1. ## proving differentiability

hi, we already know a function is

(A) differentiable at $a$ if
$\lim_{x \to a} \left[f(x) - f(a) / (x - a) \right]$ exists.

now consider f:R $\to$ R, f(x) = $x^n$, for some N E N. Using the definition of (A), how can we prove that f is differentiable and that f'(x) = $nx^{n-1}$?
help would be greatly appreciated here. thank you

2. Originally Posted by heyo12
hi, we already know a function is

(A) differentiable at $a$ if
$\lim_{x \to a} \left[f(x) - f(a) / (x - a) \right]$ exists.

now consider f:R $\to$ R, f(x) = $x^n$, for some N E N. Using the definition of (A), how can we prove that f is differentiable and that f'(x) = $nx^{n-1}$?
help would be greatly appreciated here. thank you
Hint, use the identity:
$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$

3. identity: $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$.

now that we have that identity.. we deduce $\lim_{h \to 0} \frac {(x + h)^n - (x)^n}{h} = \lim_{h \to 0} [ (x+h)^{n-1} + (x+h)^{n-2}x + ... + (x+h)x^{n-2} + x^{n-1} ] = nx^{n-1}.$

i think this is right to my knowledge.
but this is what i'm having difficulty understanding.

(1) where does the identity come from.. and how do you explain this identity's expansion?

(2) finally, how does that identity apply to the expansion of the limit, finally arriving at the right answer...

4. $\frac{f(x)-f(a)}{x-a} = \frac{x^n - a^n}{x-a} = \frac{(x-a)(x^{n-1}+x^{n-2}a+...+xa^{n-2}+a^{n-1})}{x-a}$

Now as $\lim_{x\to a}$ we have:
$a^{n-1}+a^{n-2}a+...+aa^{n-2}+a^{n-1}=na^{n-1}$

5. haha genius. i get the gist of it.

the only parts that are bugging me are:

1) on the expansion of a^n - b^n.. where does the inital (a - b) come from?

2)
Originally Posted by ThePerfectHacker
Now as $\lim_{x\to a}$ we have:
$a^{n-1}+a^{n-2}a+...+aa^{n-2}+a^{n-1}=na^{n-1}$
why does that expansion equal na^n-1?
is there something to show that when the terms add up they equal na^n-1?

and would this all be sufficient to say that we've "proved" f is differentiable?

6. 1) on the expansion of a^n - b^n.. where does the inital (a - b) come from?
Just multiply out the paranthesis and everything cancels out. Or use long division.
why does that expansion equal na^n-1?
is there something to show that when the terms add up they equal na^n-1?
Because there are $n$ terms all equal to $a^{n-1}$.

7. lol genius again. i love how u put it so simple in one line. its quite obvious now. .. how can i add some sort of rep points on u lol?

8. Originally Posted by heyo12
how can i add some sort of rep points on u lol?
You did by pressing "Thanks".