# 偏微分を含む証明

## ■問題

$z=log\sqrt{{x}^{2}+{y}^{2}}\text{\hspace{0.17em}}$ ならば ${\left(\frac{\partial z}{\partial x}\right)}^{2}+{\left(\frac{\partial z}{\partial y}\right)}^{2}=\frac{1}{{e}^{2z}}\text{\hspace{0.17em}}$ である．

## ■ヒント

${\left(\frac{\partial z}{\partial x}\right)}^{2}+{\left(\frac{\partial z}{\partial y}\right)}^{2}=\frac{1}{{e}^{2z}}\text{\hspace{0.17em}}$の右辺と左辺をそれぞれ式変形し同じ形になることを示す．

## ■解説

[1] 左辺 に関して

$u={x}^{2}+{y}^{2}\text{\hspace{0.17em}}$ とおくと

$z=\mathrm{log}{\left({x}^{2}+{y}^{2}\right)}^{\frac{1}{2}}$ $=\frac{1}{2}\mathrm{log}\left({x}^{2}+{y}^{2}\right)$ $=\frac{1}{2}\mathrm{log}u$

$\frac{dz}{du}=\frac{1}{2u}=\frac{1}{2\left({x}^{2}+{y}^{2}\right)}$　　(対数の微分を参照)

まず，$u$$x$で偏微分すると

$\frac{\partial u}{\partial x}=2x$

$\frac{\partial z}{\partial x}=\frac{dz}{du}·\frac{\partial u}{\partial x}$ $=\frac{1}{2\left({x}^{2}+{y}^{2}\right)}·2x$$=\frac{x}{{x}^{2}+{y}^{2}}$

$\frac{\partial u}{\partial y}=2y$

$\frac{\partial z}{\partial x}=\frac{dz}{du}·\frac{\partial u}{\partial y}$ $=\frac{1}{2\left({x}^{2}+{y}^{2}\right)}·2y$ $=\frac{y}{{x}^{2}+{y}^{2}}$

よって，左辺は

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

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

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

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

$=\frac{1}{{x}^{2}+{y}^{2}}$　･･････(1)

[2]右辺に関して

$z=log\sqrt{{x}^{2}+{y}^{2}}\text{\hspace{0.17em}}$ より

${e}^{2z}={e}^{2log\sqrt{{x}^{2}+{y}^{2}}}$

$={e}^{log{\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{2}}$

$={e}^{log\left({x}^{2}+{y}^{2}\right)}$

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

よって，右辺は

$\frac{1}{{e}^{2z}}=\frac{1}{{x}^{2}+{y}^{2}}$　　･･････(2)

(1)，(2) より，

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

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