Question

What is the `dy/dx` of `y^3 - 3xy = 2`?

Answer 1

I found: `(dy)/(dx)=y/(y^2-x)`

Explanation:

Here we use implicid differentiation; we trat `y` as a function of `x` and we differentiate it as well. So for example, if you have `y^2` the derivative will be `2y(dy)/(dx)` !!!!
In our case we get:
`3y^2(dy)/(dx)-3y-3x(dy)/(dx)=0`
where in the second term on the left I used the Product Rule.
Now, collect `(dy)/(dx)`:
`(dy)/(dx)=(3y)/(3(y^2-x))`
and finally:
`(dy)/(dx)=y/(y^2-x)`