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法政大学 理工学研究科 システム理工学専攻 経営システム系 2025年8月実施 微积分

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Author​

思齐塾, 祭音Myyura

Description​

f(x,y)=log⁡(x2+y2)f(x,y) = \log(x^2 + y^2) ( x,y>0x, y > 0 ) とする.

(1) 偏導関数 ∂f∂x,∂2f∂x∂y\frac{\partial f}{\partial x}, \frac{\partial^2 f}{\partial x \partial y} を求めよ.

(2) f(x,y)f(x,y) は ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 を満たすことを示せ.

题目描述​

设

f(x,y)=log⁡(x2+y2)(x,y>0).f(x,y)=\log(x^2+y^2)\qquad(x,y>0).

（1）求偏导数

∂f∂x,∂2f∂x∂y.\frac{\partial f}{\partial x}, \qquad \frac{\partial^2f}{\partial x\partial y}.

（2）证明 f(x,y)f(x,y) 满足

∂2f∂x2+∂2f∂y2=0.\frac{\partial^2f}{\partial x^2} +\frac{\partial^2f}{\partial y^2}=0.

Kai​

(1) ∂f∂x=1x2+y2⋅2x=2xx2+y2\frac{\partial f}{\partial x} = \frac{1}{x^2 + y^2} \cdot 2x = \frac{2x}{x^2 + y^2}

∂f∂y=1x2+y2⋅2y=2yx2+y2\frac{\partial f}{\partial y} = \frac{1}{x^2 + y^2} \cdot 2y = \frac{2y}{x^2 + y^2}

∂2f∂x∂y=∂∂x(2yx2+y2)=2y⋅−1(x2+y2)2⋅2x=−4xy(x2+y2)2\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{2y}{x^2 + y^2}) = 2y \cdot \frac{-1}{(x^2 + y^2)^2} \cdot 2x = -\frac{4xy}{(x^2 + y^2)^2}

(2) ∂2f∂x2=∂∂x(2xx2+y2)=2⋅(x2+y2)−x(2x)(x2+y2)2=2y2−x2(x2+y2)2\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (\frac{2x}{x^2 + y^2}) = 2 \cdot \frac{(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2} = 2\frac{y^2 - x^2}{(x^2 + y^2)^2}

∂2f∂y2=∂∂y(2yx2+y2)=2⋅(x2+y2)−y(2y)(x2+y2)2=2x2−y2(x2+y2)2\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (\frac{2y}{x^2 + y^2}) = 2 \cdot \frac{(x^2 + y^2) - y(2y)}{(x^2 + y^2)^2} = 2\frac{x^2 - y^2}{(x^2 + y^2)^2}

∂2f∂x2+∂2f∂y2=2y2−x2(x2+y2)2+2x2−y2(x2+y2)2=2y2−x2+x2−y2(x2+y2)2=0\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 2\frac{y^2 - x^2}{(x^2 + y^2)^2} + 2\frac{x^2 - y^2}{(x^2 + y^2)^2} = 2\frac{y^2 - x^2 + x^2 - y^2}{(x^2 + y^2)^2} = 0

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