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.

• I would advise against the use of $\epsilon, \delta$ unless they are necessary. These kind of proofs are mostly used for theoretical results (like for proving theorems on algebra of limits, theorems on rules of differentiation) but should be avoided when dealing with concrete problems like evaluation of a limit or a derivative. – Paramanand Singh Feb 23 '17 at 4:38

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.

• So there is really no use for proving in such a manner? – Nick Pavini Feb 23 '17 at 3:53
• @NickPavini Yes, there is no use. But, we could if we wanted to, and it is maybe good practice. – MathematicsStudent1122 Feb 23 '17 at 3:54
• oh wow! lol, I am just getting into proofs really so I like the rigorous definition as it helps to see at thoroughly.:) – Nick Pavini Feb 23 '17 at 3:56

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.

• so basically I can rely on $\epsilon-\delta$ proofs for differentiability? – Nick Pavini Feb 23 '17 at 3:48
• Usually you show the limit exists by using properties of limits, as almost in every case proving this limit by $\epsilon - \delta$ is just too tricky – positrón0802 Feb 23 '17 at 3:50
• (+1) but I would really like to see how it would be done $\epsilon-\delta$ style.:) – Nick Pavini Feb 23 '17 at 3:52

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 $$\lim_{x→c} f(x)=f(c)$$.

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – emonHR Jan 2 at 4:22