Proof of the derivative

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What's the proof for the derivative?

[tex] \frac{d}{dx} x^n = nx^{n-1} [/tex]

(once again, i've stuffed up the latex and can't fix it!)
It should read d/dx x^n = nx^(n-1)
 
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Hurkyl
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Also, look here for LaTeX help:

https://www.physicsforums.com/showthread.php?t=8997

The thing you're missing is that you have to use braces {} if you want to group a collection of things into one object (like an exponent).
 
HallsofIvy
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You prove the general power rule by using proof by induction and the product rule.

When n= 1, [itex]\frac{dx}{dx}= 1= 1x^0[/itex] so the formula is true.

Assume [itex]\frac{dx^k}{dx}= kx^{k-1}[/itex]. Then
[tex]\frac{dx^{k+1}}{dx}= \frac{d(x(x^k))}{dx}= x\frac{dx^k}{dx}+ \frac{dx}{dx}(x^k}[/tex]
[tex]= x(kx^{k-1}+ 1(x^k)= kx^k+ x^k= (k+1)x^k[/tex]
 
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what if n is not an integer, an irrational number for example. Then how can one prove it without using newton's series?

I know one can prove the rule for rational number using chain rule and implicit differentiation, but what about irrational numbers?
 
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dextercioby
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The generalized binomial formula will solve it. That's right, the one expressing Pochhammer's symbols in terms of the Gamma function.

Daniel.
 
HallsofIvy
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For n not a positive integer, use logarithmic differentiation:

If y= xn, then ln y= n log x.
[tex]\frac{1}{y}\frac{dy}{dx}= \frac{n}{x}[/tex]
[tex]\frac{dy}{dx}= \frac{ny}{x}= \frac{nx^n}{x}= nx^{n-1}[/tex]
 
matt grime
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Or you could use the fact that for non integer/rational exponents the definition of x^k is exp(k logx).
 
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that logrithmic one is clever, ill have to remember that. i always got myself going in circles with induction :(
 
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You prove the general power rule by using proof by induction and the product rule.

When n= 1, [itex]\frac{dx}{dx}= 1= 1x^0[/itex] so the formula is true.

Assume [itex]\frac{dx^k}{dx}= kx^{k-1}[/itex]. Then
[tex]\frac{dx^{k+1}}{dx}= \frac{d(x(x^k))}{dx}= x\frac{dx^k}{dx}+ \frac{dx}{dx}(x^k}[/tex]
[tex]= x(kx^{k-1}+ 1(x^k)= kx^k+ x^k= (k+1)x^k[/tex]


Hi, would you know the name of the classic proof that doesn't use induction?
I have been looking for it but dont know how it was done.

In my first calculus class at the moment and we have been asked to locate it.

thanks.
 
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theres a proof of the power rule involving the binomial theorem. It shuold be in your textbook.
 
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That is in the textbook.

But as an extra assignment he challenged us to look for the classic proof.
Said it was messy and took a good page to write up. It is most likely found in old calculus books. Thats all I've got.

thanks for the reply though,
 
HallsofIvy
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Then google on "binomial theorem" "power rule"
 
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Then google on "binomial theorem" "power rule"
thats like saying go look in a book.
No help whatsoever.
 
HallsofIvy
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thats like saying go look in a book.
No help whatsoever.
That's very insulting to the person who asked the question. Are you implying that he can't read?
 
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what if n is not an integer, an irrational number for example. Then how can one prove it without using newton's series?

I know one can prove the rule for rational number using chain rule and implicit differentiation, but what about irrational numbers?

f(x)=xc,f'(x)=?


[tex]\triangle f=(x+\triangle x)^c -x^c=\left(x+(1+\frac{\triangle x}{x})\right)^c-x^c=
x^c\left(\left(1+\frac{\triangle x}{x}\right)^c-1\right)[/tex].

[tex]\frac{\triangle f}{\triangle x}=x^{c-1}\cdot \frac{\left(1+\frac{\triangle x}{x}\right)^c-1}
{\frac{\triangle x}{x}}=x^{c-1}\cdot \frac{(1+u)^c-1}{u};u=\frac{\triangle x}{x}[/tex]


Lemma:
[tex]\lim_{u\to 0}\frac{(1+u)^c-1}{u}=c[/tex]
Proof:

Let [tex](1+u)^c-1=\phi [/tex].

Becouse the function (1+u)c is continous and defined in u=0 we conclude:
[tex]\lim_{u\to 0}(1+u)^c =1\longrightarrow \lim_{u\to 0}\phi =0[/tex].
[tex](1+u)^c=1+\phi\rightarrow c\cdot ln(1+u)=ln(1+\phi);\frac{c\cdot ln(1+u)}{ln(1+\phi)}=1
\longrightarrow \frac{(1+u)^c-1}{u}=\frac{\phi}{u}=\frac{\phi}{u}\cdot\frac{c\cdot ln(1+u)}{ln(1+\phi)}=
c\frac{\phi}{ln(1+\phi)}\cdot\frac{ln(1+u)}{u}[/tex]

[tex]\lim_{u\to 0}\frac{(1+u)^c-1}{u}=
c\cdot\lim_{\phi \to 0}\frac{1}{\frac{1}{\phi}ln(1+\phi)}\cdot \lim_{u\to 0}\frac{1}{u}ln(1+u)=
c\lim_{\phi \to 0}\frac{1}{ln(1+\phi)^{1/\phi}}\cdot \lim_{u\to 0}ln(1+u)^{1/u}=c[/tex]

(Becouse [itex]\lim_{u\to 0}(1+u)^{\frac{1}{u}}=ln\ e=1[/itex]).This concludes proof of the lemma.

Therefore:

[tex]f'(x)=\lim_{\triangle x \to 0}\frac{\triangle f}{\triangle x}=x^{c-1}\lim_{u\to 0}
\frac{(1+u)^c-1}{u}=x^{c-1}\cdot c[/tex]

QED

dextercioby said:
The generalized binomial formula will solve it. That's right, the one expressing Pochhammer's symbols in terms of the Gamma function.

Daniel.
Possible.
But why tu use nuclear weapon when one can kill prey by a bow and arrow?
 
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arildno
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what if n is not an integer, an irrational number for example. Then how can one prove it without using newton's series?

I know one can prove the rule for rational number using chain rule and implicit differentiation, but what about irrational numbers?
Define, for all x:
[tex]Exp(x)=1+\sum_{i=1}^{\infty}\frac{x^{i}}{i!}[/tex]
This can be proven to be differentiable, with Exp'(x)=Exp(x).
Also, Exp(x) is invertible, so it has an inverse function Log(x).
Now, Log'(x)=1/x, because:
[tex]x=Exp(Log(x))\to1=x*Log'(x)[/tex]
From which it follows trivially.

Now, for any x>0, we may define:
[tex]x^{c}=Exp(cLog(x)),c\in\mathbb{R}[/tex]
Thus, the result required follows easily.
 
mathwonk
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derivative means "linear part of the change in y".

so just expand (x+ h)^n and look at tbhe coefficient of the linear part of the change, i.e. of (x+h)^n - x^n. the linear term (in h) will have coefficient

nx^(n-1).

thats all. derivatives are pretty trivial for polyonm ials. they do not get tricky until you go to transcendental functions.

e.g. (x+h)^3 - x^3 = 3x^2 h + 3 xh^2 + h^3, so the linear part in h, is 3x^2 h, and the coefficient is 3x^2, which is therefore the derivative.
 
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mathwonk
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of course there is a fancy much more difficult limit definition of derivatves that can be checked to agree with this, but I think it is pretty stupid to use that approach to find the derivative of a polynomial.

that approach is only needed for sin or cos, or log or exp.
 

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