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 $0\times x^{-1}$
$\endgroup$ 15 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $\binom{0}{1} = 0$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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 $0^0=1$ (?). And then $0$ is in the domain of the function.
$\endgroup$ 11 $\begingroup$To provide more to the question:
What about the following proof?
Let $f(x) = x^0 = 1, \forall x \neq 0$,
$f^{'}(x) = \lim_{\delta x\rightarrow 0}\frac{f(x+\delta x) - f(x)}{\delta x} = \lim_{\delta x\rightarrow 0}\frac{(x+\delta x)^0 - (x)^0}{\delta x} = \lim_{\delta x\rightarrow 0} \frac{1 - 1}{\delta x} = \lim_{\delta x\rightarrow 0} \frac{0}{\delta x} = 0$
I think is correct, but the last step seems dubious.
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