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It is my first week of calculus and I am struggling with understanding something related to the power rule of exponents.

First off, we are working on derivatives. The problem that came up was the following:

$$\frac{(d)}{(dx)}(8)=0$$

As we work through the problem, it becomes

$$8(0x^{0-1})$$

And then it is explained by the professor that "per the power rule we just talked about, anything raised to the power of zero, is zero." Mathway confirms that the answer to this problem is 0. And the rules of a derivatives state the answer will always be 0.

I guess my problem is that a couple of weeks ago, I watched a video that shows that zero raised to the power of zero is 1. My calculator confirms this, Eddie Woo confirms why in a Youtube video.

So why, when it comes to derivatives, this rule is not true and if it's not true, then why is this "power rule" a rule? Shouldn't the rule say, "zero to the power of zero is always 1, except, when you're working on derivatives, then it's zero"

A little confused. I will not lie, I am not always great at mathematics concepts, so maybe it's something I missed in a previous lesson or maybe I just don't have a strong understanding of derivatives yet. I took Algebra 3 years ago and just returning back to university so it may be that I forgot some rule or that I am just bonkers. Let me know.

Neon Storm
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    It's weird to use the power rule to prove that the derivative of a constant is zero. Can you show us more of that argument? – Karl Oct 18 '23 at 04:40
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    $0^0$ is an indeterminate form. Those limits need to be handled delicately. – Sean Roberson Oct 18 '23 at 04:40
  • for real $x > 0$ we have $x^0 = 1$ but $0^x = 0.$ So, what would you like $0^0$ to be? – Will Jagy Oct 18 '23 at 04:46
  • I guess I would expect it to be 1 since that's what every calculator says. – Neon Storm Oct 18 '23 at 04:51
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    Thank you Sean, you have lead me down the right path of looking up "Indeterminate Forms." I guess I should have been taught that prior but I skipped precalculus. – Neon Storm Oct 18 '23 at 04:55
  • Anything raised to a power of $0$ is equal to $1$... not $0$ – RyRy the Fly Guy Oct 18 '23 at 05:14
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    There are two different ways to interpret $0^0$ and argue what it should be. One argument is that one should always start with the multiplicative identity, 1, when working with multiplication or exponents. Since an exponent says how many of a given base are multiplied together, $0^0$ is saying "zero 0's multiplied together". If we start with 1, 1 multiplied by no zeros remains 1, so $0^0 = 1$. On the other hand, using exponent rules, $0^0 = 0^{1-1} = \frac{0^1}{0^1} = \frac{0}{0}$, which is indeterminate. – Nate Oct 18 '23 at 05:24
  • "Anything raised to a power of 0 is equal to 1... not 0 " - that was my point in the question RyRy the Fly Guy. The answer to the derivative problem above is equal to ONE even though it's 0 to the power of 0. – Neon Storm Oct 18 '23 at 05:33
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    Thanks Nate, starting to understand. – Neon Storm Oct 18 '23 at 05:33

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The expression $8(0x^{0-1})$ has no meaning when $x = 0$, exactly because $x^{0-1} = x^{-1}$ is not defined for $x=0$. This illustrates the following phenomenon: if we use a rule without thinking about its requirements, we can arrive in absurd or nonsensical results.

We have to be careful when applying some "rules" in math. The so-called power rule is actually a theorem: when we recall that weird definition of a derivative using limits, whenever $f(x) = x^\alpha$ for $\alpha$ a real number different than $0$, we can work out the calculations to conclude that $f'(x) = \alpha x^{\alpha -1}$. In most proofs, we really need that $\alpha \ne 0$. So we have to take extra care when thinking about the power rule for $\alpha = 0$. The hypotheses require $\alpha \ne 0$ in the general case.

For $\alpha = 0$, if you take any other nonzero real number, this number raised to $0$ is going to be $1$ (not $0$ as you wrote in your question). So $384^0 = \pi^0 = (-27)^0 = 1$; as long as the base is not zero, the powers are all zero. Using the definition of the derivative at a point, we have that $\frac{d}{dx} x^0 = 0$ for $x \ne 0$. As you pointed out, we can actually set $0^0 = 1$ with no problems. In that case $0^0$ is also a constant so $\frac{d}{dx} x^0 = 0$ for $x = 0$ too!

So what is wrong? Well, the thing is that when $x \ne 0$, the number $x^{-1}$ is well defined. So the function that sends $x$ to $0\cdot x^{0-1}$ is also well defined and constant $= 0$ for all $x \ne 0$. Therefore, $\frac{d}{dx} x^{0} = 0x^{0-1}$ for $x \ne 0$. For $x = 0$, the expression $x^{-1}$ doesn't make sense, so the best we can do is $\frac{d}{dx} x^0 = 0$. But any expression that would use $x^{-1}$ doesn't make sense!

Another basic arithmetic "rule" is the following: if $ab = ac$ and $a$ is nonzero, then $b = c$. There are lots of fake proofs that $1 = 2$ which all essentially discard the requirement for $a \ne 0$. $0\cdot 3 = 0\cdot 7$, but certainly $3 \ne 7$. Always check the requirements of a theorem before applying it! :)

Lucas Henrique
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  • Just a comment about $0^0 = 1$. This definition is quite controversial among non-mathematicians because people usually overlook the hypotheses to apply theorems. For example, if you consider the limit $\lim f(x)^{g(x)}$ and $f(x), g(x) \to 0$, it is not true that $\lim f(x)^{g(x)} \to 0^0$ in general, while other limits like $\lim f(x) + g(x) \to 0 + 0$, so we have to be careful. In academia, $0^0$ is fine and is actually convienient for e.g. algebraists. Since there are not lots of uses of $0^0$ for everyday, it's pedagogically easier to refer to it as "indeterminate". – Lucas Henrique Oct 18 '23 at 05:15
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    Thanks Lucas, all of this is starting to come together. I appreciate you writing all of that out. It really helps me visualize why my thinking needs to be somewhat flexible despite there being "rules" per se. – Neon Storm Oct 18 '23 at 05:34