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, $\forall\epsilon&gt;0\hspace{0.2cm}\exists\hspace{0.2cm}\delta=\delta(\epsilon)&gt;0$ such that $0&lt;|x-a|&lt;\delta\implies|f(x)-L|&lt;\epsilon$, 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.
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 $\epsilon&gt;0$ there is a $\delta&gt;0$ such that $0&lt;|x-x_0| &lt; \delta$ implies $\left|\frac{f(x) - f(x_0)}{x - x_0} - L\right| &lt; \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.
My proof of derivative implies continuity:
Think about this: For all ϵ>0 there exists $δ&gt;0$ such that $0&lt;|x-c|&lt;δ$ implies $|\frac{f(x)−f(c)}{x−c}|&lt;ϵ$.

Start my proof: Since f is derivative at $c ∈ R$, 
There exists δ>0 such that $0&lt;|x-c|&lt;δ$ implies $|\frac{f(x)−f(c)}{x−c}|=|f(x)-f(c)|\times \frac{1}{|x-c|}$. 

We assign $ϵ=\frac{ϵ_1}{|x-c|}&gt;0$, then $|f(x)-f(c)|*\frac{1}{|x-c|}&lt;ϵ$ implies $|f(x)-f(c)|&lt;ϵ=\frac{ϵ_1}{|x-c|}$.

Hence for all $ϵ&gt;0$, There exists $δ&gt;0$ such that $0&lt;|x-c|&lt;δ$ implies $|f(x)−f(c)|&lt;ϵ$, i.e. we have $\lim_{x→c} f(x)=f(c)$.