# Proof of chain rule of differentiation for $g(f(x))$.

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 $$D(g)$$?

- 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 $$(g \circ f)'(x)$$?

• 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. May 23, 2020 at 13:02
• See proper proof here math.stackexchange.com/a/1853088/72031 May 23, 2020 at 13:04
• 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. May 23, 2020 at 23:47
• @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 $f(x) - f(x_0)$. May 24, 2020 at 8:47

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.

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.

• 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?
– Iti
May 23, 2020 at 12:51
• @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 . May 23, 2020 at 13:25