Question

Find dy/dx for 4 - xy = y^3

Answer 1

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

Explanation:

4-xy=y^3

dy/dx=?

`(d(4-xy))/(dx)=(d(y^3))/(dx)\\ (d(4))/(dx)-(d(xy))/(dx)=3y^(3-1)(dy)/(dx)\\ 0-((dx)/(dx)y+x(dy)/(dx))=3y^2(dy)/(dx)\\ -(1y+x(dy)/(dx))=3y^2(dy)/(dx)\\ -(y+x(dy)/(dx))=3y^2(dy)/(dx)\\ -y-x(dy)/(dx)=3y^2(dy)/(dx)`

Solving for dy/dx: Addind x dy/dx both sides of the equation:

`-y-x(dy)/(dx)+x(dy)/(dx)=3y^2(dy)/(dx)+x(dy)/(dx) \\ -y=3y^2(dy)/(dx)+x(dy)/(dx)`

Common factor dy/dx on the right side of the equation:

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

Dividing both sides of the equation by 3y^2+x:

`(-y)/(3y^2+x)=((3y^2+x))/(3y^2+x)(dy)/(dx)\\ -(y)/(3y^2+x)=(dy)/(dx)\\ (dy)/(dx)=-(y)/(3y^2+x)`