Doubt in quotient rule for derivatives.

Question

I'm reading Rudin, and in theorem 5.3 he states that:

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 $f$ is differentiable on an open or closed interval.

How do I interpret/solve this?

For completeness, here is the relevant definition:

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 $$f'(x) = \lim_{t \to x} \phi(t)$$

Maybe I'm just worrying too much..

Answer 1

I suppose the problem is that $[a,b]\setminus\{x:g(x)=0\}$ may not be an open or closed interval, and Rudin's definitions only encompass differentiability on such sets. If so, so much the worse for Rudin's definitions.

There is no real problem here: if $x_0\in [a,b]$ and $g(x_0)\ne0$ then there is $d>0$ such that $g$ is continuous on $I=(x_0-d,x_0+d) \cap[a,b]$. That will be an interval containing $x_0$ on which $f/g$ is defined, and on which $f/g$ will turn out to be differentiable. Whatever $(f/g)'(x_0)$ means, it should certainly equal the value of $(f/g)'$, as defined on $I$, at $x_0$.

Answer 2

Theorem: Let $g$ be a continuous function on an open interval $(a,b)$. Suppose that $c \in (a,b)$ satisfies $g(c) \neq 0$. Then there exists $d,e$ such that $a < d < c < e < b$ with the property that $g(x) \neq 0$ for every $x \in (d,e)$.

In other words, if $c$ is a point in the domain of $f/g$, then there is an open interval containing $c$ which is contained in the domain of $f/g$.