Please post the proof of derivative of power function.

Question

turingtest · Answer

$$\lim_{h \rightarrow 0}{(x+h)^n-x^n \over h}={x^n+nx^{n-1}h+{n(n-1)\over 2}x^{n-2}h^2+...+h^n-x^n \over h}$$

anonymous · Answer

ah, was on my way to doing it, but TuringTest has done it. and he's explained it clearly. dividing through by h should yield an expression where the only term with no h's is nx^(n-1), and all other terms will vanish

turingtest · Answer

yep and here it the final bit spelled out for you:$$=\lim_{h \rightarrow 0}nx^{n-1}+n(n-1)x^{n-2}h+...+h^{n-1}=nx^{n-1}$$

anonymous · Answer

Is this first principle??Please post the method you have used.

turingtest · Answer

I forgot to divide n(n-1)x^(n-2)h by 2 in the second part, but it doesn't change the answer

anonymous · Answer

yes, it's first principle. there is no more powerful proof than by using first principles.

turingtest · Answer

what do you mean first principle?
the method is expanding binomially and using the limits

anonymous · Answer

it's a proof by first principles, since we're using the original definition of derivative.

anonymous · Answer

I meant ab-initio.

turingtest · Answer

?

turingtest · Answer

that is from the basic definition of the derivative, so it is ab initio

anonymous · Answer

$$\lim_{x \rightarrow 0} f(x)-f(a)/x-a$$
This is first principle.

turingtest · Answer

I think what you mean is$$\lim_{x \rightarrow a}{f(a)-f(x)\over a-x}$$but this will not work here. Another way to start is$$\lim_{a \rightarrow b}{f(b)-f(a)\over b-a}$$but I don't remember how to do that one offhand.

In any event all those forms above are ab initio. There are multiple (yet equivalent) ways to write the limit definition.

turingtest · Answer

actually the first form in my last post is for the derivative at a specific point, so that's not what we want anyway. we want the general formula, which I have provided the proof for.

anonymous · Answer

Oh yeah I meant that,I posted wrong.So can u please explain how did u get the function on which u have applied binomial theorem or u have supposed it?

turingtest · Answer

I used the binomial expansion of
$$(x+h)^n=x^n+nx^{n-1}h+{n(n-1)\over2}x^{n-1}+...+h^n$$
I supposed nothing, the definition I originally posted has f(x+h)-f(x) in the numerator. Since f(x)=x^n, then f(x+h)=(what I have above)

turingtest · Answer

oh  messed up again, the second term in the expansion above should be $${n(n-1)\over2}x^{n-2}h^2$$

turingtest · Answer

^^^ I meant third term
(I must not have slept last night)