陰関数の2次導関数

■問題

次の関係で定義される陰関数 $3$\displaystyle{y={\varphi} $x$ }$ について $3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}}$ を求めよ．

$3$\displaystyle{3{x}^{2}+2xy+{y}^{2}=1}$

■答

$3$\displaystyle{- \frac{\hspace{2}2\hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$

■ヒント

まず， $3$\displaystyle{3{x}^{2}+2xy+{y}^{2}=1}$ より定義した $3$\displaystyle{f\left({ x,y }\right)=f\left({ x,{\phi}\left({x}\right) }\right)=0}$ となる2変数関数 $3$\displaystyle{ f\left({ x,y }\right) }$ を $3$x$ で微分する．その結果から $3$\displaystyle{ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }$ を求める．

■解説

⇒別解

与式を変形すると

$3$\displaystyle{3{x}^{2}+2xy+{y}^{2}- 1=0}$

となる．

$3$\displaystyle{f $ x,y $ =f $ x,{\varphi} \(x$ \) }$ $3$\displaystyle{=3{x}^{2}+2xy+{y}^{2}- 1}$

とおく． $3$\displaystyle{f $ x,y $ }$ を $3$\displaystyle{x=x}$ ， $3$\displaystyle{y={\phi}\left({x}\right)}$ とする合成関数と考えて，これを $3$\displaystyle{x}$ で微分すると

$3$\displaystyle{\frac{\hspace{2}d\hspace{2}}{\hspace{2} dx \hspace{2}}f $ x,y $ }$ $3$\displaystyle{=\hspace{1}{f}_{x}\frac{\hspace{2} dx \hspace{2}}{\hspace{2} dx \hspace{2}}+\hspace{1}{f}_{y}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$

合成関数の偏導関数の公式を用た．

$3$\displaystyle{=\hspace{1}{f}_{x}+\hspace{1}{f}_{y}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=0}$

よって

$3$\displaystyle{\hspace{1}{f}_{y}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$ $3$\displaystyle{=- \hspace{1}{f}_{x}}$

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2} \hspace{1}{f}_{x} \hspace{2}}{\hspace{2} \hspace{1}{f}_{y} \hspace{2}}}$ ･･････(1)

ここで

$3$\displaystyle{\hspace{1}{f}_{x}}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f $ x,y $ }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ 3{x}^{2}+2xy+{y}^{2}- 1 $ }$

偏導関数の定義より， $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}y\text{\hspace{1}}\text{\hspace{1}}}$ を定数とみなして $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}x\text{\hspace{1}}\text{\hspace{1}}}$ で微分する．

$3$\displaystyle{=6x+2y}$

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

$3$\displaystyle{\hspace{1}{f}_{y}}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}f $ x,y $ }$

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ 3{x}^{2}+2xy+{y}^{2}- 1 $ }$

偏導関数の定義より， $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}x\text{\hspace{1}}\text{\hspace{1}}}$ を定数とみなして $3$\displaystyle{\text{\hspace{1}}\text{\hspace{1}}y\text{\hspace{1}}\text{\hspace{1}}}$ で微分する．

$3$\displaystyle{=2x+2y}$

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

(1)に(2)，(3)を代入すると

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2} 2 $ 3x+y $ \hspace{2}}{\hspace{2} 2 $ x+y $ \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$ ･･････(4)

となる．

$3$\displaystyle{g $ x,y $ =g $ x,{\varphi} \(x$ \) =- \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$ ･･････(5)

とおく．

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dx \hspace{2}} $ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} $ }$

$3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dx \hspace{2}} \{ g $ x,y $ \} }$

$3$\displaystyle{g $ x,y $ }$ を $3$\displaystyle{x=x}$ ， $3$\displaystyle{y={\phi}\left({x}\right)}$ とする合成関数と考えて，これを $3$\displaystyle{x}$ で微分する．このとき合成関数の偏導関数の公式を用いる．

$3$\displaystyle{=\hspace{1}{g}_{x}+\hspace{1}{g}_{y}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$ ･･････(6)

ここで

$3$\displaystyle{\hspace{1}{g}_{x}}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}g $ x,y $ }$

(5)を代入する．

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ - \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}} $ }$

$3$\displaystyle{=- \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}} $ }$

分数関数（商）の導関数の公式を用いる．

$3$\displaystyle{=- \frac{\hspace{2} \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ 3x+y $ \cdot $ x+y $ - $ 3x+y $ \cdot \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} 3\cdot $ x+y $ - $ 3x+y $ \cdot 1 \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} 3x+3y- 3x- y \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} 2y \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$ ･･････(7)

$3$\displaystyle{\hspace{1}{g}_{y}}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}g $ x,y $ }$

(5)を代入する．

$3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ - \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}} $ }$

$3$\displaystyle{=- \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}} $ }$

分数関数（商）の導関数の公式を用いる．

$3$\displaystyle{=- \frac{\hspace{2} \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ 3x+y $ \cdot $ x+y $ - $ 3x+y $ \cdot \frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} 1\cdot $ x+y $ - $ 3x+y $ \cdot 1 \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} x+y- 3x- y \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} - 2x \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} 2x \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}}$ ･･････(8)

(6)に(7)，(8)，(4)を代入する．

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2} 2y \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}+\frac{\hspace{2} 2x \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}\cdot $ - \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}} $ }$

$3$\displaystyle{=- \frac{\hspace{2} 2y \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}- \frac{\hspace{2} 2x $ 3x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$

$3$\displaystyle{=- \{ \frac{\hspace{2} 2y \hspace{2}}{\hspace{2} { $ x+y $ }^{2} \hspace{2}}+\frac{\hspace{2} 2x $ 3x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}} \} }$

$3$\displaystyle{=- \frac{\hspace{2} 2y $ x+y $ +2x $ 3x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} 2xy+2{y}^{2}+6{x}^{2}+2xy \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} 6{x}^{2}+4xy+2{y}^{2} \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} 2 $ 3{x}^{2}+2xy+{y}^{2} $ \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$

$3$\displaystyle{3{x}^{2}+2xy+{y}^{2}=1}$ の関係から

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=- \frac{\hspace{2}2\hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$

●グラフ

点 $3$\text{P}$ をドラックして曲線 $3$3{x}^{2}+2xy+{y}^{2}=1$ 上を動かすと， $3$\displaystyle{ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }$ のグラフ(青色)， $3$\displaystyle{ \frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}} }$ のグラフ(緑色)が描かれる．

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■別解

$3$\displaystyle{3{x}^{2}+2xy+{y}^{2}=1}$ の両辺を $3$x$ で微分する．

$3$\displaystyle{6x+2y+2x\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}+2y\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=0}$ ･･････(9)

( $3$\displaystyle{\frac{\hspace{2}d\hspace{2}}{\hspace{2} dx \hspace{2}}{y}^{2}}$ についてはここを参照)

(9)を $3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$ について解く．

$3$\displaystyle{6x+2y+\left({ 2x+2y }\right)\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=0}$

$3$\displaystyle{\left({ 2x+2y }\right)\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=- 6x- 2y}$

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=- \frac{\hspace{2} 6x+2y \hspace{2}}{\hspace{2} 2x+2y \hspace{2}}=- \frac{\hspace{2} 3x+y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$ ･･････(10)

(9)を $3$\displaystyle{x}$ さらに $3$\displaystyle{x}$ で微分する．

$3$\displaystyle{6+2\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}+2\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}+2x\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}+2\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}\cdot \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}+2y\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=0}$ ･･････(11)

(11)を $3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}}$ について解く．

$3$\displaystyle{3+\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}+\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}+x\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}+{\left({ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }\right)}^{2}+y\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=0}$

$3$\displaystyle{\left({ x+y }\right)\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=- 3- 2\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}- {\left({ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }\right)}^{2}}$ ･･････(12)

(12)に(10)を代入する．

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

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

$3$\displaystyle{\left({ x+y }\right)\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2} - 3\left({ {x}^{2}+2xy+{y}^{2} }\right)+2\left({ 3{x}^{2}+4xy+{y}^{2} }\right)- \left({ 9{x}^{2}+6xy+{y}^{2} }\right) \hspace{2}}{\hspace{2} { \left({ x+y }\right) }^{2} \hspace{2}}}$

$3$\displaystyle{\left({ x+y }\right)\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=\frac{\hspace{2} - 6{x}^{2}- 4xy- 2{y}^{2} \hspace{2}}{\hspace{2} { \left({ x+y }\right) }^{2} \hspace{2}}}$

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

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

$3$\displaystyle{3{x}^{2}+2xy+{y}^{2}=1}$ の関係から

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

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最終更新日： 2025年1月20日