Question

Найти производную неявной функции.$x^{ frac{2}{3} } *ln xy+ xy=0$

Answer 1

$x^{frac{2}{3}}cdotln( xy)+xy=0,\\left(x^{frac{2}{3}}cdotln( xy)right)'=-(xy)',\\left(x^frac{2}{3}right)'ln(xy)+x^{frac{2}{3}}left(ln(xy)right)'=-left(x'y+xy'right),\\frac{2}{3}x^{frac{2}{3}-1}cdotln(xy)+x^{frac{2}{3}}cdotfrac{1}{xy}cdot(xy)'=-left(1cdot y+xy'right),\\frac{2}{3}x^{-frac{1}{3}}cdotln(xy)+frac{x^{frac{2}{3}}}{xy}cdotleft(y+xy'right)=-y-xy',\\frac{2ln(xy)}{3sqrt[3]{x}}+frac{x^{frac{2}{3}}left(y+xy'right)}{xy}=-y-xy',$

$frac{2ln(xy)}{3sqrt[3]{x}}+frac{y+xy'}{sqrt[3]{x}y}=-y-xy' |bulletsqrt[3]{x}y,\\frac{2ln(xy)cdotsqrt[3]{x}y}{3sqrt[3]{x}}+y+xy'=sqrt[3]{x}yleft(-y-xy'rigth),$

$frac{2yln(xy)}{3}+y+xy'=-y^2sqrt[3]{x}-xyy'sqrt[3]{x},\\frac{2yln(xy)}{3}+y+xy'=-y^2sqrt[3]{x}-yy'sqrt[3]{x^4},\\xy'+yy'sqrt[3]{x^4}=-y^2sqrt[3]{x}-y-frac{2yln(xy)}{3},\\y'left(x+ysqrt[3]{x^4}right)=-yleft(ysqrt[3]{x}+1+frac{2ln(xy)}{3}right),\\y'=frac{-yleft(ysqrt[3]x+1+frac{2ln(xy)}{3}right)}{x+ysqrt[3]{x^4}},\\y'=frac{-yleft(ysqrt[3]x+1+frac{2ln(xy)}{3}right)}{xleft(1+ysqrt[3]{x}right)}.$


$OTBET: y'=frac{-yleft(ysqrt[3]x+1+frac{2ln(xy)}{3}right)}{xleft(1+ysqrt[3]{x}right)}.$