Proof $f$ is Differentiable. I have been looking through the stack and can not find a direct answer to this question. I recently read that we say $f$ is defferentiable on $I$, if $f$ is differentiable at every point on an open interval $I$. It almost reminds me of the rigorous deffinition of a limt, $\forall\epsilon>0\hspace{0.2cm}\exists\hspace{0.2cm}\delta=\delta(\epsilon)>0$ such that $0<|x-a|<\delta\implies|f(x)-L|<\epsilon$, and how we can use said definition to write all kinds of $\epsilon-\delta$ proofs regarding limits. Proving continuity, proving uniqueness of limits, so on and so forth. 
My question is: What is the standard proof writing format (such as $\epsilon-\delta$ proofs for limits) for proving differentiability based of the rigorous definition of a derivative? Assuming there is a rigorous definition of a derivative similar to that of limits, please provide an example proof.
 A: The derivative is a limit. You can just apply the definition of the limit to prove differentiability rigorously. Namely, a function $f$ defined on an open $U \subset \mathbb{R}$ is differentiable at a point $x_0 \in U$ if there is a real number $L$ such that for all $\epsilon>0$ there is a $\delta>0$ such that $0<|x-x_0| < \delta$ implies $\left|\frac{f(x) - f(x_0)}{x - x_0} - L\right| < \epsilon $. 
You might wonder, why don't we ever use $\epsilon-\delta$ arguments with derivatives? Well, it's because we don't need to. The proofs of the differentiation rules (e.g., product rule, chain rule) require an $\epsilon-\delta$ argument, but after we've proven these theorems, we never need to use $\epsilon-\delta$ arguments again. 
A: The derivative of a function $f:X \to Y$ with respect to a (cluster) point $a \in X$ is defined as
$$ \lim_{x \to a} \frac{f(x)-f(a)}{x-a},$$
or, equivalently, making $h=x-a,$ as
$$ \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$
So, for proving $f$ is differentiable at $a$, you have to show that limit exist.
A: My proof of derivative implies continuity:
Think about this: For all ϵ>0 there exists $δ>0$ such that $0<|x-c|<δ$ implies $|\frac{f(x)−f(c)}{x−c}|<ϵ$.
Start my proof: Since f is derivative at $c ∈ R$, 
There exists δ>0 such that $0<|x-c|<δ$ implies $|\frac{f(x)−f(c)}{x−c}|=|f(x)-f(c)|\times \frac{1}{|x-c|}$. 
We assign $ϵ=\frac{ϵ_1}{|x-c|}>0$, then $|f(x)-f(c)|*\frac{1}{|x-c|}<ϵ$ implies $|f(x)-f(c)|<ϵ=\frac{ϵ_1}{|x-c|}$.
Hence for all $ϵ>0$, There exists $δ>0$ such that $0<|x-c|<δ$ implies $|f(x)−f(c)|<ϵ$, i.e. we have $\lim_{x→c} f(x)=f(c)$.
