偏微分とその値

■問題

次の関数について $3$\displaystyle{\hspace{1}{f}_{x}\left({ 1,- 2 }\right)}$ と $3$\displaystyle{\hspace{1}{f}_{y}\left({ 1,- 2 }\right)}$ を求めよ．

$3$\displaystyle{ f $ x,y $ ={ \tan }^{ - 1 }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} }$

■答

それぞれ， $3$\displaystyle{ \frac{\hspace{2}2\hspace{2}}{\hspace{2}5\hspace{2}} }$ ， $3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}}$

■ヒント

$3$\displaystyle{ { \tan }^{ - 1 } }$ の微分を，合成関数の微分によって行う．
偏導関数の定義を用いて偏微分する．
微分後の式に数値を代入する．

■解説

$3$\displaystyle{ u=\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}} }$ とおくと

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}f\left({ x,y }\right)=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}x \hspace{2}}{\tan }^{ - 1 }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} du \hspace{2}}{\tan }^{ - 1 }u\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}x \hspace{2}}}$

$3$\displaystyle{\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}f\left({ x,y }\right)=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}{\tan }^{ - 1 }\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2}d\hspace{2}}{\hspace{2} du \hspace{2}}{\tan }^{ - 1 }u\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}}$

となる．

$3$\displaystyle{\frac{\hspace{2}d\hspace{2}}{\hspace{2} du \hspace{2}}{\tan }^{ - 1 }u=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{u}^{2} \hspace{2}}}$

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

$3$\displaystyle{\frac{\hspace{2} {\partial}u \hspace{2}}{\hspace{2} {\partial}y \hspace{2}}=\frac{\hspace{2}{\partial}\hspace{2}}{\hspace{2} {\partial}y \hspace{2}}\frac{\hspace{2}y\hspace{2}}{\hspace{2}x\hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}x\hspace{2}}}$

よって

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

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

すなわち

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

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

となる．

$3$\displaystyle{ \hspace{1}{f}_{x}\left({ 1,- 2 }\right)=\frac{\hspace{2} - \left({ - 2 }\right) \hspace{2}}{\hspace{2} {1}^{2}+{ \left({ - 2 }\right) }^{2} \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2}2\hspace{2}}{\hspace{2} 1+4 \hspace{2}}=\frac{\hspace{2}2\hspace{2}}{\hspace{2}5\hspace{2}} }$

$3$\displaystyle{ \hspace{1}{f}_{y}\left({ 1,- 2 }\right)=\frac{\hspace{2}1\hspace{2}}{\hspace{2} {1}^{2}+{ \left({ - 2 }\right) }^{2} \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}} }$

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最終更新日： 2024年5月7日