find by implicit form x^3 - 3x^2y + 2xy^2 =12

Question

calculus chapter 2 section 2.5

Answer 1

Implicit differentiation produces a function involving the two variables, instead of just one. If we differentiate x^3-3x^2y+2xy^2=12 we get: 3x^2-(6xy+3x^2y')+(2y^2+4xyy')=0, where the brackets show the result of differentiating one term by parts. The brackets aren't necessary, but I hope they show the working. If we open the brackets and group terms we get: 3x^2-6xy+y'(4xy-3x^2)+2y^2=0. So dy/dx=y'=(6xy-3x^2-2y^2)/(4xy-3x^2) or (2y^2-6xy+3x^2)/(3x^2-4xy) or (2y^2-6xy+3x^2)/x(3x-4y).