# 偏微分の基礎

## ■問題

$z=\frac{2{x}^{2}+{y}^{3}}{{x}^{3}-3{y}^{2}}$

## ■答

$\frac{\partial z}{\partial x}=\frac{-2{x}^{4}-3{x}^{2}{y}^{3}-12x{y}^{2}}{{\left({x}^{3}-3{y}^{2}\right)}^{2}}$

$\frac{\partial z}{\partial y}=\frac{-3{y}^{4}+3{x}^{3}{y}^{2}+12{x}^{2}y}{{\left({x}^{3}-3{y}^{2}\right)}^{2}}$

## ■解説

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

$=\frac{4{x}^{4}-12x{y}^{2}-6{x}^{4}-3{x}^{2}{y}^{3}}{{\left({x}^{3}-3{y}^{2}\right)}^{2}}$

$=\frac{-2{x}^{4}-3{x}^{2}{y}^{3}-12x{y}^{2}}{{\left({x}^{3}-3{y}^{2}\right)}^{2}}$

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

$=\frac{3{x}^{3}{y}^{2}-9{y}^{4}+12{x}^{2}y+6{y}^{4}}{{\left({x}^{3}-3{y}^{2}\right)}^{2}}$

$=\frac{-3{y}^{4}+3{x}^{3}{y}^{2}+12{x}^{2}y}{{\left({x}^{3}-3{y}^{2}\right)}^{2}}$

ホーム>>カテゴリー分類>>微分>>偏微分>>問題演習>>偏微分の基礎