偏微分を含む証明

■問題

$z=f\left(\frac{y}{x}\right)\text{\hspace{0.17em}}$ ならば, $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0\text{\hspace{0.17em}}$

である．

■ヒント

$z\text{\hspace{0.17em}}$$x\text{\hspace{0.17em}},\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ でそれぞれ偏微分し，２式を連立させる．

■解説

$\frac{\partial z}{\partial x}={f}^{\prime }\left(\frac{y}{x}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(-\frac{y}{{x}^{2}}\right)$

$\left(-\frac{{x}^{2}}{y}\right)\frac{\partial z}{\partial x}={f}^{\prime }\left(\frac{y}{x}\right)$

$\frac{\partial z}{\partial y}={f}^{\prime }\left(\frac{y}{x}\right)\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(\frac{1}{x}\right)$

$x\frac{\partial z}{\partial y}={f}^{\prime }\left(\frac{y}{x}\right)$

$\left(-\frac{{x}^{2}}{y}\right)\frac{\partial z}{\partial x}\text{\hspace{0.17em}}=x\frac{\partial z}{\partial y}$

$\frac{{x}^{2}}{y}\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0$

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$

ホーム>>カテゴリー分類>>微分>>偏微分>>問題演習>>偏微分を含む証明