Is it logical to attempt differentiation of y=1? Is y=1 different from y=x^0? I am wary of getting in over my head, after initial searching it is apparent to me that I don't know how to properly phrase this question for the answer I want. I am only working at high school level but in class we learnt differentiation and how when 
$y = ax^n$, 
$ \frac{dy}{dx} = anx^{(n-1)}$.
I was wondering if this applies to when
$y = 1$,
$\frac{dy}{dx} = 1\times1^0 = 1$.
and also when
$y = 1^2$,
$\frac{dy}{dx} = 2\times1^1 = 2$.
Obviously the gradient should be 0, and when calculated in terms of x it makes sense
$y = x^0$,
$\frac{dy}{dx} = 0\times \frac{1}{x} = 0$.
I was wondering how this should be approached, since to me it implies that you CANNOT have a line without a $y$ AND $x$ term, yet $y=1$ CAN be drawn.
Is y=1 a different thing to $y=x^0$? Does $y=1$ even exist in 2D space? Or do you just have to simplify your equation before differentiating it?

$$y=x^{0}$$ is only defined if $$x\ne 0$$ but $$y=1$$ is defined for all real $x$
if $$y=1$$ then we get the first derivative as zero, since $1$ is constant
