2次の偏微分

■問題

次の関数の第2次偏導関数を求めよ．

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

■答

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}=- \frac{\hspace{2} 4y \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=\frac{\hspace{2} 4x \hspace{2}}{\hspace{2} { $ x+y $ }^{3} \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}=\frac{\hspace{2} 2\left({ x- y }\right) \hspace{2}}{\hspace{2} { \left({ x+y }\right) }^{3} \hspace{2}}}$

■ヒント

2次偏導関数 $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$ の4つを求める．

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$ を計算してから，それぞれを更に $3$x$ ， $3$y$ で偏微分する．

その際，商の微分の公式を用いる．

■解説

● $3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ の計算

$3$\displaystyle{z=\frac{\hspace{2} x- y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$ を偏導関数の定義より， $3$\displaystyle{y}$ を定数とみなして $3$\displaystyle{x}$ で微分する．

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

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

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

$3$\displaystyle{=\frac{\hspace{2} x+y- x+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}}}$ 　･･････(1)

● $3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$ の計算

$3$\displaystyle{z=\frac{\hspace{2} x- y \hspace{2}}{\hspace{2} x+y \hspace{2}}}$ を偏導関数の定義より， $3$\displaystyle{x}$ を定数とみなして $3$\displaystyle{y}$ で微分する．

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

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

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

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

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}}$ の計算

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

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

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

$3$\displaystyle{=\frac{\hspace{2} 0\text{\hspace{1}}\cdot \text{\hspace{1}}{ $ x- y $ }^{2}- 2y\text{\hspace{1}}\cdot \text{\hspace{1}}2 $ x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{4} \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} 0- 2y\text{\hspace{1} }\cdot \text{\hspace{1} }2 $ x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{4} \hspace{2}}}$

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

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

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}}$ の計算

(2)を更に，偏導関数の定義より， $3$\displaystyle{x}$ を定数とみなして $3$\displaystyle{y}$ で微分する．

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

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

$3$\displaystyle{=\frac{\hspace{2} 0\text{\hspace{1}}\cdot \text{\hspace{1}}{ $ x+y $ }^{2}- $ - 2x $ \text{\hspace{1}}\cdot \text{\hspace{1}}2 $ x+y $ \text{\hspace{1}}\cdot \text{\hspace{1}}1 \hspace{2}}{\hspace{2} { $ x+y $ }^{4} \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} 0- $ - 2x $ \text{\hspace{1} }\cdot \text{\hspace{1} }2 $ x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{4} \hspace{2}}}$

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

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

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}}$ の計算

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

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ }$

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

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

$3$\displaystyle{=\frac{\hspace{2} 2{ $ x+y $ }^{2}- 2y\text{\hspace{1} }\cdot \text{\hspace{1} }2 $ x+y $ \hspace{2}}{\hspace{2} { $ x+y $ }^{4} \hspace{2}}}$

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

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

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

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

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

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$ の計算

(2)を更に，偏導関数の定義より， $3$\displaystyle{y}$ を定数とみなして $3$\displaystyle{x}$ で微分する．

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }$

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

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

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

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

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

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

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

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

(3)，(4)より

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$

となり，2次偏導関数は偏微分する順序には無関係であることが確かめられた．

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最終更新日： 2023年8月30日

2次の偏微分

■問題

■答

■ヒント

■解説

● の計算

● の計算

● の計算

● の計算

● の計算

● の計算

● $3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$ の計算

● $3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$ の計算

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}}$ の計算

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}}$ の計算

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}y{\partial}x \hspace{2}}}$ の計算

● $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}x{\partial}y \hspace{2}}}$ の計算