Найти производную неявной функции.[tex] x^{ \frac{2}{3} } *ln xy+ xy=0[/tex]

Ответы 1

$x^{\frac{2}{3}}\cdot\ln( xy)+xy=0,\\\\\left(x^{\frac{2}{3}}\cdot\ln( xy)ight)'=-(xy)',\\\\\left(x^\frac{2}{3}ight)'\ln(xy)+x^{\frac{2}{3}}\left(\ln(xy)ight)'=-\left(x'y+xy'ight),\\\\\frac{2}{3}x^{\frac{2}{3}-1}\cdot\ln(xy)+x^{\frac{2}{3}}\cdot\frac{1}{xy}\cdot(xy)'=-\left(1\cdot y+xy'ight),\\\\\frac{2}{3}x^{-\frac{1}{3}}\cdot\ln(xy)+\frac{x^{\frac{2}{3}}}{xy}\cdot\left(y+xy'ight)=-y-xy',\\\\\frac{2\ln(xy)}{3\sqrt[3]{x}}+\frac{x^{\frac{2}{3}}\left(y+xy'ight)}{xy}=-y-xy',$ $\frac{2\ln(xy)}{3\sqrt[3]{x}}+\frac{y+xy'}{\sqrt[3]{x}y}=-y-xy'\ |\bullet\sqrt[3]{x}y,\\\\\frac{2\ln(xy)\cdot\sqrt[3]{x}y}{3\sqrt[3]{x}}+y+xy'=\sqrt[3]{x}y\left(-y-xy'igth),$ $\frac{2y\ln(xy)}{3}+y+xy'=-y^2\sqrt[3]{x}-xyy'\sqrt[3]{x},\\\\\frac{2y\ln(xy)}{3}+y+xy'=-y^2\sqrt[3]{x}-yy'\sqrt[3]{x^4},\\\\xy'+yy'\sqrt[3]{x^4}=-y^2\sqrt[3]{x}-y-\frac{2y\ln(xy)}{3},\\\\y'\left(x+y\sqrt[3]{x^4}ight)=-y\left(y\sqrt[3]{x}+1+\frac{2\ln(xy)}{3}ight),\\\\y'=\frac{-y\left(y\sqrt[3]x+1+\frac{2\ln(xy)}{3}ight)}{x+y\sqrt[3]{x^4}},\\\\y'=\frac{-y\left(y\sqrt[3]x+1+\frac{2\ln(xy)}{3}ight)}{x\left(1+y\sqrt[3]{x}ight)}.$ $OTBET:\ \ y'=\frac{-y\left(y\sqrt[3]x+1+\frac{2\ln(xy)}{3}ight)}{x\left(1+y\sqrt[3]{x}ight)}.$
- Автор:
  tomasgallegos
- 5 лет назад
- 0
Добавить свой ответ

Еще вопросы