Question #ac40e

2 Answers
Feb 5, 2018

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

Explanation:

#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)#

Feb 5, 2018

#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