Question

How do you differentiate `2x-y=y^2-x^2/y`?

Answer 1

Use implicit differentiation. (Please see the explanation)

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

Explanation:

The process is called implicit differentiation.

Multiply both sides by -y:

`-2xy + y^2 = -y^3 + x^2`

Move all of the terms containing y to the left side:

`y^3 + y^2 - 2xy = x^2`

Differentiate remembering that `(df(y))/dx = ((df(y))/dy)(dy/dx)`:

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

Note: Differentiation of the term `-2xy` required the use of the product rule `(gh)' = g'h + gh'` where `g = -2x , g' = -2, h = y and h' = dy/dx`

Move all of the terms NOT containing `dy/dx` to the right:

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

Factor out `dy/dx`:

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

Divide both sides by `(3y^2 + 2y - 2x)`:

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