = \ln y \cdot x y^{x-1} + y^{x} \cdot \frac{1}{y} = x y^{x-1} \ln y + y^{x-1} $$ 也可以写作： $$ \frac{\partial^{2} z}{\partial x \partial y} = y^{x-1} (x \ln y + 1) $$
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**难度评级**：★★☆☆☆ （主要考察基本偏导计算与链式法则，步骤直接，无复杂技巧）