Question

What is the implicit derivative of `1= xy^3-3xy`?

Answer 1

It is `dy/dx=(y*(y^2-3))/(3x(1-y^2))`

Explanation:

It is

#(1')=x'*y^3+x*3*y^2*y'-3x'y-3xy'=> 0=y^3+3xy^2*y'-3y-3xy'=> y'(3x-3xy^2)=y^3-3y=> dy/dx=(y^3-3y)/(3x-3xy^2)#

where `()'=d/dx`