Two variable definition of derivative Let $f:(0,1)\rightarrow \mathbb R$ be a real valued map from the unit interval. Let 
$$A:=\left \{a\in (0,1):\exists    f^*(a)=\lim_{x\ne y,\,(x,y)\to(a,a)} \frac{f(x)-f(y)}{x-y}\right \}$$
It is true that if $a\in A$, then $f$ is differentiable at $a$. I am trying to show that it is false to claim that if $f$ is differentiable at $a$, then $a\in A$. 
Is there any easy example of a function $f:(0,1)\rightarrow \mathbb R$ satisfying


*

*$f(x)$ is differentiable for every $x \in (0,1)$.

*There exists $x^* \in (0,1)$ such that $x^* \not \in A$.


so that $A$ is a proper subset of $(0,1)$?
 A: Yes, the existence of $f^*(a)$ is a much stronger condition than the existence of $f'(a)$. For example
Theorem If $f:\Bbb R\to\Bbb R$ is differentiable then $A$ is the set of points where $f'$ is continuous.
Proof: Suppose that $f'$ is not continuous at $a$. There exists a sequence $x_n\to a$ so that $$f'(x_n)\not\to f'(a).$$ By definition of the derivative there exists $y_n$ with $0<y_n-x_n<1/n$ such that $$\left|f'(x_n)-\frac{f(x_n)-f(y_n)}{x_n-y_n}\right|<\frac1n.$$So $(x_n,y_n)\to(a,a)$ but $$\frac{f(x_n)-f(y_n)}{x_n-y_n}\not\to f'(a);$$as mentioned in the OP this shows that in fact the limit $f^*(a)$ does not exist.
Suppose now that $f'$ is continuous at $a$. Let $\epsilon>0$, and choose $\delta>0$ so that $$|f'(x)-f'(a)|<\epsilon\quad(|x-a|<\delta).$$
Now if $|x-a|<\delta$, $|y-a|<\delta$ and $x\ne y$ then the Mean Value Theorem shows that $$\left|\frac{f(x)-f(y)}{x-y}|-f'(a)\right|<\epsilon.$$
So $f^*(a)=f'(a)$. QED
So any differentiable function with a discontinuous derivative is an example of the sort you're looking for.
A: $f(x) = \sin x$ differentiable everywhere, but $0 \not \in A$. If $0$ were in $A$, then $(x, 0) \to (0,0)$ implies $f^*(0)=1$ but $(x,-x) \to (0,0)$ implies the limit does not exist. If you like, you can shift the function to define on $(0,1)$, but the idea is the same.
