# 偏微分の基礎

## ■問題

$z={\mathrm{sin}}^{-1}3{x}^{2}{y}^{3}$

## ■答

$\frac{\partial z}{\partial x}=\frac{6x{y}^{3}}{\sqrt{1-9{x}^{4}{y}^{6}}}$

$\frac{\partial z}{\partial y}=\frac{9{x}^{2}{y}^{2}}{\sqrt{1-9{x}^{4}{y}^{6}}}$

## ■ヒント

${sin}^{-1}$ の微分を，合成関数の微分によって行う．

## ■解説

$u=3{x}^{2}{y}^{3}$ とおくと，

$z={\mathrm{sin}}^{-1}u$

$\frac{\partial z}{\partial u}=\frac{1}{\sqrt{1-{u}^{2}}}$

$u=3{x}^{2}{y}^{3}$ を代入する．

$\frac{\partial z}{\partial u}=\frac{1}{\sqrt{1-{\left(3{x}^{2}{y}^{3}\right)}^{2}}}$

$=\frac{1}{\sqrt{1-9{x}^{4}{y}^{6}}}$

$\frac{\partial u}{\partial x}=6x{y}^{3}$

$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial x}$

$=\frac{1}{\sqrt{1-9{x}^{4}{y}^{6}}}\cdot 6x{y}^{3}$

$=\frac{6x{y}^{3}}{\sqrt{1-9{x}^{4}{y}^{6}}}$

$\frac{\partial u}{\partial y}=9{x}^{2}{y}^{2}$

$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\cdot \frac{\partial u}{\partial y}$

$=\frac{1}{\sqrt{1-{\left(3{x}^{2}{y}^{3}\right)}^{2}}}\cdot 9{x}^{2}{y}^{2}$

$=\frac{9{x}^{2}{y}^{2}}{\sqrt{1-9{x}^{4}{y}^{6}}}$

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