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{y={e}^{ x+y }}$

■答

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

■ヒント

まず， $3$\displaystyle{y={e}^{ x+y }}$ 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{{e}^{ x+y }- y=0}$

となる．

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

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いた．

$3$\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}}$

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}} $ {e}^{ x+y }- y $ }$

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

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{={e}^{x}\cdot {e}^{y}}$

$3$\displaystyle{={e}^{ x+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}} $ {e}^{ x+y }- y $ }$

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

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{={e}^{x}\cdot {e}^{y}- 1}$

$3$\displaystyle{={e}^{ x+y }- 1}$ 　･･････(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} \hspace{1}{f}_{x} \hspace{2}}{\hspace{2} \hspace{1}{f}_{y} \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2} {e}^{ x+y } \hspace{2}}{\hspace{2} {e}^{ x+y }- 1 \hspace{2}}}$ 　･･････(4)

となる．

$3$\displaystyle{g $ x,y $ =g $ x,{\varphi} \(x$ \) =- \frac{\hspace{2} {e}^{ x+y } \hspace{2}}{\hspace{2} {e}^{ x+y }- 1 \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} {e}^{ x+y } \hspace{2}}{\hspace{2} {e}^{ x+y }- 1 \hspace{2}} $ }$

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

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

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

$3$\displaystyle{{e}^{x}}$ の微分の公式を用いる．

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

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

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

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

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

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

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

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

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

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

$3$\displaystyle{{e}^{x}}$ の微分の公式を用いる．

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

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

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

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

$3$\displaystyle{=\frac{\hspace{2} {e}^{ x+y } \hspace{2}}{\hspace{2} { $ {e}^{ x+y }- 1 $ }^{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{=\hspace{1}{g}_{x}+\hspace{1}{g}_{y}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$

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

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

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

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

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

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

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

■別解

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

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

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}={e}^{ x+y }\frac{\hspace{2}d\hspace{2}}{\hspace{2} dx \hspace{2}}\left({ 1+\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }\right)}$

$3$\displaystyle{y={e}^{ x+y }}$ より

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=y\left({ 1+\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }\right)}$ 　･･････(9)

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

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=y+y\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}}$

$3$\displaystyle{\left({ 1- y }\right)\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=y}$

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}=\frac{\hspace{2}y\hspace{2}}{\hspace{2} 1- y \hspace{2}}}$ 　･･････(10)

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

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}\left({ 1+\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }\right)+y\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}}$ 　･･････(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{\left({ 1- y }\right)\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}\left({ 1+\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }\right)}$

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- y \hspace{2}}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}}\left({ 1+\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dx \hspace{2}} }\right)}$ 　･･････(12)

(12)に(10)を代入する．

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- y \hspace{2}}\cdot \frac{\hspace{2}y\hspace{2}}{\hspace{2} 1- y \hspace{2}}\left({ 1+\frac{\hspace{2}y\hspace{2}}{\hspace{2} 1- y \hspace{2}} }\right)}$

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

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{x}^{2} \hspace{2}}=\frac{\hspace{2}y\hspace{2}}{\hspace{2} { \left({ 1- y }\right) }^{3} \hspace{2}}}$

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最終更新日： 2023年9月15日