How do you find #(dy)/(dx)# given #-2y^3-3y+4=2x^3#?

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Answer 1 · 2016-10-27T04:18:42

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

#-2y^3-3y+4=2x^3#

Differentiating, we get
#(-2y^3)'-(3y)'+(4)'=(2x^3)'#

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

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

Dividing by 3

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

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