Power Rule + Chain Rule Question

[Kelsier42]

Hey I've been messing around with bringing powers down and I've discovered some... interesting things.
So first as I'm sure we all have done countless times, the general power rule for finding the derivative of a function to any degree.

Clean, simple, we've done it a million times. Got it. I started wondering as I toyed with this rule, does it apply to constants? The math checks out... except it doesn't.
All constants are just a constant to the power of 1 right? So if we do a little something like this,

Hey what do you know the chain + power rule and any constant's derivative is 0, just like we all know.
Except it starts to fall apart when you take the derivative of 0.
0 is to the first power, so doing the same power rule + chain we get startling answers.

Not only is that indeterminate, its an infinite amount of indeterminates. And that's like, bad mojo man.
So yeah I've confused myself and I want clarification on how this rule functions. The derivative of 0 is 0 yeah?
Any pieces of insight would be greatly appreciated, thanks for your time!

 

[Dr.Peterson]

The power rule applies when the base is x, not a constant. A constant is not x^1; it's Cx^0.

 

[ksdhart2]

I'm not sure I understand your reasoning on the supposed application of the chain rule. In particular, I'm confused by this term:

\(\displaystyle y = x^2\)

\(\displaystyle y^{\prime} = 2x^{1} \cdot {\color{red}1}\)

Where does this \(\displaystyle {\color{red}1}\) come from? Best as I can tell from your other "examples", it appears to be because there's a 1 in the exponent, but that train of logic completely breaks down if we look at a different example:

\(\displaystyle y = x^3\)

\(\displaystyle y^{\prime} \stackrel{?}{=} 3x^{2} \cdot 2\)

\(\displaystyle y^{\prime} \neq 6x^{2}\)

Can you explain your thinking more clearly?

 

[HallsofIvy]

I'm not sure I understand your reasoning on the supposed application of the chain rule. In particular, I'm confused by this term:

\(\displaystyle y = x^2\)

\(\displaystyle y^{\prime} = 2x^{1} \cdot {\color{red}1}\)

Where does this \(\displaystyle {\color{red}1}\) come from? Best as I can tell from your other "examples", it appears to be because there's a 1 in the exponent, but that train of logic completely breaks down if we look at a different example:

No, it's not because "there's a 1 in the exponent". He's applying the chain rule: The derivative of (a(x))^2 is 2a(x) times the derivative of a(x). Here a(x)= x so the derivative of x is 1.
It's a bit of "overkill" since typically one learns that the derivative of \(\displaystyle x^2\) is 2x before one learns the chain rule!

\(\displaystyle y = x^3\)

\(\displaystyle y^{\prime} \stackrel{?}{=} 3x^{2} \cdot 2\)

\(\displaystyle y^{\prime} \neq 6x^{2}\)

Can you explain your thinking more clearly?

 

[ksdhart2]

No, it's not because "there's a 1 in the exponent". He's applying the chain rule: The derivative of (a(x))^2 is 2a(x) times the derivative of a(x). Here a(x)= x so the derivative of x is 1.
It's a bit of "overkill" since typically one learns that the derivative of \(\displaystyle x^2\) is 2x before one learns the chain rule!

Okay, yeah, I guess that makes sense. The whole thing's still really stupid though, in my opinion. Plus, even with this idea in mind, I can't see any reason why it would end up with an infinite chain. Letting \(a(x) = 0\), and thus \(a^{\prime}(x) = 0\), we have:

\(\displaystyle y = a(x) = 0^1\)

\(\displaystyle y^{\prime} = 1 \times a(x)^0 \times a^{\prime}(x) = 1 \times 0^0 \times 0\)

No infinite chain to be found, but still a heaping load of nonsense, because \(0^0\) is undefined.

 

[HallsofIvy]

Yes, it is a "heaping load of nonsense" because only a fool would apply the "power rule" to the constant function f(x)= 0.

 

[JeffM]

Okay, yeah, I guess that makes sense. The whole thing's still really stupid though, in my opinion. Plus, even with this idea in mind, I can't see any reason why it would end up with an infinite chain. Letting \(a(x) = 0\), and thus \(a^{\prime}(x) = 0\), we have:

\(\displaystyle y = a(x) = 0^1\)

\(\displaystyle y^{\prime} = 1 \times a(x)^0 \times a^{\prime}(x) = 1 \times 0^0 \times 0\)

No infinite chain to be found, but still a heaping load of nonsense, because \(0^0\) is undefined.

Technically, isn't [MATH]0^0[/MATH] undefined?

Furthermore,

[MATH]f(x) = x^0 \text { if } x \ne 0 \implies f(x) = 1 \implies f'(x) = 0 \ne x^{-1}.[/MATH]
So I don't know why we are even talking about the power rule with respect to [MATH]x^0.[/MATH]