Question

What is `(dy)/dx` if `x^2y-xy^2=1`?

Answer 1

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

Explanation:

Here you need Implicit Differentiation. Your function is not easily written as `y=f(x)` so what you do is to derive `y` as well as a function of `x`;

for example, if you have:
`y^2`
then differentiating you get:
`2y(dy)/(dx)` where the `(dy)/(dx)` takes into account that your `y` is still a function of `x`.

In our case we have:

`2xy+x^2(dy)/(dx)-y^2-2xy(dy)/(dx)=0`

where we used the Product Rule and the fact that the constant gives zero in differentiation.

Rearranging we isolate `(dy)/(dx)`:

`(dy)/(dx)[x^2-2xy]=y^2-2xy`
and:
`(dy)/(dx)=(y^2-2xy)/(x^2-2xy)`