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When I read about the proof of chain rule of differentiation in a real analysis book

$(g\circ f)'(x_0)= \lim\limits_{x \to x_0}\frac{g(f(x))-g(f(x_0))}{x-x_0}$ $=\lim\limits_{x \to x_0}\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}.\frac{f(x)-f(x_0)}{x-x_0}$

They don't explain the case $f(x)=f(x_0)$.

I know the reason that this proof works only when $f(x)\neq f(x_0)$. But to prove the case of $f(x)=f(x_0)$, how we find some $\delta$ such that $f(x)\neq f(x_0)$ $\forall x\in N'_{\delta}x_0$.What is the rigorous proof for it?

Can we prove this rule if $f(x)$ is a constant function $\forall x \in D(f)$? I think we can't prove the chain rule in this case because then $f(x)$ has only one value and it can't be in $D(g)$?

[Edit]- After reading all the answers, I have also tried to do it in new way. Please check it if any step is wrong.

As f is differentiable at $x_0$. $\forall \delta_2>0, \exists\delta_1>0$ s.t. $\forall x \in |x-x_0|<\delta_1 \implies |\frac{f(x)-f(x_0)}{x-x_0}|<\delta_2$ Also $g(u)$ is differentiable at $g(f(x_0))$. So $\forall \epsilon>0, \forall u \in |u-f(x_0)|<\delta_2\delta_1 \implies |\frac{g(u)-g(f(x_0)}{u-f(x_0)}|<\epsilon/\delta_2$ $\exists \delta_3>0$ s.t $\forall x \in |x-x_0|<\delta_3, f(x) \in D(g).$ Now take $\delta = min{\delta_1,\delta_3}$ $\forall x \in |x-x_0|<\delta \implies f(x)\in D(g)$ and $|f(x)-f(x_0)|<\delta_2 |x-x_0|$ $\implies x \in D(g \circ f) and |\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}|<\epsilon/\delta_2$ $\implies |\frac{g(f(x))-g(f(x_0))}{x-x_0}|<\epsilon$

So, $g(f(x))$ is differentiable at $x_0$. Now we can use the highlighted proof in the post without concerning division by zero to find the value of $(g \circ f)'(x)$?

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    $\begingroup$ I think you need to change your real analysis book. Analysis books should be written with more care compared to typical trash available for calculus. $\endgroup$
    – Paramanand Singh
    May 23, 2020 at 13:02
  • $\begingroup$ See proper proof here math.stackexchange.com/a/1853088/72031 $\endgroup$
    – Paramanand Singh
    May 23, 2020 at 13:04
  • $\begingroup$ Your original question was to clarify doubts on an (erroneous) proof in your real analysis book. After you received answers, you completely changed your question by more or less copying Christian Blatter's answer and asking whether it is correct. I don't think that this an appropriate approach. If you receive answers, you should give response by accepting and up- or downvoting (see math.stackexchange.com/help/accepted-answer, math.stackexchange.com/help/why-vote). In case of a downvote it would be desirable if you explain why you do not think the answer is useful. $\endgroup$
    – Paul Frost
    May 23, 2020 at 23:43
  • $\begingroup$ I think people who spend their time in answering your questions deserve at least any reaction. And please do not change your questions by edits producing something completely new. Ask a new question in that case. $\endgroup$
    – Paul Frost
    May 23, 2020 at 23:47
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    $\begingroup$ @lti There is no obligation for voting, you need not do anything. But in my opinion it is a matter of good communication to give feedback to people who answer your questions. Concerning your edit: It has again the same problem. The expression $|\frac{g(f(x))-g(f(x_0))}{f(x)-f(x_0)}|$ is undefined if $f(x) = f(x_0)$. You can use $\epsilon$-$\delta$-arguments in the spirit of Christian Blatter's answer using the function $m$, but you have to avoid dividing by $f(x) - f(x_0)$. $\endgroup$
    – Paul Frost
    May 24, 2020 at 8:47

2 Answers 2

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You need the following Lemma that makes differentiation denominator free:

If $f:\>I\to{\mathbb R}$ s differentiable at $x_0\in I$ then there is a function $m:\>I\to{\mathbb R}$ which is continuous at $x_0$ such that $m(x_0)=f'(x_0)$ and $$f(x)-f(x_0)=m(x)\>(x-x_0)\qquad(x\in I)\ .\tag{1}$$ Conversely: If $(1)$ holds with $m$ continuous at $x_0$ then $f$ is differentiable at $x_0$, and $f'(x_0)=m(x_0)$.

In your case, let $f(x_0)=:y_0$. Then there is an $m_f$ for $f$ at $x_0$ and an $m_g$ for $g$ at $y_0$, and one has $$\eqalign{g\bigl(f(x)\bigr)-g\bigl(f(x_0)\bigr)&=g\bigl(f(x)\bigr)-g(y_0)\cr &=m_g\bigl(f(x)\bigr)\bigl(f(x)-y_0\bigr) =m_g\bigl(f(x)\bigr)\bigl(f(x)-f(x_0)\bigr)\cr&=m_g\bigl(f(x)\bigr)\>m_f(x)\>(x-x_0)\ .\cr}$$ The chain rule now follows with the converse part of the Lemma.

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The proof does definitely not work. As you say, if $f$ is constant, then you have $f(x) - f(x_0) = 0$ for all $x$. But even if $f$ is not constant you may have $f(x_n) - f(x_0) = 0$ for suitable sequences $(x_n)$ in $D(f) \setminus \{x_0\}$ such that $x_n \to x_0$. An example is $f(x) = x^2\sin(1/x)$ for $x \ne 0$, $f(0) = 0$. In that case $f(1/n\pi) = 0 = f(0)$ for all $n$.

For a correct proof you should consult another textbook.

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  • $\begingroup$ So then if we can't find such $\delta>0$ for which $|x-x_0|<\delta \implies f(x)\neq f(x_0)$. Then how we generalize this rule? $\endgroup$
    – Iti
    May 23, 2020 at 12:51
  • $\begingroup$ @lti No, we can't find $\delta$. The chain rule is of course a valid theorem, but another proof is needed. Have a look into another textbook or read en.wikipedia.org/wiki/Chain_rule . $\endgroup$
    – Paul Frost
    May 23, 2020 at 13:25

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