Question

Find dy/dx, given x^2y^3 + 2 - 3y = x^2

would anyne know how to solve this equation step by step? im having trouble understanding how to find dy/dx

Answer 1

start with the product rule to get \[2xy^3+3x^2y^2y'-3y'=2x\]

Answer 2

if that is not clear, let me know which step

Answer 3

solve for \(y'\) to finish it

Answer 4

My answer came out to \[\frac{2x-2xy^3 }{ 3y^2x^2-3 }\] yet the actual answer is \[\frac{ -2x-2xy^3 }{ 3y^2x^2-3 }\] and im not sure where that negative came from? do you mind explaining how?