# 合成関数の偏微分

## ■問題

$z=xtany\text{\hspace{0.17em}},\text{\hspace{0.17em}}x={sin}^{-1}2t\text{\hspace{0.17em}},\text{\hspace{0.17em}}y={cos}^{-1}2t\text{\hspace{0.17em}}$ のとき， $\frac{dz}{dt}\text{\hspace{0.17em}}$ を求めよ．

## ■答

$=\frac{1}{t}+\frac{{\mathrm{sin}}^{-1}2t}{2{t}^{2}\left(1-4{t}^{2}\right)}$

## ■解説

$\frac{dx}{dt}=\frac{{\left(2t\right)}^{\prime }}{\sqrt{1-{\left(2t\right)}^{2}}}=\frac{2}{\sqrt{1-4{t}^{2}}}$

$\frac{dy}{dt}=-\frac{{\left(2t\right)}^{\prime }}{\sqrt{1-{\left(2t\right)}^{2}}}=-\frac{2}{\sqrt{1-4{t}^{2}}}$

${z}_{x}=\frac{\partial z}{\partial x}=tany$

${z}_{y}=\frac{\partial z}{\partial y}=x\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\frac{1}{{cos}^{2}y}=\frac{x}{{cos}^{2}x}$

$\frac{dz}{dt}$$={z}_{x}\frac{dx}{dt}+{z}_{y}\frac{dy}{dt}$

$=\mathrm{tan}y\left(\frac{2}{\sqrt{1-4{t}^{2}}}\right)+\frac{x}{{\mathrm{cos}}^{2}y}\left(-\frac{2}{\sqrt{1-4{t}^{2}}}\right)$

$=\frac{2}{\sqrt{1-4{t}^{2}}}\left\{\frac{\mathrm{sin}\left({\mathrm{cos}}^{-1}2t\right)}{\mathrm{cos}\left({\mathrm{cos}}^{-1}2t\right)}+\frac{{\mathrm{sin}}^{-1}2t}{{\left(2t\right)}^{2}}\right\}$

$=\frac{2}{\sqrt{1-4{t}^{2}}}\left\{\frac{\sqrt{1-{\mathrm{cos}}^{2}\left({\mathrm{cos}}^{-1}2t\right)}}{2t}+\frac{{\mathrm{sin}}^{-1}2t}{4{t}^{2}}\right\}$

$=\frac{2}{\sqrt{1-4{t}^{2}}}\left\{\frac{\sqrt{1-{\left(2t\right)}^{2}}}{2t}+\frac{{\mathrm{sin}}^{-1}2t}{1-{\left(2t\right)}^{2}}\right\}$

$=\frac{1}{t}+\frac{{\mathrm{sin}}^{-1}2t}{2{t}^{2}\left(1-4{t}^{2}\right)}$

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