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mikau

Member

Registered: 2005-08-22

Posts: 1,504

Proof the derivative of x^n = nx^(n-1)

I can't speak for everyone here but my mathbook used binomial series to prove the derivative of x^n = nx^(n-1) but this only proved it for positive integer values of n. The book said "This proof is only valid for positive integer values of n, however the formula holds true for all real values of n". I find this sort of incomplete proof unfullfilling and I've been curious as to why it holds true for values of n such as 1/2. However I was pondering about it a few minutes ago and I think I came up with an absurdly simple proof to show the derivative of x^n = nx^(n-1) for all real values of n using logarithmic differentiation. If your book used binomial series as well, you may find this more satisfying.

Let y = x^n where n is some real number.

ln y = n ln x   took natural log of both sides

1/y dy = n 1/x dx   differentiated

dy/dx = n y/x         rearanged

dy/dx= n (x^n)/x      subsituted for y

dy/dx = n x^(n-1)

And thats it! Absurdly simple isn't it?

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A logarithm is just a misspelled algorithm.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Proof the derivative of x^n = nx^(n-1)

If you're taking the natural log of both sides, then that means that x is restricted to being positive. It doesn't have to be an integer though, so your proof is still more general than your textbook's.

I'd imagine that your book showed you the specialised proof because it hadn't covered logs by that point, and then when it did cover logs it would be silly to say "oh yeah and you can use these to prove that thing that we proved back in that chapter about half a book ago but more generally and stuff".

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Why did the vector cross the road?
It wanted to be normal.

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Proof the derivative of x^n = nx^(n-1)

True, but according to my book, logarithmic differentiation restricts the domain to positive numbers, but "free's up" the domain again if all terms containing the natural log function are elliminated in the differentiation process, as is the case here.

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A logarithm is just a misspelled algorithm.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Proof the derivative of x^n = nx^(n-1)

Yeah, your proof is restricted to the base nonnegative, while the proof by binomial expension is restricted to exponent natural numbers.

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X'(y-Xβ)=0

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Proof the derivative of x^n = nx^(n-1)

Very well, then lets prove it for negative numbers without using binomial theorem. We'll take into account we already proved it for postive real values of n.

Let n be some negative real number, and u be come positive real number such that n = -u

y = x^n

y = x^-u

y = 1/x^u

using the derivative of a quotient theorem. The derivative of a quotient is the denominator times the derivative of the numerator over the denominator squared, minus the numerator times the derivative of the denominator over the denominator squared. The first half dissapears as the derivative of the numerator is 0, the second half requires we find the derivative of x^u, u is a positive number so we can find its derivative is u x^(u -1) as previously proved.

dy/dx =   ( -ux^u-1 )/ ( x^2u )

dy/dx  = -u x^(-u - 1)

n = -u, substitue:

dy/dx = n x^(n-1)  QED.

Last edited by mikau (2006-05-09 04:43:31)

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A logarithm is just a misspelled algorithm.

Alkirin

Member

Registered: 2007-11-05

Posts: 1

Re: Proof the derivative of x^n = nx^(n-1)

Question.

I just wrote up an equation and subsequent proof by induction for the derivitive of x^1/2 for any (n+1)th  derivitive.

That is, you can plug in any number for N (say, 5) and get the Nth derivitive for x^1/2

My question is...has this already been published?  If not, I would like to publish it.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Proof the derivative of x^n = nx^(n-1)

Alkirin, generally speaking if you find any derivative, it's already been known.  General formula or not.

For a general proof, you can use the product rule for integers, the quotient rule for rationals, and the fact that the rationals are dense in the reals for real values.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Proof the derivative of x^n = nx^(n-1)

Post 5 is creative, how nice it would be if a textbook include this piece of proof!

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X'(y-Xβ)=0

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Proof the derivative of x^n = nx^(n-1)

why, thank you!

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A logarithm is just a misspelled algorithm.