1. ## differantiation proof question..

if f(x) is differentiable on x_0
prove that

??

using the given i could
say that lim [f(x+x_0) - f(x0)]/[x-x_0] exist

2. $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

now let $\displaystyle u = x+h \Rightarrow x = u-h$

So $\displaystyle f'(u-h) = \lim_{h \to 0} \frac{f(u-h+h)-f(u-h)}{h} = \lim_{h \to 0} \frac{f(u)-f(u-h)}{h}$

Since $\displaystyle h \to 0, f'(u-h) \to f'(u)$

Therefore $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h}$

$\displaystyle f'(x)+f'(x) = 2f'(x) = \lim_{h \to 0}\left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right)$

$\displaystyle 2f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{h}$

Hence, $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h}$

3. how did you know to pick those variables??

4. and why you didnt use the point x_0
that they presented
there is another definition to limit
$\displaystyle f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

is there a way to use this definition like you did before
what variables to pick??

5. Originally Posted by transgalactic
and why you didnt use the point x_0
that they presented
there is another definition to limit
$\displaystyle f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

is there a way to use this definition like you did before
what variables to pick??
Does your problem require you to use that definition of $\displaystyle f'(x)$?

6. Originally Posted by transgalactic
and why you didnt use the point x_0
that they presented
there is another definition to limit
$\displaystyle f'(x) = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$

is there a way to use this definition like you did before
what variables to pick??
A simple substitution will change your defition to Chiph's. Now what substitution is that?

7. if ill take x-x_0 =h

8. correct