The derivative of $x^0$ For some reason I have not been able to find a straight answer to this.
We know that $\frac{d}{dx}x^n=nx^{n-1}$
And this is true for $n=-1$ and $n=1$ $\implies$ $\frac{d}{dx}x^{-1}=-1x^{-2}$ and $\frac{d}{dx}x^1=1$
We also know that $\frac{d}{dx}C=0$ where $C$ is a constant.
Suppose that $f(x)=x^0$.
Obviously any number to the power of zero is $1$, i.e. $x^0=1$, and $\frac{d}{dx}1=0$, but $x$ is not a constant. So, $$\frac{d}{dx}x^0=x^{-1}$$
Is this true? My thought is possibly. Based on the fact that if $\frac{d}{dx}x^1=1$ and obviously any value to the power of one is equal to that value. I.e. $x^1$ simplifies to be $C$ a constant but $\frac{d}{dx}x^1\not=0$, and we know that $\frac{d}{dx}C=0$. So is it true that $f'(x)=x^{-1}$? Hopefully this is not way more simple than I am making it.
UPDATE:
I obviously made an error by saying that $\frac{d}{dx}x^0=x^{-1}$ It actually would evaluate directly as $0\times x^{-1}$

You're incorrectly applying the power rule. We have that $\frac{d}{dx}x^n = nx^{n-1}$. For $x^0$, $n = 0$. So $\frac{d}{dx}x^0 = 0x^{-1} = 0$.
$x^1$ does not simplify to a constant. $x^1$ is just $x$, which is anything but constant.
