合成関数の2次偏導関数

■問題

$3$\displaystyle{z=f $ x,y $ ,x=u\cos {\theta}- v\sin {\theta},}$ $3$\displaystyle{y=u\sin {\theta}+v\cos {\theta}}$ のとき

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{x}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{y}^{2} \hspace{2}}=\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}\text{\hspace{1} }\text{\hspace{1} }}$

となることを示せ．

■ヒント

右辺を変形して左辺を導く． $3$\displaystyle{x}$ ， $3$\displaystyle{y}$ を $3$\displaystyle{u}$ ， $3$\displaystyle{v}$ でそれぞれ2回偏微分する．求めた式を合成関数の2次偏導関数の公式に代入する．

■解説

$3$\displaystyle{x}$ を $3$\displaystyle{u}$ で偏微分すると

$3$\displaystyle{\frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}u \hspace{2}}=\cos {\theta}}$

これを更に $3$\displaystyle{u}$ で偏微分すると

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}x \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ \frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ =\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ \cos {\theta} $ =0}$

同様の手順で $3$\displaystyle{y}$ を $3$\displaystyle{u}$ で偏微分をすると

$3$\displaystyle{\frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}u \hspace{2}}=\sin {\theta}}$

更に， $3$\displaystyle{y}$ を $3$\displaystyle{u}$ で偏微分する．

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ \frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ =\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ \sin {\theta} $ =0}$

よって， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}}$ は合成関数の2次偏導関数の公式より

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}=\hspace{1}{f}_{ xx }{ $ \frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ }^{2}+2\hspace{1}{f}_{ xy }\frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}u \hspace{2}}\frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}u \hspace{2}}+\hspace{1}{f}_{ yy }{ $ \frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}u \hspace{2}} $ }^{2}+\hspace{1}{f}_{x}\frac{\hspace{2} {{\partial}}^{2}x \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}+\hspace{1}{f}_{y}\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}}$

$3$\displaystyle{=\hspace{1}{f}_{ xx }{ $ \cos {\theta} $ }^{2}+2\hspace{1}{f}_{ xy }\cos {\theta}\sin {\theta}+\hspace{1}{f}_{ yy }{ $ \sin {\theta} $ }^{2}+\hspace{1}{f}_{x}\text{\hspace{1} }\cdot \text{\hspace{1} }0+\hspace{1}{f}_{y}\text{\hspace{1} }\cdot \text{\hspace{1} }0}$

$3$\displaystyle{=\hspace{1}{f}_{ xx }{\cos }^{2}{\theta}+2\hspace{1}{f}_{ xy }\sin {\theta}\cos {\theta}+\hspace{1}{f}_{ yy }{\sin }^{2}{\theta}}$ 　･･････(1)

更に， $3$\displaystyle{v}$ についても同様に偏微分すると

$3$\displaystyle{\frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}v \hspace{2}}=- \sin {\theta}}$

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}x \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}v \hspace{2}} $ \frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}v \hspace{2}} $ =\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}v \hspace{2}} $ - \sin {\theta} $ =0}$

$3$\displaystyle{\frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}v \hspace{2}}=\cos {\theta}}$

よって， $3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}}$ は合成関数の2次偏導関数の公式より

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}=\hspace{1}{f}_{ xx }{ $ \frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}v \hspace{2}} $ }^{2}+2\hspace{1}{f}_{ xy }\frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}v \hspace{2}}\frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}v \hspace{2}}}$ $3$\displaystyle{+\hspace{1}{f}_{ yy }{ $ \frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}v \hspace{2}} $ }^{2}}$ $3$\displaystyle{+\hspace{1}{f}_{x}\frac{\hspace{2} {{\partial}}^{2}x \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}}$ $3$\displaystyle{+\hspace{1}{f}_{y}\frac{\hspace{2} {{\partial}}^{2}y \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}}$

$3$\displaystyle{=\hspace{1}{f}_{ xx }{ $ - \sin {\theta} $ }^{2}}$ $3$\displaystyle{+2\hspace{1}{f}_{ xy } $ - \sin {\theta} $ \cos {\theta}}$ $3$\displaystyle{+\hspace{1}{f}_{ yy }{ $ \cos {\theta} $ }^{2}}$ $3$\displaystyle{+\hspace{1}{f}_{x}\text{\hspace{1} }\cdot \text{\hspace{1} }0}$ $3$\displaystyle{+\hspace{1}{f}_{y}\text{\hspace{1} }\cdot \text{\hspace{1} }0}$

$3$\displaystyle{=\hspace{1}{f}_{ xx }{\sin }^{2}{\theta}- 2\hspace{1}{f}_{ xy }\sin {\theta}\cos {\theta}}$ $3$\displaystyle{+\hspace{1}{f}_{ yy }{\cos }^{2}{\theta}}$ 　･･････(2)

(1)，(2)より

$3$\displaystyle{\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{u}^{2} \hspace{2}}+\frac{\hspace{2} {{\partial}}^{2}z \hspace{2}}{\hspace{2} {\partial}{v}^{2} \hspace{2}}}$

$3$\displaystyle{= $ \hspace{1}{f}_{ xx }{ \cos }^{2}{\theta}+2\hspace{1}{f}_{ xy }\sin {\theta}\cos {\theta}+\hspace{1}{f}_{ yy }{ \sin }^{2}{\theta} $ }$ $3$\displaystyle{+ $ \hspace{1}{f}_{ xx }{ \sin }^{2}{\theta}- 2\hspace{1}{f}_{ xy }\sin {\theta}\cos {\theta}+\hspace{1}{f}_{ yy }{ \cos }^{2}{\theta} $ }$

$3$\displaystyle{=\hspace{1}{f}_{ xx }{\cos }^{2}{\theta}+2\hspace{1}{f}_{ xy }\sin {\theta}\cos {\theta}}$ $3$\displaystyle{+\hspace{1}{f}_{ yy }{\sin }^{2}{\theta}}$ $3$\displaystyle{+\hspace{1}{f}_{ xx }{\sin }^{2}{\theta}- 2\hspace{1}{f}_{ xy }\sin {\theta}\cos {\theta}}$ $3$\displaystyle{+\hspace{1}{f}_{ yy }{\cos }^{2}{\theta}}$

$3$\displaystyle{= $ { \cos }^{2}{\theta}+{ \sin }^{2}{\theta} $ \hspace{1}{f}_{ xx }}$ $3$\displaystyle{+ $ { \sin }^{2}{\theta}+{ \cos }^{2}{\theta} $ \hspace{1}{f}_{ yy }}$

$3$\displaystyle{=\hspace{1}{f}_{ xx }+\hspace{1}{f}_{ yy }}$

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

となる．したがって

ホーム>>カテゴリー分類>>微分>>偏微分>>問題演習>>合成関数の2次偏導関数

学生スタッフ作成

最終更新日： 2023年9月1日