# 全微分

## ■問題

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

## ■答

$\frac{\partial }{\partial x}f\left(x,y\right)$$=\frac{\partial }{\partial x}\left(\frac{xy}{2x+y}\right)$

$=\frac{\frac{\partial }{\partial x}xy·\left(2x+y\right)-xy·\frac{\partial }{\partial x}\left(2x+y\right)}{{\left(2x+y\right)}^{2}}$

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

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

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

$\frac{\partial }{\partial y}f\left(x,y\right)$$=\frac{\partial }{\partial y}\left(\frac{xy}{2x+y}\right)$

$=\frac{\frac{\partial }{\partial y}xy·\left(2x+y\right)-xy·\frac{\partial }{\partial y}\left(2x+y\right)}{{\left(2x+y\right)}^{2}}$

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

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

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

$df\left(x,y\right)$$=\frac{\partial }{\partial x}f\left(x,y\right)dx+\frac{\partial }{\partial y}f\left(x,y\right)dy$

$=\frac{{y}^{2}}{{\left(2x+y\right)}^{2}}dx+\frac{2{x}^{2}}{{\left(2x+y\right)}^{2}}dy$

$=\frac{{y}^{2}dx+2{x}^{2}dy}{{\left(2x+y\right)}^{2}}$

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