# 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 $0\times x^{-1}$

• $\frac{d}{dx}x^0=0x^{-1}$. You forgot that it is $nx^{n-1}$. Commented Feb 25, 2017 at 1:49

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.

• oh wow! you are totally right! hand to face Commented Feb 25, 2017 at 1:43

This is actually quite an interesting question: To provide some context, I'll include an example of a possible proof of $f(x) = x^{n} \Rightarrow f^{'}(x) = nx^{n-1}$

From the definition: $f^{'}(x) = \lim_{\delta x\rightarrow 0}\frac{f(x+\delta x) - f(x)}{\delta x}$

$\Rightarrow f^{'}(x) = \lim_{\delta x\rightarrow 0}\frac{(x+\delta x)^{n} - x^{n}}{\delta x}$

$= \lim_{\delta x\rightarrow 0}\frac{x^{n} + \binom{n}{1}x^{n-1}\delta x + \binom{n}{2} x^{n-2}\delta x^{2}+...+\delta x^{n} - x^{n}}{\delta{x}}$

$\Rightarrow f^{'}(x) = \binom{n}{1}x^{n-1}$

Now if we take the case $n=0$, it is clear to see that $f^{'}(x) = 0$ since $\binom{0}{1} = 0$

• Nice example (+1) and thank you.:) Commented Feb 25, 2017 at 2:06

Symbols aren't magic.

$f(x) = x^0$ means $f(x) = 1; x \ne 0;f(0)$ undefined.

$f'(x) = 0; x \ne 0$. Because $f$ is a constant function. That's all there is to it.

If one want to be clever, or so called clever.

$f(x) = x^k; k = 0$ so $f'(x) = kx^{k-1} = 0*x^{-1} = 0; x \ne 0$

so everything is consistent. But it's not magic.

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.

• $0$ is not in the domain of a derivative? Commented Feb 25, 2017 at 1:45
• @Nick Pavini: it is not in the domain of the function $x^0$, 'cause $0^0$ is not defined. Commented Feb 25, 2017 at 1:47
• @NickPavini no, it is in the domain of the derivative, but not in the result using power rule. Commented Feb 25, 2017 at 1:47
• Well, $0^0$ is defined, by a lot of people and I use that $0^0=1$ implicitly all the time. Commented Feb 25, 2017 at 1:56
• @Thomas Andrews: $0^0$ is not defined by a lot of people. Commented Feb 25, 2017 at 2:07

To provide more to the question:

Let $f(x) = x^0 = 1, \forall x \neq 0$,
$f^{'}(x) = \lim_{\delta x\rightarrow 0}\frac{f(x+\delta x) - f(x)}{\delta x} = \lim_{\delta x\rightarrow 0}\frac{(x+\delta x)^0 - (x)^0}{\delta x} = \lim_{\delta x\rightarrow 0} \frac{1 - 1}{\delta x} = \lim_{\delta x\rightarrow 0} \frac{0}{\delta x} = 0$