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 $0\times x^{-1}$

If you follow your thought you should get $n=0$ and then
$$\frac{d}{dx}x^0=0\cdot x^{0-1}=0$$
EDIT
After Thomas's comment I realized that in fact we can define $0^0=1$ (https://www.quora.com/What-is-0-0-the-zeroth-power-of-zero-1/answers/1373648?). And then $0$ is in the domain of the function.
