Proof of the Power Rule - Stuck at the End

Astrum
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Homework Statement


Prove that the following

f'(x)=nx^{n-1} if f(x)=x^{n}

Homework Equations


Binomial theorem, definition of the derivative

The Attempt at a Solution



f'(x)=lim_{h\rightarrow0}\frac{f((x+h)^{n})-f(x)}{h}

We need to expand the (x+h)^2 term now

\sum^{n}_{k=0}{n\choose k} x^{n-k}h^{k}={n\choose 0}x^n+{n\choose 1}x^{n-1}h+...+{n\choose k}x^{n-k}h^{k}

So, we sub this for the f((x+h)^n) term:

lim_{h \rightarrow 0}\frac{{n\choose 0}x^n+{n\choose 1}x^{n-1}h+...+{n\choose k}x^{n-k}h^{k}-x}{h}

{n\choose k}=\frac{n!}{(n-k)!k!}

This now simplifies to---

lim_{h\rightarrow0}\frac{x^{n}}{h}+nx^{n-1}-lim_{h\rightarrow0}\frac{x}{h}

The first and third terms create only a one sided limit, and they both go to infinity, I'm not sure where I went wrong...

I could just look up the proof, but I'm trying to do it by myself, so I'm only looking for a hint.

I feel like I'm really close, because I have the answer there, I just don't know why those other two terms are messing it up.

There are only two possibilities, either I've gone in the completely wrong direction, or I made a silly mistake somewhere.
 
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Your definition of f'(x) is a little strange.

Since f(x) = x^n, then

f'(x) = lim (h -> 0) [(x+h)^n - x^n) / h]

Try expanding now using the Binomial Theorem.
 
SteamKing said:
Your definition of f'(x) is a little strange.

Since f(x) = x^n, then

f'(x) = lim (h -> 0) [(x+h)^n - x^n) / h]

Try expanding now using the Binomial Theorem.

:bugeye:

Thanks, I didn't catch that... how silly.

So, that means those two terms cancel out, and we're left with the correct one.
 
SteamKing said:
Your definition of f'(x) is a little strange.

Since f(x) = x^n, then

f'(x) = lim (h -> 0) [(x+h)^n - x^n) / h]

Try expanding now using the Binomial Theorem.

Astrum said:
:bugeye:

Thanks, I didn't catch that... how silly.

So, that means those two terms cancel out, and we're left with the correct one.

What do you mean by "we're left with the correct one"? There are two terms that do cancel but there are a lot of others and you have to deal with the h's. Have you actually worked it out?
 
LCKurtz said:
What do you mean by "we're left with the correct one"? There are two terms that do cancel but there are a lot of others and you have to deal with the h's. Have you actually worked it out?

Yes, I skipped steps out of laziness (meaning, I did them on paper, but I didn't feel like taking the time to write them out here).

The h on the bottom cancels out and all the hs on the top go to 0.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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