# Epsilon Delta definition of a Derivative

The derivative at a specific point $$c$$ is represented as a limit by: $$f'(c) = \lim_{x\to c} \frac{f(x) - f(c)}{x - c}$$

It's clear to me that the epsilon delta definition of a derivative at a point $$c$$ would be:

$$\forall \epsilon > 0 ~\exists \delta > 0 \forall x: \\ 0 < |x-c| < \delta \rightarrow |\frac{f(x)-f(c)}{x-c} - L| < \epsilon$$

What's unclear to me is how to formally represent the derivative as a function of $$x$$, rather than only at point $$c$$. Basically, how would we represent this limit formally (the $$\Delta x$$ is the part tripping me up):

$$f'(x) = \lim_{\Delta x\to0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$

You do know the definition of limit, right? So, just apply it. We can argue that if the derivative at $$x$$ is $$L$$, then $$\forall \epsilon > 0 ~\exists \delta > 0 \;\forall \Delta x: \\ 0<|\Delta x| < \delta \implies \left|\frac{f(x + \Delta x) - f(x)}{\Delta x}-L\right| < \epsilon$$ Also, the derivative as a function, $$f^\prime (x)$$ is simply a function which takes a point in, and spits out the derivative of $$f$$ at that point. So, you can also define definition of a derivative at a point $$c$$ and collect all those derivatives under a function.

• Ok thanks.. and delta x just represents a distance from whatever point we're considering Dec 17, 2021 at 7:51
• @bgcode yes, you can say that. Or you may also say that $\Delta x$ is just a (small) real number- in the context of $\mathbb R$, both of them are same, doesn't really matter! Dec 17, 2021 at 15:52
• This seems to me to be doubly defective: (a) The $\forall x$ is wrong, it should be $\forall\Delta x$; (b) The condition yields that $f'(x)=0$, You need to write a proper $\epsilon$-$\delta$ definition of the phrase "$f$ is differentiable at $x$ with derivative $L$" Dec 17, 2021 at 15:53
• @ancientmathematician corrected it. Thanks for pointing it out. I must have been daydreaming when I wrote that :| Dec 17, 2021 at 15:56
• Sayan: you may want to add $\forall \Delta x$ then? Dec 17, 2021 at 23:22

You may write the last formula as :

$$f'(c) = \lim_{\Delta x\to0} \frac{f(c + \Delta x) - f(c)}{c+\Delta x-c}$$

The last formula appears more intuitive when looking at $$f'$$ as a function and not at $$f'(x)$$, which is the value of the derivative at $$x$$. Indeed the limit of the variable does not depend on the point (i.e. we have $$\Delta x \to 0$$ and not $$x \to c$$).

Formally, if you had to write an $$\varepsilon-\delta$$ proof of a formula for a derivative, for example if $$f(x)=x^n$$, then $$f'(x)=nx^{n-1}$$, you have to prove it independently for any $$x$$. So you may start your proof by "let $$x \in \mathbb{R}$$, then ...". But once this formula has been proven, then you can easily compute the derivative of any polynomial.

Note

I am unsure I have well spotted your misunderstanding so I gave a quite general answer. I hope it helps anyway.

• Thanks, I guess my misunderstanding was regarding $\Delta x$ and thinking it might change when we use epsilon-delta. giving both of you upvotes Dec 17, 2021 at 7:52
• Glad to hear we have been useful then. ;) Dec 17, 2021 at 7:53