Theorem 5.5, Rudin's Principles of Mathematical analysis says:

Suppose $f$ is continuous on $\color{red}{[a,b]}$,$ f'(x)$ exists at some point $x\in [a,b], g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the point $f(x)$. If $$h(t)=g(f(t)) \quad (a\le t \le b)$$ then $h$ is differentiable at $x$, and $$h'(x)=g'(f(x))f'(x)$$

I believe, I have understood the proof. But why is continuity at $[a,b]$ required? To me differentiablity at $x$ of $f$ and the same at $f(x)$ of $g$ is the only required conditions.

  • $\begingroup$ You don’t show a hypothesis that $a<b$. Is the proof correct when $a=b$? $\endgroup$
    – Steve Kass
    Dec 3, 2014 at 6:51
  • $\begingroup$ @SteveKass, differentiability at an isoleted point does not makes sense. $\endgroup$ Dec 3, 2014 at 7:48
  • $\begingroup$ Martín-Blas: The function $f(x)=\begin{cases} x^2 & x\mbox{ is rational} \\ 0 & x\mbox{ is irrational} \\ \end{cases} $ is differentiable at $x=0$ and nowhere else. books.google.com/books?id=b05c370fLdsC&pg=PA143 $\endgroup$
    – Steve Kass
    Dec 3, 2014 at 22:59
  • $\begingroup$ @SteveKass: But $0$ is not an isolated point. $\endgroup$
    – Frog
    Dec 4, 2014 at 5:51
  • $\begingroup$ Martín-Blas: Ah, I see. If $h$ is defined at only one point, you’re right that the theorem makes no sense. I wondered if the proof was valid under the assumption only that f was continuous and differentiable at a single $x$-value, as opposed to continuous on a larger interval, but the proof couldn’t work with $a=b$, since $h$ would only be defined at a point, and $h'$ wouldn’t make sense. Thanks for noting this. $\endgroup$
    – Steve Kass
    Dec 4, 2014 at 6:09

3 Answers 3


You are right, in the proof of Rudin only is used the continuity of $f$ at $x$ (consequence of differentiability without more hypothesis). In another sources (Wikipedia, Cartan...) only the differentiability of $f$ at $x$ and of $g$ at $f(x)$ is required.

  • 2
    $\begingroup$ One can also compare to Rudin's Theorem 9.15 (the Chain Rule for vector-valued functions), where there is no assumption of continuity. $\endgroup$ Dec 3, 2014 at 7:51

In the first edition (1953) Rudin writes at the start of the proof:

First of all, Theorems 4.10 and 4.19 show that $R$ is an interval so that it makes sense to talk about the derivative of $g$ (we have defined the derivative only for functions defined on intervals and segments).

The premise is missing in the other editions.

Note1. The theorem assumes that "$g$ is defined on the range $R$ of $f$".

Note2. Rudin calls $[a,b]$ an interval, $]a,b[$ a segment .


Basically, what happens in Rudin's theorem 5.5 is this.

We start constructing an expression for a limit. \begin{align} \frac{g(f(t)) - g(f(x))}{\underbrace{f(t) - f(x)}_\text{comment *}} = g'(f(x)) \quad (A) \end{align}

Comment (*): here we use the assumption of continuity of $f(x)$, because we can find $\delta >0$ to place $t$ in the $\delta$-neighborhood of $x$. In other words, so that the denominator will converge to zero as we start to push $t$ toward $x$.

Now we construct the main expression for the limit.

\begin{align} \lim_{t \to x} \frac{g(f(t)) - g(f(x))}{t-x} \quad \stackrel{\text{use (A) above}}{=} \lim_{t \to x} \quad \frac{g'(f(x)) \cdot \left[ f(t) -f(x) \right]}{t-x} = g'(f(x)) \cdot f'(x) \end{align}

as required


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