Question

How do you implicitly differentiate `x+xy-2x^3 = 2`?

Answer 1

Let us define `f(x,y) = x + xy - 2x^3 = 2`.

`(df)/(dx) = (df)/(dy)*(dy)/(dx)`, and so every time you differentiate (`(df)/(dy)`) a part of the function that has `y`, you have to multiply by `(dy)/(dx)`, the derivative of `y` with respect to `x`, for the overall derivative (`(df)/(dx)`) to still be with respect to `x` (even though you're differentiating `y`).

`(df)/(dx) = (d(x))/(dx) + [x*(dy)/(dx) + y*(d(x))/(dx)] - (d(2x^3))/(dx) = cancel((d(0))/(dx))`
(has product rule)

`1 + [x(dy)/(dx) + y] - 6x^2 = 0`

Simple algebra from here:
`x(dy)/(dx)= 6x^2 - 1 - y`

`(dy)/(dx) = (6x^2 - y - 1)/x`