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2次の偏微分

■問題

$3$\displaystyle{z=f $ x,y $ }$ ， $3$\displaystyle{x=2{t}^{2}- 3}$ ， $3$\displaystyle{y={t}^{2}+3t+7}$ のとき， $3$\displaystyle{\frac{\hspace{2} {d}^{2}z \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}}$ を求めよ．

■答

$3$\displaystyle{16{t}^{2}\hspace{1}{f}_{ xx }+8t $ 2t+3 $ \hspace{1}{f}_{ xy }+\hspace{1}{f}_{ yy }{ $ 2t+3 $ }^{2}+4\hspace{1}{f}_{x}+2\hspace{1}{f}_{y}}$

■ヒント

$3$\displaystyle{x}$ ， $3$\displaystyle{y}$ を $3$\displaystyle{t}$ で2回微分する．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{t}$ で微分すると，

$3$\displaystyle{\frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}}=4t}$

これを更に $3$\displaystyle{t}$ で微分すると，

$3$\displaystyle{\frac{\hspace{2} {d}^{2}x \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ \frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}} $ =\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ 4t $ =4}$

同様の手順で $3$\displaystyle{y}$ を $3$\displaystyle{t}$ で微分をすると

$3$\displaystyle{\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}}}=2t+3$

更に， $3$\displaystyle{y}$ を $3$\displaystyle{t}$ で微分する．

$3$\displaystyle{\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}=\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}} $ =\frac{\hspace{2}d\hspace{2}}{\hspace{2} dt \hspace{2}} $ 2t+3 $ =2}$

以上より

$3$\displaystyle{\frac{\hspace{2} {d}^{2}z \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}=\hspace{1}{f}_{ xx }{ $ \frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}} $ }^{2}+2\hspace{1}{f}_{ xy }\frac{\hspace{2} dx \hspace{2}}{\hspace{2} dt \hspace{2}}\frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}}+\hspace{1}{f}_{ yy }{ $ \frac{\hspace{2} dy \hspace{2}}{\hspace{2} dt \hspace{2}} $ }^{2}+\hspace{1}{f}_{x}\frac{\hspace{2} {d}^{2}x \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}+\hspace{1}{f}_{y}\frac{\hspace{2} {d}^{2}y \hspace{2}}{\hspace{2} d{t}^{2} \hspace{2}}}$

$3$\displaystyle{=\hspace{1}{f}_{ xx }{ $ 4t $ }^{2}+2\hspace{1}{f}_{ xy }4t $ 2t+3 $ +\hspace{1}{f}_{ yy }{ $ 2t+3 $ }^{2}+4\hspace{1}{f}_{x}+2\hspace{1}{f}_{y}}$

$3$\displaystyle{=\hspace{1}{f}_{ xx }16{t}^{2}+8t $ 2t+3 $ \hspace{1}{f}_{ xy }+\hspace{1}{f}_{ yy }{ $ 2t+3 $ }^{2}+4\hspace{1}{f}_{x}+2\hspace{1}{f}_{y}}$

$3$\displaystyle{=16{t}^{2}\hspace{1}{f}_{ xx }+8t $ 2t+3 $ \hspace{1}{f}_{ xy }+\hspace{1}{f}_{ yy }{ $ 2t+3 $ }^{2}+4\hspace{1}{f}_{x}+2\hspace{1}{f}_{y}}$

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最終更新日： 2013年8月14日

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