How do you find #(dy)/(dx)# given #4x^2+3xy^2-6x^2y=y^3#?
1 Answer
Sep 1, 2016
#dy/dx=(-8x-3y^2+12xy)/(6xy-6x^2-3y^2)#
Explanation:
Given -
#4x^2+3xy^2-6x^2y=y^3#
#8x+[(3xdy/dx2y)+(y^2. 3)]-[(6x^2 .dy/dx)+(y. 12x)]=dy/dx3y^2#
#8x+[6xydy/dx+3y^2]-[6x^2dy/dx+12xy]=dy/dx3y^2#
#8x+6xydy/dx+3y^2-6x^2dy/dx-12xy=dy/dx3y^2#
#8x+6xydy/dx+3y^2-6x^2dy/dx-12xy-dy/dx3y^2=0#
#6xydy/dx-6x^2dy/dx-dy/dx3y^2=-8x-3y^2+12xy#
#dy/dx(6xy-6x^2-3y^2)=-8x-3y^2+12xy#
#dy/dx=(-8x-3y^2+12xy)/(6xy-6x^2-3y^2)#
