陰関数の2次導関数

■問題

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

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

■答

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

■ヒント

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

■解説

⇒別解

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

とおく． $f (x, y)$ $f (x, y)$ を $x = x$ $x = x$ ， $y = ϕ (x)$ $y = ϕ (x)$ とする合成関数と考えて，これを $x$ $x$ で微分すると

$\frac{d}{d x} f (x, y)$ $\frac{d}{d x} f (x, y)$ $= f_{x} \frac{d x}{d x} + f_{y} \frac{d y}{d x}$ $= f_{x} \frac{d x}{d x} + f_{y} \frac{d y}{d x}$

合成関数の偏導関数の公式を用いた．

$= f_{x} + f_{y} \frac{d y}{d x}$ $= f_{x} + f_{y} \frac{d y}{d x}$

$= 0$ $= 0$

よって

$f_{y} \frac{d y}{d x}$ $f_{y} \frac{d y}{d x}$ $= - f_{x}$ $= - f_{x}$

$\frac{d y}{d x}$ $\frac{d y}{d x}$ $= - \frac{f_{x}}{f_{y}}$ $= - \frac{f_{x}}{f_{y}}$ 　･･････(1)

ここで

$f_{x}$ $f_{x}$ $= \frac{\partial}{\partial x} f (x, y)$ $= \frac{\partial}{\partial x} f (x, y)$

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

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

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

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

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

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

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

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

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

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

$f_{y}$ $f_{y}$ $= \frac{\partial}{\partial y} f (x, y)$ $= \frac{\partial}{\partial y} f (x, y)$

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

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

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

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

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

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

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

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

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

(1)に(2)，(3)を代入すると

$\frac{d y}{d x}$ $\frac{d y}{d x}$ $= - \frac{f_{x}}{f_{y}}$ $= - \frac{f_{x}}{f_{y}}$ $= - \frac{\frac{x + y}{x^{2} + y^{2}}}{(- \frac{x - y}{x^{2} + y^{2}})}$ $= - \frac{\frac{x + y}{x^{2} + y^{2}}}{(- \frac{x - y}{x^{2} + y^{2}})}$ $= \frac{\frac{x + y}{x^{2} + y^{2}}}{\frac{x - y}{x^{2} + y^{2}}}$ $= \frac{\frac{x + y}{x^{2} + y^{2}}}{\frac{x - y}{x^{2} + y^{2}}}$ $= \frac{x + y}{x - y}$ $= \frac{x + y}{x - y}$ 　･･････(4)

となる．

$g (x, y) = g (x, ϕ (x)) = \frac{x + y}{x - y}$ $g (x, y) = g (x, ϕ (x)) = \frac{x + y}{x - y}$ 　･･････(5)

とおく．

$\frac{d^{2} y}{d x^{2}}$ $\frac{d^{2} y}{d x^{2}}$ $= \frac{d}{d x} (\frac{d y}{d x})$ $= \frac{d}{d x} (\frac{d y}{d x})$

$= \frac{d}{d x} {g (x, y)}$ $= \frac{d}{d x} {g (x, y)}$

合成関数の偏導関数の公式を用いる．

$= g_{x} + g_{y} \frac{d y}{d x}$ $= g_{x} + g_{y} \frac{d y}{d x}$ 　･･････(6)

ここで

$g_{x}$ $g_{x}$ $= \frac{\partial}{\partial x} g (x, y)$ $= \frac{\partial}{\partial x} g (x, y)$

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

$= \frac{\frac{\partial}{\partial x} (x + y) \cdot (x - y) - (x + y) \cdot \frac{\partial}{\partial x} (x - y)}{{(x - y)}^{2}}$ $= \frac{\frac{\partial}{\partial x} (x + y) \cdot (x - y) - (x + y) \cdot \frac{\partial}{\partial x} (x - y)}{{(x - y)}^{2}}$

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

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

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

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

$g_{y}$ $g_{y}$ $= \frac{\partial}{\partial y} g (x, y)$ $= \frac{\partial}{\partial y} g (x, y)$

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

$= \frac{\frac{\partial}{\partial y} (x + y) \cdot (x - y) - (x + y) \cdot \frac{\partial}{\partial y} (x - y)}{{(x - y)}^{2}}$ $= \frac{\frac{\partial}{\partial y} (x + y) \cdot (x - y) - (x + y) \cdot \frac{\partial}{\partial y} (x - y)}{{(x - y)}^{2}}$

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

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

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

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

(6)に(7)，(8)，(4)を代入する．

$\frac{d^{2} y}{d x^{2}}$ $\frac{d^{2} y}{d x^{2}}$ $= g_{x} + g_{y} \frac{d y}{d x}$ $= g_{x} + g_{y} \frac{d y}{d x}$

$= - \frac{2 y}{{(x - y)}^{2}} + \frac{2 x}{{(x - y)}^{2}} \cdot (\frac{x + y}{x - y})$ $= - \frac{2 y}{{(x - y)}^{2}} + \frac{2 x}{{(x - y)}^{2}} \cdot (\frac{x + y}{x - y})$

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

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

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

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

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

■別解

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

(9)の両辺を $x$ $x$ で微分する．

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

$\frac{1}{\sqrt{x^{2} + y^{2}}} \cdot \frac{1}{2 \sqrt{x^{2} + y^{2}}} (2 x + 2 y \frac{d y}{d x})$ $\frac{1}{\sqrt{x^{2} + y^{2}}} \cdot \frac{1}{2 \sqrt{x^{2} + y^{2}}} (2 x + 2 y \frac{d y}{d x})$ $- \frac{x^{2}}{x^{2} + y^{2}} (- \frac{y}{x^{2}} + \frac{1}{x} \frac{d y}{d x}) = 0$ $- \frac{x^{2}}{x^{2} + y^{2}} (- \frac{y}{x^{2}} + \frac{1}{x} \frac{d y}{d x}) = 0$ 　･･････(10)

(10)を $\frac{d y}{d x}$ $\frac{d y}{d x}$ について解く．

$\frac{x}{x^{2} + y^{2}} + \frac{y}{x^{2} + y^{2}} \frac{d y}{d x} + \frac{y}{x^{2} + y^{2}}$ $\frac{x}{x^{2} + y^{2}} + \frac{y}{x^{2} + y^{2}} \frac{d y}{d x} + \frac{y}{x^{2} + y^{2}}$ $- \frac{x}{x^{2} + y^{2}} \frac{d y}{d x} = 0$ $- \frac{x}{x^{2} + y^{2}} \frac{d y}{d x} = 0$

$\frac{x + y}{x^{2} + y^{2}} - \frac{x - y}{x^{2} + y^{2}} \frac{d y}{d x} = 0$ $\frac{x + y}{x^{2} + y^{2}} - \frac{x - y}{x^{2} + y^{2}} \frac{d y}{d x} = 0$

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

$x + y - (x - y) \frac{d y}{d x} = 0$ $x + y - (x - y) \frac{d y}{d x} = 0$ 　･･････(11)

$\frac{d y}{d x} = \frac{x + y}{x - y}$ $\frac{d y}{d x} = \frac{x + y}{x - y}$ 　･･････(12)

(11)を $x$ $x$ さらに $x$ $x$ で微分する．

$1 + \frac{d y}{d x} - (1 - \frac{d y}{d x}) \frac{d y}{d x}$ $1 + \frac{d y}{d x} - (1 - \frac{d y}{d x}) \frac{d y}{d x}$ $- (x - y) \frac{d^{2} y}{d x^{2}} = 0$ $- (x - y) \frac{d^{2} y}{d x^{2}} = 0$ 　･･････(13)

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

$1 + {(\frac{d y}{d x})}^{2} - (x - y) \frac{d^{2} y}{d x^{2}} = 0$ $1 + {(\frac{d y}{d x})}^{2} - (x - y) \frac{d^{2} y}{d x^{2}} = 0$

$\frac{d^{2} y}{d x^{2}} = \frac{1}{x - y} {1 + {(\frac{d y}{d x})}^{2}}$ $\frac{d^{2} y}{d x^{2}} = \frac{1}{x - y} \{1 + {(\frac{d y}{d x})}^{2}\}$ 　･･････(14)

(14)に(12)を代入する．

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

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

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

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最終更新日： 2023年9月15日

陰関数の2次導関数

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