# 合成関数の偏微分

## ■問題

$z=f\left(x,y\right)$, $x=u\mathrm{cos}\alpha -v\mathrm{sin}\alpha$, $y=u\mathrm{sin}\alpha +v\mathrm{cos}\alpha$ならば

${\left(\frac{\partial z}{\partial x}\right)}^{2}+{\left(\frac{\partial z}{\partial y}\right)}^{2}$ $={\left(\frac{\partial z}{\partial u}\right)}^{2}+{\left(\frac{\partial z}{\partial v}\right)}^{2}$

となることを示せ．

## ■ヒント

### ●参考

$x=u\mathrm{cos}\alpha -v\mathrm{sin}\alpha$, $y=u\mathrm{sin}\alpha +v\mathrm{cos}\alpha$の変換を行列を使って表すと

$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}u\mathrm{cos}\alpha -v\mathrm{sin}\alpha \\ u\mathrm{sin}\alpha +v\mathrm{cos}\alpha \end{array}\right)$$=\left(\begin{array}{cc}\mathrm{cos}\alpha & -\mathrm{sin}\alpha \\ \mathrm{sin}\alpha & \mathrm{cos}\alpha \end{array}\right)\left(\begin{array}{c}u\\ v\end{array}\right)$

$\left(\begin{array}{c}u\\ v\end{array}\right)={\left(\begin{array}{cc}\mathrm{cos}\alpha & -\mathrm{sin}\alpha \\ \mathrm{sin}\alpha & \mathrm{cos}\alpha \end{array}\right)}^{-1}\left(\begin{array}{c}x\\ y\end{array}\right)$

$=\left(\begin{array}{cc}\mathrm{cos}\alpha & \mathrm{sin}\alpha \\ -\mathrm{sin}\alpha & \mathrm{cos}\alpha \end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

$=\left(\begin{array}{cc}\mathrm{cos}\left(-\alpha \right)& -\mathrm{sin}\left(-\alpha \right)\\ \mathrm{sin}\left(-\alpha \right)& \mathrm{cos}\left(-\alpha \right)\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

## ■解答

$z\text{\hspace{0.17em}}$$u\text{\hspace{0.17em}}$合成関数の偏微分をすると

$\frac{\partial z}{\partial v}={f}_{x}\frac{\partial x}{\partial v}+{f}_{y}\frac{\partial y}{\partial v}$ $={f}_{x}\mathrm{cos}\alpha +{f}_{y}\mathrm{sin}\alpha$

となる．

$\frac{\partial z}{\partial v}={f}_{x}\frac{\partial x}{\partial v}+{f}_{y}\frac{\partial y}{\partial v}$

$={f}_{x}\left(-\mathrm{sin}\alpha \right)+{f}_{y}\mathrm{cos}\alpha$

$={f}_{y}\mathrm{cos}\alpha -{f}_{x}\mathrm{sin}\alpha$

したがって

${\left(\frac{\partial z}{\partial u}\right)}^{2}+{\left(\frac{\partial z}{\partial v}\right)}^{2}$

$={\left({f}_{x}cos\alpha +{f}_{y}sin\alpha \right)}^{2}+{\left({f}_{y}cos\alpha -{f}_{x}sin\alpha \right)}^{2}$

$=\left({f}_{x}^{2}{cos}^{2}\alpha +2{f}_{x}{f}_{y}sin\alpha cos\alpha +{f}_{y}^{2}{sin}^{2}\alpha \right)$$+\left({f}_{y}^{2}{cos}^{2}\alpha -2{f}_{x}{f}_{y}sin\alpha cos\alpha +{f}_{x}^{2}{sin}^{2}\alpha \right)$

$={f}_{x}^{2}{cos}^{2}\alpha +2{f}_{x}{f}_{y}sin\alpha cos\alpha +{f}_{y}^{2}{sin}^{2}\alpha$$+{f}_{y}^{2}{cos}^{2}\alpha -2{f}_{x}{f}_{y}sin\alpha cos\alpha +{f}_{x}^{2}{sin}^{2}\alpha$

$=\left({sin}^{2}\alpha +{cos}^{2}\alpha \right){f}_{x}^{2}+\left({sin}^{2}\alpha +{cos}^{2}\alpha \right){f}_{y}^{2}$

$={f}_{x}^{2}+{f}_{y}^{2}$

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

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2023年8月29日