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早稲田大学 創造理工学研究科 経営システム工学専攻 2014年8月実施 微积分

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Author​

思齐塾, 祭音Myyura

Description​

関数(function) f(x,y)=cos⁡yxf(x,y) = \cos{\frac{y}{x}} に対して、次式の関係が成り立つことを示せ。

x∂f(x,y)∂x+y∂f(x,y)∂y=x2∂2f(x,y)(∂x)2+2xy∂2f(x,y)∂x∂y+y2∂2f(x,y)(∂y)2x\frac{\partial f(x,y)}{\partial x} + y\frac{\partial f(x,y)}{\partial y} = x^2\frac{\partial^2 f(x,y)}{(\partial x)^2} + 2xy\frac{\partial^2 f(x,y)}{\partial x\partial y} + y^2\frac{\partial^2 f(x,y)}{(\partial y)^2}

题目描述​

对于函数

f(x,y)=cos⁡yx(x≠0),f(x,y)=\cos\frac{y}{x}\qquad(x\neq0),

证明

x∂f(x,y)∂x+y∂f(x,y)∂y=x2∂2f(x,y)(∂x)2+2xy∂2f(x,y)∂x∂y+y2∂2f(x,y)(∂y)2.x\frac{\partial f(x,y)}{\partial x} +y\frac{\partial f(x,y)}{\partial y} =x^2\frac{\partial^2f(x,y)}{(\partial x)^2} +2xy\frac{\partial^2f(x,y)}{\partial x\partial y} +y^2\frac{\partial^2f(x,y)}{(\partial y)^2}.

Kai​

Let's verify the given relation.

First, compute the first-order partial derivatives:

∂f∂x=∂∂xcos⁡(yx)=−sin⁡(yx)⋅(−yx2)=yx2sin⁡(yx)\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \cos(\frac{y}{x}) = -\sin(\frac{y}{x}) \cdot (-\frac{y}{x^2}) = \frac{y}{x^2}\sin(\frac{y}{x})

∂f∂y=∂∂ycos⁡(yx)=−sin⁡(yx)⋅(1x)=−1xsin⁡(yx)\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \cos(\frac{y}{x}) = -\sin(\frac{y}{x}) \cdot (\frac{1}{x}) = -\frac{1}{x}\sin(\frac{y}{x})

Next, compute the second-order partial derivatives:

∂2f∂x2=∂∂x(yx2sin⁡(yx))=y[−2x3sin⁡(yx)+1x2cos⁡(yx)(−yx2)]=−2yx3sin⁡(yx)−y2x4cos⁡(yx)\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (\frac{y}{x^2}\sin(\frac{y}{x})) = y[\frac{-2}{x^3}\sin(\frac{y}{x}) + \frac{1}{x^2}\cos(\frac{y}{x})(-\frac{y}{x^2})] = \frac{-2y}{x^3}\sin(\frac{y}{x}) - \frac{y^2}{x^4}\cos(\frac{y}{x})

∂2f∂x∂y=∂∂x(−1xsin⁡(yx))=−[−1x2sin⁡(yx)+1xcos⁡(yx)(−yx2)]=1x2sin⁡(yx)+yx3cos⁡(yx)\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial}{\partial x} (-\frac{1}{x}\sin(\frac{y}{x})) = -[-\frac{1}{x^2}\sin(\frac{y}{x}) + \frac{1}{x}\cos(\frac{y}{x})(-\frac{y}{x^2})] = \frac{1}{x^2}\sin(\frac{y}{x}) + \frac{y}{x^3}\cos(\frac{y}{x})

∂2f∂y2=∂∂y(−1xsin⁡(yx))=−1xcos⁡(yx)⋅1x=−1x2cos⁡(yx)\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (-\frac{1}{x}\sin(\frac{y}{x})) = -\frac{1}{x}\cos(\frac{y}{x})\cdot \frac{1}{x} = -\frac{1}{x^2}\cos(\frac{y}{x})

Now, substitute these into the given equation:

x∂f∂x+y∂f∂y=x(yx2sin⁡(yx))+y(−1xsin⁡(yx))=yxsin⁡(yx)−yxsin⁡(yx)=0x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = x(\frac{y}{x^2}\sin(\frac{y}{x})) + y(-\frac{1}{x}\sin(\frac{y}{x})) = \frac{y}{x}\sin(\frac{y}{x}) - \frac{y}{x}\sin(\frac{y}{x}) = 0

x2∂2f∂x2+2xy∂2f∂x∂y+y2∂2f∂y2=x2(−2yx3sin⁡(yx)−y2x4cos⁡(yx))+2xy(1x2sin⁡(yx)+yx3cos⁡(yx))+y2(−1x2cos⁡(yx))x^2\frac{\partial^2 f}{\partial x^2} + 2xy\frac{\partial^2 f}{\partial x\partial y} + y^2\frac{\partial^2 f}{\partial y^2} = x^2(\frac{-2y}{x^3}\sin(\frac{y}{x}) - \frac{y^2}{x^4}\cos(\frac{y}{x})) + 2xy(\frac{1}{x^2}\sin(\frac{y}{x}) + \frac{y}{x^3}\cos(\frac{y}{x})) + y^2(-\frac{1}{x^2}\cos(\frac{y}{x}))

=−2yxsin⁡(yx)−y2x2cos⁡(yx)+2yxsin⁡(yx)+2y2x2cos⁡(yx)−y2x2cos⁡(yx)=0= \frac{-2y}{x}\sin(\frac{y}{x}) - \frac{y^2}{x^2}\cos(\frac{y}{x}) + \frac{2y}{x}\sin(\frac{y}{x}) + \frac{2y^2}{x^2}\cos(\frac{y}{x}) - \frac{y^2}{x^2}\cos(\frac{y}{x}) = 0

Since both sides are equal to 0, the given relation holds.

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