陰関数の2次導関数

■問題

次の関係で定義される陰関数 ${\displaystyle y=\varphi \left(x\right)}$ について ${\displaystyle \frac{d^{2}y}{d{x}^2}}$ を求めよ．

${\displaystyle log\sqrt{x^2+y^2}-{tan}^{-1}\frac{y}{x}=0}$

■答

${\displaystyle \frac{2\left(x^2+y^2\right)}{{\left(x-y\right)}^3}}$

■ヒント

まず，${\displaystyle \log \sqrt{x^2+y^2}-{\tan }^{-1}\frac{y}{x}=0}$ より定義した${\displaystyle f\left(x,y\right)=f\left(x,\varphi \left(x\right)\right)=0}$ となる2変数関数 ${\displaystyle f\left(x,y\right)}$を$x$ で微分する．その結果から${\displaystyle \frac{dy}{dx}}$ を求める．

■解説

⇒別解

${\displaystyle f\left(x,y\right)=f\left(x,\varphi \left(x\right)\right)}$${\displaystyle =log\sqrt{x^2+y^2}-{tan}^{-1}\frac{y}{x}}$

とおく．${\displaystyle f\left(x,y\right)}$ を${\displaystyle x=x}$ ，${\displaystyle y=\varphi \left(x\right)}$ とする合成関数と考えて，これを${\displaystyle x}$ で微分すると

${\displaystyle \frac{d}{dx}f\left(x,y\right)}$${\displaystyle =f_x\frac{dx}{dx}+f_y\frac{dy}{dx}}$

合成関数の偏導関数の公式を用いた．

${\displaystyle =f_x+f_y\frac{dy}{dx}}$

${\displaystyle =0}$

よって

${\displaystyle f_y\frac{dy}{dx}}$${\displaystyle =-f_x}$

${\displaystyle \frac{dy}{dx}}$${\displaystyle =-\frac{f_x}{f_y}}$　･･････(1)

ここで

${\displaystyle f_x}$${\displaystyle =\frac{\partial }{\partial x}f\left(x,y\right)}$

${\displaystyle =\frac{\partial }{\partial x}\left(log\sqrt{x^2+y^2}-{tan}^{-1}\frac{y}{x}\right)}$

${\displaystyle =\frac{\partial }{\partial x}\left\{log{\left(x^2+y^2\right)}^{\frac{1}{2}}-{tan}^{-1}y{x}^{-1}\right\}}$

${\displaystyle =\frac{\partial }{\partial x}\left\{\frac{1}{2}log\left(x^2+y^2\right)-{tan}^{-1}y{x}^{-1}\right\}}$

偏導関数の定義より， ${\displaystyle y}$ を定数とみなして ${\displaystyle x}$ で微分する．

${\displaystyle =\frac{x}{x^2+y^2}-\frac{1}{1+{\left(\frac{y}{x}\right)}^2}·\left(-\frac{y}{x^2}\right)}$

${\displaystyle =\frac{x}{x^2+y^2}+\frac{\frac{y}{x^2}}{1+{\left(\frac{y}{x}\right)}^2}}$

${\displaystyle =\frac{x}{x^2+y^2}+\frac{\frac{y}{x^2}}{1+\frac{y^2}{x^2}}}$

${\displaystyle =\frac{x}{x^2+y^2}+\frac{y}{x^2+y^2}}$

${\displaystyle =\frac{x+y}{x^2+y^2}}$　･･････(2)

${\displaystyle f_y}$${\displaystyle =\frac{\partial }{\partial y}f\left(x,y\right)}$

${\displaystyle =\frac{\partial }{\partial x}\left(log\sqrt{x^2+y^2}-{tan}^{-1}\frac{y}{x}\right)}$

${\displaystyle =\frac{\partial }{\partial x}\left\{log{\left(x^2+y^2\right)}^{\frac{1}{2}}-{tan}^{-1}\frac{1}{x}y\right\}}$

${\displaystyle =\frac{\partial }{\partial x}\left\{\frac{1}{2}log\left(x^2+y^2\right)-{tan}^{-1}\frac{1}{x}y\right\}}$

偏導関数の定義より， ${\displaystyle x}$ を定数とみなして ${\displaystyle y}$で微分する．

${\displaystyle =\frac{1}{2}·\frac{1}{x^2+y^2}·2y-\frac{1}{1+{\left(\frac{y}{x}\right)}^2}·\frac{1}{x}}$

${\displaystyle =\frac{y}{x^2+y^2}-\frac{\frac{1}{x}}{1+\frac{y^2}{x^2}}}$

${\displaystyle =\frac{y}{x^2+y^2}-\frac{x}{x^2+y^2}}$

${\displaystyle =\frac{y-x}{x^2+y^2}}$

${\displaystyle =-\frac{x-y}{x^2+y^2}}$　･･････(3)

(1)に(2)，(3)を代入すると

${\displaystyle \frac{dy}{dx}}$${\displaystyle =-\frac{f_x}{f_y}}$${\displaystyle =-\frac{\frac{x+y}{x^2+y^2}}{\left(-\frac{x-y}{x^2+y^2}\right)}}$${\displaystyle =\frac{\frac{x+y}{x^2+y^2}}{\frac{x-y}{x^2+y^2}}}$${\displaystyle =\frac{x+y}{x-y}}$　･･････(4)

となる．

${\displaystyle g\left(x,y\right)=g\left(x,\varphi \left(x\right)\right)=\frac{x+y}{x-y}}$　･･････(5)

とおく．

${\displaystyle \frac{d^{2}y}{d{x}^2}}$${\displaystyle =\frac{d}{dx}\left(\frac{dy}{dx}\right)}$

${\displaystyle =\frac{d}{dx}\left\{g\left(x,y\right)\right\}}$

合成関数の偏導関数の公式を用いる．

${\displaystyle =g_x+g_y\frac{dy}{dx}}$　･･････(6)

ここで

${\displaystyle g_x}$${\displaystyle =\frac{\partial }{\partial x}g\left(x,y\right)}$

${\displaystyle =\frac{\partial }{\partial x}\left(\frac{x+y}{x-y}\right)}$

偏導関数の定義より， ${\displaystyle y}$ を定数とみなして ${\displaystyle x}$ で微分する．
分数関数の微分の公式を用いる．

${\displaystyle =\frac{1·\left(x-y\right)-\left(x+y\right)·1}{{\left(x-y\right)}^2}}$

${\displaystyle =\frac{x-y-x-y}{{\left(x-y\right)}^2}}$

${\displaystyle =-\frac{2y}{{\left(x-y\right)}^2}}$　･･････(7)

${\displaystyle g_y}$${\displaystyle =\frac{\partial }{\partial y}g\left(x,y\right)}$

${\displaystyle =\frac{\partial }{\partial y}\left(\frac{x+y}{x-y}\right)}$

${\displaystyle =\frac{1·\left(x-y\right)-\left(x+y\right)·\left(-1\right)}{{\left(x-y\right)}^2}}$

偏導関数の定義より， ${\displaystyle x}$ を定数とみなして ${\displaystyle y}$ で微分する．
分数関数の微分の公式を用いる．

${\displaystyle =\frac{x-y+x+y}{{\left(x-y\right)}^2}}$

${\displaystyle =\frac{2x}{{\left(x-y\right)}^2}}$　･･････(8)

(6)に(7)，(8)，(4)を代入する．

${\displaystyle \frac{d^{2}y}{d{x}^2}}$${\displaystyle =g_x+g_y\frac{dy}{dx}}$

${\displaystyle =-\frac{2y}{{\left(x-y\right)}^2}+\frac{2x}{{\left(x-y\right)}^2}·\left(\frac{x+y}{x-y}\right)}$

${\displaystyle =\frac{-2y}{{\left(x-y\right)}^2}+\frac{2x\left(x+y\right)}{{\left(x-y\right)}^3}}$

${\displaystyle =\frac{-2y\left(x-y\right)+2x\left(x+y\right)}{{\left(x-y\right)}^3}}$

${\displaystyle =\frac{-2xy+2y^2+2x^2+2xy}{{\left(x-y\right)}^3}}$

${\displaystyle =\frac{2x^2+2y^2}{{\left(x-y\right)}^3}}$

${\displaystyle =\frac{2\left(x^2+y^2\right)}{{\left(x-y\right)}^3}}$

■別解

${\displaystyle \log \sqrt{x^2+y^2}-{\tan }^{-1}\frac{y}{x}=0}$　･･････(9)

(9)の両辺を$x$で微分する．

${\displaystyle \frac{1}{\sqrt{x^2+y^2}}\cdot \frac{d}{dx}\sqrt{x^2+y^2}}$${\displaystyle -\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\frac{d}{dx}\left(\frac{y}{x}\right)=0}$

${\displaystyle \frac{1}{\sqrt{x^2+y^2}}\cdot \frac{1}{2\sqrt{x^2+y^2}}\left(2x+2y\frac{dy}{dx}\right)}$${\displaystyle -\frac{x^2}{x^2+y^2}\left(-\frac{y}{x^2}+\frac{1}{x}\frac{dy}{dx}\right)=0}$　･･････(10)

(10)を${\displaystyle \frac{dy}{dx}}$ について解く．

${\displaystyle \frac{x}{x^2+y^2}+\frac{y}{x^2+y^2}\frac{dy}{dx}+\frac{y}{x^2+y^2}}$${\displaystyle -\frac{x}{x^2+y^2}\frac{dy}{dx}=0}$

${\displaystyle \frac{x+y}{x^2+y^2}-\frac{x-y}{x^2+y^2}\frac{dy}{dx}=0}$

(9)より${\displaystyle x^2+y^2\ne 0}$である．よって，両辺を${\displaystyle x^2+y^2}$ で割る．

${\displaystyle x+y-\left(x-y\right)\frac{dy}{dx}=0}$　･･････(11)

${\displaystyle \frac{dy}{dx}=\frac{x+y}{x-y}}$　･･････(12)

(11)を${\displaystyle x}$ さらに${\displaystyle x}$ で微分する．

${\displaystyle 1+\frac{dy}{dx}-\left(1-\frac{dy}{dx}\right)\frac{dy}{dx}}$${\displaystyle -\left(x-y\right)\frac{d^{2}y}{d{x}^2}=0}$　･･････(13)

(13)を${\displaystyle \frac{d^{2}y}{d{x}^2}}$ について解く．

${\displaystyle 1+{\left(\frac{dy}{dx}\right)}^2-\left(x-y\right)\frac{d^{2}y}{d{x}^2}=0}$

${\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{1}{x-y}\left\{1+{\left(\frac{dy}{dx}\right)}^2\right\}}$　･･････(14)

(14)に(12)を代入する．

${\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{1}{x-y}\left\{1+{\left(\frac{x+y}{x-y}\right)}^2\right\}}$

${\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{1}{x-y}\cdot \frac{{\left(x-y\right)}^2+{\left(x+y\right)}^2}{{\left(x-y\right)}^2}}$

${\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{2\left(x^2+y^2\right)}{{\left(x-y\right)}^3}}$

　

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最終更新日： 2023年9月15日