Question

How can I implicitly differentiate `(x-y)^3=0`?

Answer 1

If you insist on using implicit differentiation, you'd write:

`(x-y)^3=0`

`d/(dx)((x-y)^3)=d/(dx)(0)`

`3(x-y)^2 d/(dx)(x-y) = 0`

`3(x-y)^2 (1-dy/dx) = 0`

So `3(x-y)^2 - 3(x-y)^2dy/dx = 0`

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

And `(x-y)^2 = (x-y)^2dy/dx`.

Now we can say, either `x-y=0`, or we can divide by `x-y` to get `dy/dx = 1`

The difficulty is that the only solutions to `(x-y)^3=0` are `x-y=0`
so `y=x` and `dy/dx =1`. Why use implicit differentiation?

Answer 2

`dy/dx=1`

Rewrite it as `x^3 -3x^2y+3xy^2 -y^3`=0

Now differentiate term by term with respect to x,

`3x^2-6xy -3x^2dy/dx +3y^2 +6xydy/dx -3y^2 dy/dx`=0

`-3(x^2 -2xy+y^2)dy/dx= -3x^2 +6xy -3y^2`

`dy/dx=1`