Question #ac40e

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Answer 1 · 2018-02-05T15:33:06

$y'=-(2x-y)/(2y-x)$

$x^2+y^2=xy$

After taking derivative both sides,

$2x+2y*y'=y+xy'$

$2x-y=(x-2y)*y'$

$y'=(2x-y)/(x-2y)=-(2x-y)/(2y-x)$

Answer 2 · 2018-02-05T15:33:41

$x^2+y^2=xy$

differentiating with respect to $x$ ,

$dx^2/dx+dy^2/dx=(d(xy))/dx$

applying product rule for the derivative of $xy$

$2x + 2y dy/dx = dx/dx*y + dy/dx*x$

$2x + 2y dy/dx = y + xdy/dx$

$2x -y = xdy/dx-2y dy/dx$

$2x -y = (x-2y) dy/dx$

$dy/dx=(2x -y)/ (x-2y)$

-Sahar