# 偏微分を含む証明

## ■問題

$z=f\left({x}^{2}-{y}^{2}\right)\text{\hspace{0.17em}}$ ならば $y\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0\text{\hspace{0.17em}}$

である．

## ■ヒント

$z$$x$$y$ でそれぞれ偏微分し，２式を連立させる．

## ■解説

$\frac{\partial z}{\partial x}={f}^{\prime }\left({x}^{2}-{y}^{2}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2x\text{\hspace{0.17em}}$

$\frac{1}{2x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial x}={f}^{\prime }\left({x}^{2}-{y}^{2}\right)$

$\frac{\partial z}{\partial y}={f}^{\prime }\left({x}^{2}-{y}^{2}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(-2y\right)$

$\left(-\frac{1}{2y}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial y}={f}^{\prime }\left({x}^{2}-{y}^{2}\right)$

$\frac{1}{2x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial x}\text{\hspace{0.17em}}=\left(-\frac{1}{2y}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial y}$

$\frac{1}{2x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial x}+\frac{1}{2y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{\partial z}{\partial y}=0$

$y\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0$

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