I am recently learning the Rigorous definition of the derivative so now I come across a problem here is the definition of derivative and definition of differentiability
A real function $f$ is said to be differentiable at a point $a \in \mathbb{R}$ if and only if $f$ is defined on some open interval $I$ containing $a$ and
$$f^{\prime}(a) :=\lim_{h\to 0 } \frac{f(a+h)-f(a)}{h}$$
exists. In this case $f^{\prime}(a)$ is called the derivative of $f$ at $a$
and
Let $I$ be an interval. A function $f:I \rightarrow \mathbb{R} $ is said to be differentiable on $I$ if and only if$$f^{\prime}(a) :=\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ exists and is finite for every $a \in I $
I feel first definition also can be used for differentiability
Let $I$ be an interval. A function $f:I \rightarrow \mathbb{R} $ is said to be differentiable on $I$ if and only if $$f^{\prime}(x) :=\lim_{h\to 0 } \frac{f(x+h)-f(x)}{h}$$ exists and is finite for every $x \in I $
My problem is what is the difference between these two? Why there is two formula for one thing?
Best Answer
There are two definitions because you're defining two different things:
$f$ is differentiable at the point $x$.
$f$ is differentiable on the interval $I$.
If I were writing the text, I'd define the first one, and then say that $f$ is differentiable on $I$ if for every point $b \in I$, $f$ is differentiable at $b$. I don't know why your author didn't do this. One peculiarity (which may depend on how the author defined limits) is that it's possible (in the given definition) for the function to be differentiable on a closed interval $I$ without being differentiable at the endpoints. Why? Because maybe there's no open interval around the endpoint that's entirely contained in the domain of $f$. [If having such an interval is an essential part of the author's definition of limit, then it should probably have been omitted from the definition of derivative.]
As an example, consider the function $$ f:[0, 1] \to [0, 1] : x \mapsto x. $$ This happens to take on the same values as $$ g: \Bbb R \to \Bbb R : x \mapsto x $$ at every point $x$ of the unit interval, but $f$ is undefined outside that unit interval nonetheless.
As an example of this subtle distinction:
For this function, the limit in the second definition is well-defined** at $x = 0$, for instance, but the function is not differentiable at $x = 0$ (according to the first definition) because $f$ is not defined on any open interval containing $x = 0$.
**at least, it's well-defined if the definition of "limit" is the one I'm thinking of; it may be that your author uses something different.
NB: Most folks would require "and is finite" as part of the definition of differentiability as well; it's not clear why your author omitted that.