Calculus Proof Question.....Can you do it?

Question

anonymous · Answer

Show that $$\frac{ d }{ dx } x^{n} = n x^{n-1} $$
Through the definition of a derivative...
$$\lim_{h \rightarrow 0} \frac{(x+h)^{n} - x^{n}}{h}$$
I have spent a few hours, just trying to figure out where to go from there.

anonymous · Answer

use the binomial expansion for (x+h)^n.

anonymous · Answer

See, now that'd be cool...if I knew the binomial expansion. I've really been trying not to use any resources. I knew there must be some other way to express that quantity, and I've spent hours trying to figure that out. Can you explain what it is?

zzr0ck3r · Answer

I would google it. It is a general method for expanding (a+b)^n

anonymous · Answer

Even with that....I am now only here...
$$\lim_{h \rightarrow 0} \frac{\sum_{k=0}^{n}\left(\begin{matrix}n \ k\end{matrix}\right) x^{n-k}h^{k} - x^{n}}{h} $$

jhannybean · Answer

There was a similar post about the expansion of $ (a+b)^n$ cept it had something to do with programming the expansion into the calculator for $n \ge 2$

jhannybean · Answer

is that supposed to be a fraction, @Eleven17 ? $\large \left( \frac{n}{k}\right)$ like that?

anonymous · Answer

It's supposedly called the binomial coefficient. But this isn't necessarily the easiest way to solve the problem, I broke down and looked it up and of course it's a million times easier than I Could have ever imagined it be.

jhannybean · Answer

Could you explain it? :) I'd like to learn the expansion as well.

nincompoop · Answer

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anonymous · Answer

Well I just looked up the expansion on wikipedia, but then I looked up the proof myself and it involves something a little different. Because the two terms are both being raised to the power n, you can do a certain factoring, that will allow you to cancel out the h's, and it ultimately becomes embarassingly easy.
$$a^{n}-{b}^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2....b^{n-1})$$
Using that little rule/identity/whatever, you can easily subsitute a for x+h and b for x to reveal that 
$$a^{n}-{b}^n=(x+h-x)((x+h)^{n-1}+(x+h)^{n-2}x+(x+h)^{n-3}x^2....x^{n-1}$$
But all that matters is that you can cancel those h's out. So you can directly sub in for the other h's,

zzr0ck3r · Answer

:)

jhannybean · Answer

it kind of looks like a a cubic expansion lol.

$$\large a^3 + a^2b +ab^2 + b^3$$ kind of...something like that. ... haha. First thing i could relate it to.

zzr0ck3r · Answer

it would be that if n=3