Proof for alternative definition of the derivative I read somewhere recently that you can define the derivative as follows:
$$f'(x) = \lim_{h, k \to 0^+} \frac{f(x +h) - f(x - k)}{h + k}$$ 
I have been trying to prove this for about 2 hours, and can't seem to get it done. How should I proceed?
Edit:
Assume $f$ is differentiable for all $x \in (x - \delta, x + \delta)$ for some $\delta > 0$.
 A: It is not hard to show that if the limit given in the post exists, then the derivative exists, and is equal to the given limit.  Just let $k=h$, and rewrite the expression as 
$$\frac{1}{2}\frac{f(x+h)-f(x)}{h}+\frac{1}{2}\frac{f(x-h)-f(x)}{-h}
.$$
For the converse, suppose $f'(x)$ exists. Rewrite our expression as 
$$\frac{h}{h+k}\frac{f(x+h)-f(x)}{h} +\frac{k}{h+k}\frac{f(x-k)-f(x)}{-k}.\tag{$1$}$$ 
Choose an $\epsilon \gt 0$. Because $f'(x)$ exists, there is a $\delta$ such that if $|t|\lt \delta$, then 
$$\left|\frac{f(x+t)-f(x)}{t}-f'(x)\right|\lt \epsilon.$$ 
It follows that if $h$ and $k$ are $\lt \delta$, then the differential quotients in $(1)$ are each within $\epsilon$ of $f'(x)$. Thus $(1)$ is equal to 
$$\frac{h}{h+k}(f'(x)\pm\epsilon_1)+\frac{k}{h+k}(f'(x)\pm\epsilon_2),$$
where $\epsilon_1$ and $\epsilon_2$ are non-negative quantities $\lt \epsilon$. 
But $h/(h+k)$ and $k/(h+k)$ are both less than $1$. (This little fact is key: we could run into trouble using two points on the same side of $x$.)
It follows that if $h$ and $k$ are less than $\delta$, then $(1)$ differs from $f'(x)$ by less than $\epsilon$. So the limit as $h$, $k$ independently go to $0^+$ exists.   
