Is uniform differentiablity symmetric? For all $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x \in \mathbb R$,
$$\left \lvert \frac{f(x)-f(y)}{x-y} - f'(y)\right \rvert \leq \varepsilon$$
when $|x-y| < \delta$.
Note: Initially, I had incorrectly written the following, as Theo Bendit and Anne Bauval helped correct:
For all $\varepsilon > 0$, for all $x \in \mathbb R$, there exists $\delta > 0$ such that
$$\left \lvert \frac{f(x)-f(y)}{x-y} - f'(y)\right \rvert \leq \varepsilon$$
when $|x-y| < \delta$.
End note.
Does this imply that $$\left \lvert \frac{f(x)-f(y)}{x-y} - f'(x)\right \rvert \leq \varepsilon$$?
Intuitively, it seems it must: The first equation is stating that the slope of the secant is arbitrarily close to the derivative at one point of the secant.  This should apply to both points of the secant -  there's no way to distinguish one point $(y)$ from the other $(x)$.
Yet, algebraically, I haven't been able to prove it.  Is my assertion true? Can it be proven algebraically? Is it true if we replace $f'$ with arbitrary function $g$? Or does the proof somehow depend on the nature of derivative? What am I missing?
 A: Just to be explicit about the logical form of the definition, it is:
For all $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x,y \in \mathbb R$
$$\left \lvert \frac{f(x)-f(y)}{x-y} - f'(y)\right \rvert \leq \varepsilon$$
if $|x-y| < \delta$.
By uniformly swapping the variables $x$ and $y$, this is obviously equivalent to:
For all $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x,y \in \mathbb R$
$$\left \lvert \frac{f(y)-f(x)}{y-x} - f'(x)\right \rvert \leq \varepsilon$$
if $|x-y| < \delta$.
But
$$\left \lvert \frac{f(y)-f(x)}{y-x} - f'(x)\right \rvert = \left \lvert \frac{-(f(x)-f(y))}{-(x-y)} - f'(x)\right\rvert = \left \lvert \frac{f(x)-f(y)}{x-y} - f'(x)\right \rvert.$$
So, the original definition is equivalent to:
For all $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x,y \in \mathbb R$
$$\left \lvert \frac{f(x)-f(y)}{x-y} - f'(x)\right \rvert \leq \varepsilon$$
if $|x-y| < \delta$.
A: Now your assertion is true, even if you replace $f'$ by any function $g,$ and $\frac{f(x)-f(y)}{x-y}$ by any other symmetric function $F(x,y).$ Even more generally, if
$$\forall(x,y)\in\Bbb R^2\quad(|x-y|<\delta\implies h(F(x,y),y)\leq \varepsilon)$$
then (by change of notations)
$$\forall(y,x)\in\Bbb R^2\quad(|y-x|<\delta\implies h(F(y,x),x)\leq \varepsilon),$$
which is equivalent to
$$\forall(x,y)\in\Bbb R^2\quad(|x-y|<\delta\implies h(F(x,y),x) \leq \varepsilon)$$
by symmetry of $F,$ of the distance, and of $\forall x\forall y.$
