How do you find #x^2y- xy^3 = 2-2x^3# implicitly?

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Answer 1 · 2015-08-07T11:51:32

#dy/dx=(y^3-6x^2-2xy)/(x^2-3xy^2)#

Differentiate term by term (using the product rule):

#(d(x^2y))/dx=2xy+x^2dy/dx#

#(d(xy^3))/dx=y^3+3xy^2dy/dx#

#(d(2))/dx=0#

#(d(2x^3))/dx=6x^2#

This gives:
#2xy+x^2dy/dx-(y^3+3xy^2dy/dx)=-6x^2#

Rearranging terms to make #dy/dx# the subject we have:

#(x^2-3xy^2)dy/dx=y^3-6x^2-2xy#

=>

#dy/dx=(y^3-6x^2-2xy)/(x^2-3xy^2)#