If $f$ is twice differentiable, $\big(f(y) - f(x)\big)/(y-x)$ is differentiable Suppose $f: \mathbb{R} \to \mathbb{R}$ is a $C^{1}$ function. Then, define a new function $F: \mathbb{R}^{2} \to \mathbb{R}$ by:
$$
F(x,y) = \begin{cases}
  \displaystyle \frac{f(y) - f(x)}{y - x} &\text{ if } x \neq y \\
  \displaystyle f'(x) &\text{ otherwise.}
\end{cases}
$$
Then, if $f''(x)$ exists, $F$ is differentiable. I can prove that $F$ is differentiable if $x \neq y$, since under these conditions $F_{x}$ and $F_{y}$ are $C^{1}$. So it's left to prove $F$ is also differentiable if $x = y$. 
At first, I conjectured that, for example, $F_{x} (a,a)$ would be $f''(a)/2$, but I'm having a hard time proving it. I started using the definition $\lim_{h \to 0} (F(a+h, a) - F(a,a)) / h$ and applying the MVT found $\bar{a}$ between $a$ and $a + h$ s.t. this difference quotient is:
$$
\frac{1}{h}(f'(\bar{a}) - f'(a))
$$
so I tried dividing and multiplying by $\bar{a} - a$, thinking it would be possible to prove that $\lim_{h \to 0} (\bar{a} - a)/h = 1$, but so far I've only been able to bound it above by $1$.
Is it true? Does this conjecture even makes sense? I'm lost in thinking about any other candidates for the differential in these points. Any help would be appreciated! 
 A: When $f$ is $C^2$, the following representation immediately shows the $C^1$-regularity of $F$:
$$ F(x, y) = \int_{0}^{1} f'(x+(y-x)t) \, \mathrm{d}t. $$
When $f$ is only assumed to be twice-differentiable, we know that $x\mapsto F(x,y)$ is differentiable at least away from $y$. To establish the differentiability at $x=y$, write $x=y+h$ and note:
$$ \frac{F(y+h,y)-F(y,y)}{h} = \frac{f(y+h) - f(y) - f'(y)h}{h^2}. $$
Now by the L'Hôpital's rule, we get
$$ \left. \frac{\partial F}{\partial x} \right|_{(y,y)} = \lim_{h\to 0} \frac{f(y+h) - f(y) - f'(y)h}{h^2}
= \lim_{h\to 0} \frac{f'(y+h) - f'(y)}{2h}
= \frac{f''(y)}{2}. $$
A: We can make a change of variables $t=y/2-x/2$, $s=y/2+x/2$. This change is $C^\infty$, diffeomorphism, the determinant of Jacobian is constant.
Then let's define $$G(s,t) = F(x(s,t),y(s,t)) = 2\frac{f(s+t)-f(s-t)}{t},\quad G(s,0)=4f'(s)$$
$$G:\Bbb R^2\to \Bbb R.$$
Easy to see that on the set $t\ne 0$ $G$ is $C^1$ because $f$ is $C^1$. On the whole $\Bbb R^2$ $G$ is continuous.
Now let's study the gradient:
$$\partial_s G = 2\frac{f'(s+t)-f'(s-t)}{t}.$$ If in addition $f\in C^2$, then this derivative is continuous on $\Bbb R^2$ (removable singularity in $t=0$).
$$\partial_t G = 2\frac{f'(s+t)+f'(s-t)}{t}-2\frac{f (s+t)-f (s-t)}{t^2}.$$
L'Hopital's rule allows to say that this function is continuous on $\Bbb R^2$.
So the conclusion is that if $f\in C^2$ then $G\in C^1$ and, therefore, $F\in C^1$.
If you need help with any part of this reasoning, ask in comments.
