Prove f'(a) is undefined when f(a) is undefined I've been asking myself how to prove that if $f(a)$ is undefined for a function, then $f'(a)$ is also undefined. I can't seem to find a proof that is generalised for any $f$. Here is what I got:
What I got
Let $f$ be any function undefined at  $a$, then
$f(a)\  \nexists$
Now, $f'(a) = \lim_{h\to 0} \frac{f(a+h) + f(a)}h$
Where I'm confused


*

*Do I only need to know that $f(a)$ is undefined to complete the proof?

*Is there a better way to write it?


I'm sorry I know it's kind of a simple question but it's been bugging me for a while and I couldn't find the answer anywhere and I'm really trying to work on being more rigorous in my understanding of maths.
Thank you!
 A: By definition, to prove that a function is differentiable at $x=a $, you need compute the limit
$$\lim_{h\to 0}\frac {f (a+h)-f (a)}{h} $$
For this, you should know the value of $f (a) $.
A: Here's my rendition of the standard definition of the derivative, fully written out:

Suppose that $f$ is a function whose values are real numbers, and which is defined at a real number $a$, and on a neighborhood of $a$. The derivative of $f$ at $a$, $f'(a)$, is the limit
$$\lim_{h \to 0} \frac{f(a + h) + f(a)}{h},$$
if this limit exists.

If we have a function $f$ which is not defined at a real number $a$ (in other words, $f(a)$ is not defined), then the conditions for the definition of $f'(a)$ are not satisfied, and so the definition of $f'(a)$ does not apply (in other words, $f'(a)$ is not defined).
A: It's probably easier to think about this using contraposition. Your statement is equivalent to : If $f$ is differentiable at $a$, then $f$ is defined at $a$
To prove the equivalent statement, note that differentiability implies continuity, which implies $f$ being well-defined at point of differentiability.
