A challenging calculus question about differentiation Let $O$ be open. We say $f :O \subset R \to R $ is strong differentiable at $a \in O$ if 
$$ \lim_{(x_1,x_2) \to (a,a) ,\;\;\; x_1 \neq x_2 } \frac{ f(x_2) - f(x_1) }{x_2- x_1} = f^S(a) $$ 
exists and is finite. We call $f^S(a)$ the strong derivative of $f$ at $a$
Problem:
 Say $O \subset A $ is open. $f$ is strong differentiable on $O$ iff $f'(x)$ is continuous on $O$.

Assume $f'$ continuous, then by the mean value theorem, for $\xi$ between $x$ and $y$ we have
$$
\left| \frac{f(x)-f(y)}{x-y} -f'(a) \right| = |f'(\xi)-f(a)|.
$$
Since $f'$ is continuous and $|\xi-a| \leq \max\{ |x-a|, |y-a| \}$ we conclude that $f$ is strongly differentiable.
For the other direction fix $a\in O$,  $\varepsilon>0$ and pick $\delta>0$ so that, whenever $|a-x|<\delta$ and $|h|<\delta$, the following holds:
$$
\left| \frac{f(a+h)-f(a)}{h}-f'(a)\right| < \varepsilon, \qquad \left|f'(a)- \frac{f(x+h)-f(x)}{h} \right|< \varepsilon,
$$
that this is possible is guaranteed by the strong differentiability of $f$. Using the triangle inequality and these estimates we get
$$
\left| \frac{f(a+h)-f(a)}{h}- \frac{f(x+h)-f(x)}{h} \right| \leq 2 \varepsilon, \qquad \text{whenever } |x-a|,\ | h|<\delta.
$$
Let $h\to 0$ to obtain $|f'(a)-f'(x)|\leq 2 \varepsilon$. This prove that $f'$ is continuous.
