- #1

member 587159

Let ##f## be defined (and real-valued) on ##[a,b]##. For any ##x \in [a,b]##, form the quotient ##\phi(t) = \frac{f(t) - f(x)}{t-x}\quad (a < t <b, t \neq x)## and define ##f'(x) = \lim_{t \to x} \phi(t)##, if the limit exists.

An analoguous definition is given for functions defined on an open interval.

Q1:

__Why do we restrict ourselves to defining derivatives on intervals?__I would propose the following definition:

Let ##f: E \subseteq \mathbb{R} \to \mathbb{R}## be a function and ##x## a limit point of ##E##.

Define $$\phi: E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t) - f(x)}{t-x}$$. Then we define $$f'(x) = \lim_{t \to x} \phi(t)$$ if the limit exists.

Q2: The quotient rule in Rudin is written in the following way:

Suppose ##f,g## are real functions on ##[a,b]## that are differentiable at ##x \in [a,b]##. Then ##f/g## is differentiable at ##x##, provided that ##g(x) \neq 0## and

##\left(\frac{f}{g}\right)'(x) = \frac{g(x)f'(x) - g'(x)f(x)}{g^2(x)}##

Following Rudin's convention, the domain of ##f/g## are those point ##x## of ##[a,b]## where ##g(x) \neq 0##.

__However, Rudin's definition of differentiability only mentions the cases where a function is defined on an open or closed interval, while ##f/g## may be defined on a domain that it is not an interval.__

How should this be interpreted?

How should this be interpreted?

I see two solutions:

(1) We can restrict the function to an interval where it is non-zero (but the theorem does not mention this!)

(2) We use the alternative definition I gave in Q1.

Any thoughts?