How do you implicitly differentiate #-y=xy+2sqrt(x-y^3) #?

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Answer 1 · 2017-08-18T13:40:29

#dy/dx=-(x-y^3)^(1/2)/(x(x-y^3)^(1/2)+(1-3y^2)+(x-y^3)^(1/2)#

#-dy/dx##=xdy/dx+y+#...

#2sqrt(x+y^3)=##2(x-y^3)^(1/2)# derivative of that equal to

#(x-y^3)^(-1/2)*(1-3y^2dy/dx)#

with some math

get #dy/dx# in one side and all in the other side

#dy/dx=-(x-y^3)^(1/2)/(x(x-y^3)^(1/2)+(1-3y^2)+(x-y^3)^(1/2)#

I need someone to check my answer or explain further, and I will be thankful.