Does $f \in C^1$ implies a difference quotient $d(x,h):=\frac{f(x+h)-f(x)}{h}$ can be extended to a continuous function on $\mathbb{R}^2$? I'm a student learning mathematical analysis. I'm trying to find a (nontrivial) [condition] that if $A \subseteq \mathbb{R}$ is dense and $f:A \to \mathbb{R}$ is a function satisfying [condition], then there is an extension $\tilde{f} \in C^1(\mathbb{R})$ extending $f$. I thought that it is useful if I can express a condition $C^1$ as a continuity of a single function. (Definition of $C^1$ needs infinitely many continuous functions to define a derivative of $f$, I mean, continuity of $d(h)=\frac{f(x+h)-f(x)}{h}$ for every $x$.)
My question is that, is $f \in C^1(\mathbb{R})$ equivalent to the existence of continuous extension of $d(x,h):=\frac{f(x+h)-f(x)}{h}$ on $\mathbb{R}^2$?
Of course, continuity of some extension of $d$ implies coordinate-wise continuity, so this gives well-definedness and continuity of $f'$. But, I don't know the converse is also true.
I tried to prove this statement by using general properties of the extension of $d$ like continuity on $(\mathbb{R} \times \mathbb{R}) \setminus (\mathbb{R} \times \{0\})$ and coordinate-wise continuity, but I failed. Actually, I guess that
$$g(x,y)=
\begin{cases}
0 & (x,y) \notin \mathbb{R}_{>0} \times \mathbb{R}_{>0}\\
1 & \frac{1}{2}x<y<2x, (x,y) \in \mathbb{R}_{>0} \times \mathbb{R}_{>0}\\
\frac{2x}{y} & 2x \le y, (x,y) \in \mathbb{R}_{>0} \times \mathbb{R}_{>0}\\
\frac{2y}{x} & 2y \le x, (x,y) \in \mathbb{R}_{>0} \times \mathbb{R}_{>0}
\end{cases}
$$
is continuous on $(\mathbb{R} \times \mathbb{R}) \setminus (\mathbb{R} \times \{0\})$ and coordinate-wisely continuous, but by considering the origin and the diagonal $\{(x,x): x \in \mathbb{R}\}$, $g$ is not continuous. Can I have any good idea on this problem?
Thanks for reading.
 A: I am not sure if I understood your question correctly but if you want to show that $f\in C^1(\mathbb{R})$ implies continuity of the function 
\begin{equation}
d(x,h)=\begin{cases}\frac{f(x+h)-f(x)}{h} \text{ if } h\neq0 \\
f'(x) \text{ if } h=0
\end{cases}
\end{equation}
then this is possible. As you wrote in your question the only points where this is nontrivial are points of the form $(x,0)$. Choose an arbitrary sequence $(x_n,h_n)\to (x,0)$ then we get the inequality
\begin{align*}
|d(x_n,h_n)-d(x,0)|&=|\frac{f(x_n+h_n)-f(x_n)}{h_n}-f'(x)|\\
&=|\frac{1}{h_n}\int_0^1 \frac{d}{ds}f(x_n+sh_n)ds-f'(x)|\\
&=|\int_0^1 f'(x_n+sh_n)-f'(x)ds|
\end{align*}
Given any $\varepsilon>0$, we can choose $n$ large enough such that the last integral is less then $\varepsilon$, since $f'$ is continuous.
A: Suppose $f$ is continuously differentiable, and define
$$D : \Bbb{R}^2 \to \Bbb{R} : (x, h) \mapsto \begin{cases} d(x, h) & \text{if } h \neq 0 \\ f'(x) & \text{if } h = 0.\end{cases}$$
I claim that $D$ is continuous. Note that $d$ is continuous on its open domain by virtue of $f$ being continuous, hence if we pick a point $(x_0, h_0)$ with $h_0 \neq 0$, then $D$ is continuous at $(x_0, h_0)$. Thus, we must show continuity at points of the form $(x_0, 0)$.
Note that $|f'|$ is continuous on the compact interval $I = [x_0 - 1, x_0 + 1]$, and hence achieves a maximum $M$. In particular, if $x, y \in I$, then mean value theorem implies that there exists some $c \in I$ such that
$$|f(x) - f(y)| = \left|\frac{f(x) - f(y)}{x - y}\right| |x - y| = |f'(c)||x - y| \le M|x - y|.$$
In particular, $f$ is uniformly continuous (Lipschitz continuous, even) on the compact domain $I \times I$. Thus, a unique continuous extension exists on $I \times I$, which must be $D$, by the definition of the derivative.
