# 偏微分とその値

## ■問題

$f\left(x,y\right)=\frac{3{x}^{2}y+2x{y}^{2}}{{x}^{2}-{y}^{2}}$

## ■答

それぞれ，$\frac{8}{9}$$\frac{7}{9}$

## ■解説

$\frac{\partial }{\partial x}f\left(x,y\right)=\frac{\left(6xy+2{y}^{2}\right)\left({x}^{2}-{y}^{2}\right)-\left(3{x}^{2}y+2x{y}^{2}\right)\cdot 2x}{{\left({x}^{2}-{y}^{2}\right)}^{2}}$

$=\frac{6{x}^{3}y-6x{y}^{3}+2{x}^{2}{y}^{2}-2{y}^{4}-6{x}^{3}y-4{x}^{2}{y}^{2}}{{\left({x}^{2}-{y}^{2}\right)}^{2}}$

$=-\frac{2{x}^{2}{y}^{2}+6x{y}^{3}+2{y}^{4}}{{\left({x}^{2}-{y}^{2}\right)}^{2}}$

$-\frac{2{x}^{2}{y}^{2}+6x{y}^{3}+2{y}^{4}}{{\left({x}^{2}-{y}^{2}\right)}^{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．

${f}_{x}\left(x,y\right)=-\frac{2\cdot 1\cdot {\left(-2\right)}^{2}+6\cdot 1\cdot {\left(-2\right)}^{3}+2\cdot {\left(-2\right)}^{4}}{\left\{1-{\left(-2\right)}^{2}\right\}}$

$=-\frac{2\cdot 4+6\cdot \left(-8\right)+2\cdot 16}{{\left(1-4\right)}^{2}}$

$=-\frac{8-48+32}{9}=\frac{8}{9}$

$\frac{\partial }{\partial y}f\left(x,y\right)=\frac{\left(3{x}^{2}+4xy\right)\left({x}^{2}-{y}^{2}\right)-\left(3{x}^{2}y+2x{y}^{2}\right)\cdot \left(-2y\right)}{{\left({x}^{2}-{y}^{2}\right)}^{2}}$

$=\frac{3{x}^{4}-3{x}^{2}{y}^{2}+4{x}^{3}y-4x{y}^{3}+6{x}^{2}{y}^{2}+4x{y}^{3}}{{\left({x}^{2}-{y}^{2}\right)}^{2}}$

$=\frac{3{x}^{4}+4{x}^{3}y+3{x}^{2}{y}^{2}}{{\left({x}^{2}-{y}^{2}\right)}^{2}}$

$\frac{3{x}^{4}+4{x}^{3}y+3{x}^{2}{y}^{2}}{{\left({x}^{2}-{y}^{2}\right)}^{2}}$$x=1\text{\hspace{0.17em}},\text{\hspace{0.17em}}y=-2\text{\hspace{0.17em}}$ を代入する．

${f}_{y}\left(1,-2\right)=\frac{3\cdot {1}^{4}+4\cdot {1}^{3}\cdot \left(-2\right)+3\cdot {1}^{2}\cdot {\left(-2\right)}^{2}}{{\left\{1-{\left(-2\right)}^{2}\right\}}^{2}}$

$=\frac{3-8+12}{9}=\frac{7}{9}$

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