# 2次の偏微分

## ■問題

$z=\mathrm{log}\left(x-y\right)$

## ■答

$\frac{{\partial }^{2}z}{\partial {x}^{2}}=-\frac{1}{{\left(x-y\right)}^{2}}$$\frac{{\partial }^{2}z}{\partial {y}^{2}}=-\frac{1}{{\left(x-y\right)}^{2}}$$\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{1}{{\left(x-y\right)}^{2}}$

## ■解説

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

$=\frac{1}{x-y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}1$

$=\frac{1}{x-y}$

$={\left(x-y\right)}^{-1}$

これを更に$\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ で偏微分すると，

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

$=\frac{\partial }{\partial x}\left\{{\left(x-y\right)}^{-1}\right\}$

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

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

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

$=-\frac{1}{{\left(x-y\right)}^{2}}$

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

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

$=-\frac{1}{x-y}$

$=-{\left(x-y\right)}^{-1}$

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

$=\frac{\partial }{\partial y}\left\{-{\left(x-y\right)}^{-1}\right\}$

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

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

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

$=-\frac{1}{{\left(x-y\right)}^{2}}$

また，２次偏導関数は偏微分する順序には無関係であるから， $\text{\hspace{0.17em}}\text{\hspace{0.17em}}z\text{\hspace{0.17em}}\text{\hspace{0.17em}}$$\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ で偏微分した後 $\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ で偏微分すると，

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

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

$=\frac{\partial }{\partial y}\left\{{\left(x-y\right)}^{-1}\right\}$

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

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

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

$=\frac{1}{{\left(x-y\right)}^{2}}$　　･･････(1)

$\text{\hspace{0.17em}}\text{\hspace{0.17em}}z\text{\hspace{0.17em}}\text{\hspace{0.17em}}$$\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ で偏微分した後 $\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ で偏微分すると，

$\frac{\partial z}{\partial y}=-{\left(x-y\right)}^{-1}$

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

$=\frac{\partial z}{\partial x}\left\{-{\left(x-y\right)}^{-1}\right\}$

$=-\left(-1\right){\left(x-y\right)}^{-2}\cdot 1$

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

$=\frac{1}{{\left(x-y\right)}^{2}}$　　･･････(2)

(1)，(2)より

$\frac{{\partial }^{2}z}{\partial y\partial x}=\frac{{\partial }^{2}z}{\partial x\partial y}$

となり，2次偏導関数は偏微分する順序には無関係であることが確かめられた．

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