Talk:Power rule

Curiosity

Consider x^2. At x=3, the derivative is 6. If graphed, the curve is plotted through x=1[1], x=2[4], x=3[9], x=4[16] etc. Derivative at x=3 is 6, the average of the prior tangent x=2 to x=3 (a line 4y to 9y), which is 5, and x=3 to x=4 (a line 9y to 16y), which is 7 ... (5+7)/2 = 6. This makes sense, since the x^2 derivative at x=3 represents an angle midway between x=2 and x=4. It's the average of prior and latter tangents.

Now consider x^3. At x=3, the derivative is 27. For x^3, x=2 is 8, x=3 is 27, and x=4 is 64. The tangent from 8 to 27 is 19; the tangent from 27 to 64 is 37. The average of prior and latter tangents is 28. Isn't is odd that x^3, which increases faster than x^2, has derivatives that are less than the prior and latter tangent averages?

Sure, there are algebraic proofs. But they rely, as all know, on using (h) to represent a very small value at first, then changing its value to zero later. Sometimes common sense makes its demands. Please explain why the power rule produces results that seem conspicuously illogical. BrianMCoyle

(I have no idea how many years ago the above was written, but I'm answering it anyway.) y = x^2, or any quadratic function with a positive x^2 term, increases faster and faster as you go further along. So if you take any three equally spaced x-values, y = x^2 at the middle value will be less than the average of the other two y-values.

Since the derivative of x^3 is a quadratic (namely 3x^2), it has the same property. It's precisely because it increases faster than linear growth that the derivative at any point is less than the average of the derivatives at equally spaced points on either side. Common sense and mathematical logic are in perfect agreement. 2.24.117.101 (talk) 01:20, 8 May 2015 (UTC)

Proof for complex plane

AxelBoldt edited the proof for this page. However, while his proof is more standard, it only applies for r a positive integer, whereas my proof can be extended to the whole complex plane. Essentially write x^n as exp(nln(x)) and differentiate, since the derivatives of e^x and ln(x) respectively are trivial to prove in the complex plane.

But the title of the article speaks of polynomials. If r is not a positive integer, then x^r is not a polynomial funtion. 193.48.101.101 15:03 19 Jun 2003 (UTC)

But wouldn't it be beneficial to have a proof that is not only non standard (which means that students, for example, would be exposed to multiple proofs which could widen their understanding) but can also be easily generalized to other applications then just the obvious (which would assist people who are looking for these other applications, but an unsure where to look)? Dented42 (talk) 02:25, 24 January 2010 (UTC)

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The derivative of e^x in the complex plane is trivial if you define e^x as a power series, and then differentiate the power series term by term. In order to do the latter, you need to know the derivative of x^k for positive integers k. AxelBoldt 16:24 Dec 10, 2002 (UTC)

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Is it too much to expect User:Hawthorn to explain his edit? Pizza Puzzle

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I explained my edit comprehensively on your talk page. Is it too much to expect you to explain why you have decided to reinsert a whole bunch of trash since then. I'll clean up the page one more time, but I have not got the time or patience to come back and do it over and over again. My regular job (as a mathematician!) takes up too much of my time. Please THINK before you reinsert rubbish on this page. Don't insert irrelevant stuff. Don't insert incorrect stuff. Don't put in strange headings. Stick the topic (polynomials). And make at least some effort to be clear and concise and not waffle. Sheesh!

• Deleted rubbish

(If f(x) = x^r, for some real number r; f '(x) = r x^r-1)

It is unnecessary - an example. It is very bad mathematical writing to clutter up a definition with examples. Put them after the definition. Furthermore "for some real number r" is just rubbish. In the definition r ranges from 0 to n so you are contradicting the definition within itself - bad bad idea. Furthermore there are problems with talking about a^r for a general r which you have completely ignored. Think about $(-1)^{\frac {1}{2}}$ for example. Tell me what $(-1)^{\pi }$ means? Unless you want to talk about this stuff - and you shouldn't need to on a page about polynomials - then stay away from anything other than positive integer exponents.

• Deleted more rubbish

(except when r = -1)

Since r ranges from 0 to n in the definition, it can never be equal to -1. So the comment is completely unnecessary.

• The heading ```differentiation is unnecessary. You are giving examples of use of the formulae. It doesn't need a separate heading. It should follow straight on from the formulae.
• Only a small number of examples need be given. Not endlessly many. And if you are going to give examples, then you need examples of integration as well as differentiation.
• Deleted more strange stuff

These results can be verified with an understanding of Newton's difference quotient and the binomial theorem. One can also derive the General Power Rule via the Chain Rule. A more complex definition of the GPR, for some real number r and some differentiable function f(x), is: f '(x) = r[f(x)]^{r - 1}(f '(x)) = rf(x)^{r - 1}f '(x). For example, if f(x) was 3x¹; then f '(x) = 1 · f(x)⁰ · 3 = 3.

The first sentence is unnecessary since a proof is given later. Your more complex GPR is just a direct consequence of the chain rule. In any case you are back to your `for any real r' habits again, completely ignoring the problems that arise for anything other than positive integer r. This stuff shouldn't be on this page. This page is ostensibly about differentiation of polynomials. These are not polynomials. They don't need to be here.

• Deleted Informal Proof section.

==Informal Proof==

Consider the derivative of x^r; where r is a positive integer, greater than 1. This derivative (as h approaches 0) equals: [(x + h)^r - x^r] / h (the difference quotient); which (as h approaches 0) equals: x^r + rx^{r - 1}(h) + {[r(r - 1)x^{r - 2}] / 2}(h)² + ... + (h)^r - x^r} / h (a binomial expansion); which equals (as h approaches 0): x^r + rx^{r - 1} - x^r; which equals rx^{r - 1}.

My question is, why is this even here. There is a good proof below. That proof is pretty much the same as this one. Why repeat it? Better to give one nice simple proof, not a whole bunch. This one is not as clear as the proper one which follows. You talk about negative r again, which is unnecessary on a page about polynomials. The page is about polynomials, so why are you even mentioning other than positive integer exponents.

• Fixed heading on Proof. The section gives a proof of the derivative formula. It isn't about linearity. I don't understand why you want to put this stupid heading here.

OK. If you want to actually improve the page ...

• It could do with a section of links to related topics. In particular it needs more links to the other calculus pages.
• The C(n,k) notation in the proof should be cleared up, and replaced with more standard ${}^{n}C_{k}$ notation.
• A proof for the integration result could be added.
• If you want to talk more about negative exponents and so on, put them in the generalisations section at the end. But keep this very short and brief, and try to give mostly links to other math pages.

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• The heading ```differentiation is unnecessary. You are giving examples of use of the formulae. It doesn't need a separate heading. It should follow straight on from the formulae.
  • By using a heading, the reader can easily skip to a portion of the text which they are looking for.
• Only a small number of examples need be given. Not endlessly many. And if you are going to give examples, then you need examples of integration as well as differentiation.
  • While the number of examples should be limited, there is no reason that one should be required to write examples for both integration and differentiation, this is a work in progress.
• These results can be verified with an understanding of Newton's difference quotient and the binomial theorem. is unnecessary since a proof is given later.
  • It is not unnecessary. The later proof does not link to Newton, his quotient, or the binomial theorem.
• Your more complex GPR is just a direct consequence of the chain rule. This stuff shouldn't be on this page.
  • Perhaps so, but it would be more productive if you would move the text to chain rule, instead of deleting it.
• This page is ostensibly about differentiation of polynomials. These are not polynomials. They don't need to be here.
  • A good point, I came to this page from a link which was not ostensibly about polynomials. Perhaps a new page should be created.
• Deleted Informal Proof section. My question is, why is this even here. There is a good proof below. That proof is pretty much the same as this one.
  • The proofs differ notably in appearance and terminology, a reader may not understand the terms for one proof, but might understand those for the other.

Pizza Puzzle

We don't need limits to differentiate polynomials, so why are you using them?

michaelliv 22:54, 6 July 2007 (UTC)

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Power law or polynomials?

The true importance of this rule is for when n is a real number. The proof contained not only uses the binomial theorem but is also useless for when n is real. What about this http://math.wikia.com/wiki/Elementary_power_rule_of_derivatives/Proof ? I know the article is titled Calculus with polynomials so perhaps the name should also be changed.

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The page Differentiation_rules explicitly lists a generalized form of the power rule that is hinted at Generalization, but is not explicitly stated. shouldn't it be listed here as well? Dented42 (talk) 02:35, 24 January 2010 (UTC)

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I have read the proof of the power differentiation, and have slightly simplified the explanation, so that people understand that while 0^n=0, 0^0=1 (ie. the limit as h--> 0 for that final term will not cancel that one term, giving us the proof. Duderman1010 12:23, 20 June 2010 (AEST)

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The proof so shown applies only for an positive-integer value of n. Further expansion is necessary for

1. the trivial case of n=0 (because c for any real x other than 0 is the constant c, and the derivative of the real constant c is then f'x = n = 0; for x = 0. the limit as x→0 of x⁰ is also a constant, so in such a case f'x = n = 0

2. n as a negative integer, in which case n = -m but m is a positive integer, and f'(x) can easily be solved with the quotient rule, in which case f(x) = 1/x^m, and

f'(x) =  [0-mxm]/(xm)^2 = -mx-m-1

But m= -n, so

f'(x) = nx^n-1, and with these further proofs, the polynomial rule can be extended to all integers.

Now for rational numbers, any rational number n is understood to be the expression p/q in which p and q are integers but q≠0.

Then f(x) = x^p/q. The laws of exponents allow

[f(x)]^q = x^p

Differentiation of both sides of the equation is then possible, and eventually

q[z^q-1]f'(x) = px^p-1 (note that the chain rule puts f'(x) in the equation after mutual differentiation), and

f'(x) = (p/q)x^p/q-1

and of course f'(x) = nx^n-1 through substitution.

For n irrational but real, n is between two rational numbers p and q and the form for rational numbers is reasonably close to a rational solution.

There. We have added appropriate rigor.Pbrower2a (talk) 00:16, 18 June 2011 (UTC)

Proof

Frankly, the proof in this section appears unsightly and unwieldy to me. If we're only interested in proving it for integer n, then an inductive proof using the product rule seems to make more sense, not to mention it can be done in just a few lines, and is (in my opinion, at least) considerably easier to follow and more intuitive than the one presented in the article. Here's my proposal:

We plan to prove that ${\frac {d}{dx}}x^{n}=nx^{n-1}$. We begin by noting the trivial case:

${\frac {d}{dx}}x^{0}={\frac {d}{dx}}1=0=0\cdot x^{-1}$

• Removed above remark from main body of article. The power rule is not applicable when n = 0 and x = 0, as it yields the undefined form 0/0. To differentiate f(x) = 1 at x = 0, one needs to use the more fundamental result of the derivative of a constant function. Slider142 (talk) 07:13, 24 April 2012 (UTC)

Now let's suppose that for some natural number $k$, we can show that ${\frac {d}{dx}}x^{k}=kx^{k-1}$. Then we can use the product rule to show that:

${\frac {d}{dx}}x^{k+1}={\frac {d}{dx}}x\cdot x^{k}=x\cdot {\frac {d}{dx}}x^{k}+x^{k}{\frac {d}{dx}}x=x\cdot k\cdot x^{k-1}+x^{k}\cdot 1=(k+1)x^{k}$

So, by the principle of mathematical induction, we've established the above formula for all nonnegative integer values of $n$. 74.94.233.237 (talk) 03:37, 11 August 2011 (UTC)

Can I add a proof when n is any positive integer, the power rule is true by induction and product rule (which is the proof shown above by User 74.94.233.237) to the page? by Raycheng200 (talk )contribs) 05:52, 12 June 2016 (UTC)

Rename to power rule

I don't think the subject of this article is very notable compared to the power rule which has a good history. Currently power rule redirects to here, I propose to move this article to there instead and just have the polynomials part be an extension or use of the power rule Dmcq (talk) 09:01, 15 February 2012 (UTC) Rachel bean

Generalization to rational exponents by implicit differentiation

Here’s a simpler way to generalize the power rule to rational exponents using implicit differentiation that’s more straightforward than the proof in the article, which considers different forms of rational exponents separately and applies the chain rule to combine them. 49.147.83.13 (talk) 22:59, 27 May 2020 (UTC)

Upon proving that the power rule holds for integer exponents, the rule can be extended to rational exponents.

Let $y=x^{r}=x^{p/q}$, where $p,q\in \mathbb {Z}$ so that $r\in \mathbb {Q}$

Then,

$y^{q}=x^{p}$

$qy^{q-1}{\frac {dy}{dx}}=px^{p-1}$

Solving for $dy/dx$,

${\frac {dy}{dx}}={\frac {px^{p-1}}{qy^{q-1}}}.$

Since $y=x^{p/q}$,

${\frac {d}{dx}}x^{p/q}={\frac {px^{p-1}}{qx^{p-p/q}}}.$

Applying laws of exponents,

${\frac {d}{dx}}x^{p/q}={\frac {p}{q}}x^{p-1}x^{-p+p/q}={\frac {p}{q}}x^{p/q-1}.$

Thus, letting $r=p/q$, we can conclude that ${\frac {d}{dx}}x^{r}=rx^{r-1}$ when $r$ is a rational number.