What is the implicit derivative of #1= ysqrt(x-y)+ xy #?

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Answer 1 · 2018-01-28T14:08:47

#y'=(y^2-2xy^2+2y)/(3y^2+2x^2y+3y^2-2xy-2x)#

#1=ysqrt(x-y)+xy#

#ysqrt(x-y)=1-xy#

#y^2*(x-y)=(1-xy)^2#

#y^2*x-y^3=x^2*y^2-2xy+1#

After differentiating both sides,

#y^2+2xy*y'-3y^2*y'=2xy^2+2x^2y*y'-2y-2xy'#

#y^2-2xy^2+2y=(3y^2+2x^2y+3y^2-2xy-2x)*y'#

#y'=(y^2-2xy^2+2y)/(3y^2+2x^2y+3y^2-2xy-2x)#