# 偏微分とその値

## ■問題

$f\left(x,y\right)={sin}^{-1}\frac{x}{y}$

## ■答

それぞれ，$-\frac{1}{\sqrt{3}}$$-\frac{1}{2\sqrt{3}}$

## ■ヒント

${sin}^{-1}$ の微分を，合成関数の微分によって行う．

## ■解説

$u=\frac{x}{y}\text{\hspace{0.17em}}$ とおくと

$f\left(x,y\right)={sin}^{-1}u$

$\frac{d}{du}f\left(x,y\right)$$=\frac{1}{\sqrt{1-{u}^{2}}}$

$u=\frac{x}{y}\text{\hspace{0.17em}}$ を代入する．

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

$=\frac{1}{\sqrt{1-\frac{{x}^{2}}{{y}^{2}}}}$

$=\frac{1}{\sqrt{\frac{{y}^{2}-{x}^{2}}{{y}^{2}}}}$

$=\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}$

$\frac{\partial u}{\partial x}$$=\frac{1}{y}$

$\frac{d}{du}f\left(x,y\right)×\frac{\partial u}{\partial x}$ $=\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}×\left(\frac{1}{y}\right)$$=\frac{1}{y}\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}$

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

$\frac{1}{y}\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}$$x=1\text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=-2\text{\hspace{0.17em}}$ を代入する．

${f}_{x}\left(1,-2\right)$ $=\sqrt{\frac{{\left(-2\right)}^{2}}{{\left(-2\right)}^{2}-{1}^{2}}}×\left(\frac{1}{-2}\right)$ $=\frac{2}{\sqrt{3}}×\left(\frac{1}{-2}\right)$ $=-\frac{1}{\sqrt{3}}$

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

$\frac{d}{du}f\left(x,y\right)×\frac{\partial u}{\partial y}$$=\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}×\left(-\frac{x}{{y}^{2}}\right)$ $=-\frac{x}{{y}^{2}}\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}$

$\frac{\partial }{\partial y}f\left(x,y\right)=-\frac{x}{{y}^{2}}\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}$

$-\frac{x}{{y}^{2}}\sqrt{\frac{{y}^{2}}{{y}^{2}-{x}^{2}}}$$\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=-2\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ を代入する．

${f}_{y}\left(1,-2\right)=-\frac{1}{{\left(-2\right)}^{2}}\sqrt{\frac{{\left(-2\right)}^{2}}{{\left(-2\right)}^{2}-{1}^{2}}}$ $=-\frac{1}{2\sqrt{3}}$

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