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THEOREM 9(THE CHAIN RULE)
If $g$ is differentiable at $a$, and $f$ is differentiable at $g(a)$, then $f\circ g$ is differentiable at $a$, and $$(f\circ g)^{'}(a)=f^{'}(g(a))\cdot g^{'}(a).$$

My proof of this theorem is here:

(1) First consider the case in which for any positive real number $\epsilon$, there exists $h$ such that $0<|h|<\epsilon$ and $g(a+h)-g(a)=0$.
In this case, $\lim_{h\to 0} \frac{g(a+h)-g(a)}{h}=0$ since $g'(a)$ exists and there exists $h$ such that $0<|h|<\epsilon$ and $g(a+h)-g(a)=0$ for any positive real number $\epsilon$.
Let $\phi(h):= \frac{f(g(a+h))-f(g(a))}{g(a+h)-g(a)}$ if $g(a+h)-g(a)\neq 0$.
Let $\phi(h):=f'(g(a))$ if $g(a+h)-g(a) = 0$.
Then $\phi$ is continuous at $h=0$ since $f$ is differentiable at $g(a)$.
$\frac{f(g(a+h))-f(g(a))}{h} = \phi(h) \cdot \frac{g(a+h)-g(a)}{h}\to f'(g(a))\cdot 0 = f'(g(a))\cdot g'(a) \,\,(h\to 0)$.

(2) Second consider the case in which there exists a positive real number $\epsilon$ such that $0<|h|<\epsilon\implies g(a+h)-g(a)\neq 0.$
In this case $\frac{f(g(a+h))-f(g(a))}{h} = \frac{f(g(a+h))-f(g(a))}{g(a+h)-g(a)} \cdot \frac{g(a+h)-g(a)}{h}\to f'(g(a))\cdot g'(a) \,\,(h\to 0)$.

I don't like the function $\phi$ because it is artificial.
Can we prove the chain rule without using an artificial trick?

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    $\begingroup$ If I attempt to provide an argument that $\phi$ is "natural" and not "aritificial", will this be acceptable? I'd start by saying that it models the "very natural" derivative of a function "with respect to another" in the form of $\frac{d(f \circ g)}{dg}(x)$. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Nov 20, 2022 at 6:38
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    $\begingroup$ Do the linear approximation proof that you learned in the multivariable setting. $\endgroup$
    – Ted Shifrin
    Commented Nov 20, 2022 at 7:13
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    $\begingroup$ The chain rule is super intuitive. A differentiable function is locally linear, which means that if the input to $f$ changes from $x$ to $x+\Delta x$ then the output changes by approximately $f’(x) \Delta x$. Now suppose $f = g \circ h$ and the input changes from $x$ to $x + \Delta x$. The output of $h$ changes by approximately $h’(x) \Delta x$, and so the output of $g$ changes by approximately $g’(h(x)) h’(x) \Delta x$. This reveals that $f’(x) = g’(h(x)) h’(x)$. Keep track of the error terms in these approximations to obtain a rigorous proof. $\endgroup$
    – littleO
    Commented Nov 20, 2022 at 7:33
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    $\begingroup$ There is simply no need of $\phi$ in first case (whether the use is natural or not I can't say, that's personal taste of author). For values of $h$ with $g(a+h) =g(a) $ we have the ratio $(f(g(a+h)) - f(g(a))) /h=0$ and for other values of $h$ the ratio can be written as a product of two ratios one of which is near $f'(g(a))$ and the other one is near $0$ so that $(f\circ g) '(a) =0$. See math.stackexchange.com/a/1853088/72031 $\endgroup$
    – Paramanand Singh ♦
    Commented Nov 21, 2022 at 2:15
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    $\begingroup$ @littleO Please consider posting that as an answer! $\endgroup$
    – Stef
    Commented Nov 21, 2022 at 13:13

2 Answers 2

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Start with a reformulation of differentiability avoiding quotients:

A function $f: I\to\mathbb R$ on a set $I\subseteq \mathbb R$ is differentiable at $a\in I$ if and only if there is $\varphi:I\to \mathbb R$ which is continuous at $a$ and satisfies $f(x)-f(a)=\varphi(x)(x-a)$. Then $\varphi(a)=f'(a)$.

If $f$ is differentiable at $a$ with correspondig function $\varphi$ and $g:f(I)\to\mathbb R$ is differentiable at $f(a)$ with corresponding function $\gamma$, we get $$g(f(x))-g(f(a))=\gamma(f(x)) (f(x)-f(a))=\gamma(f(x))\varphi(x)(x-a).$$ Since compostions and products of continuous functions are continuous we get that $g\circ f$ is differentiable at $a$ with corresponding function $\gamma(f(x))\varphi(x)$ whose value at $a$ is $g'(f(a))f'(a)$.

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    $\begingroup$ And just to make it super obvious to OP and future readers, the function $\phi$ is given by $\phi(x)=\frac{f(x)-f(a)}{x-a}$ if $x\neq a$, and $\phi(a):=f'(a)$. This is a very intuitive thing to try because when $x\neq a$, it is the usual difference quotient, while at $x=a$, we're just "filling in" the removable discontinuity of the difference quotient and making a continuous function (at $a$) out of it (it is removable if and only if $f$ is differentiable at $a$ of course). $\endgroup$
    – peek-a-boo
    Commented Nov 20, 2022 at 8:31
  • $\begingroup$ Does this proof generalise to higher dimensions? $\endgroup$
    – user1551
    Commented Nov 21, 2022 at 12:49
  • $\begingroup$ Very good point, @user1551. Unfortunately, it does not so easily. Dealing with the affine-linear approximations provided by the definition of differentiability in higher dimensions requires some more care. $\endgroup$
    – Jochen
    Commented Nov 21, 2022 at 13:44
  • $\begingroup$ @user1551 For $\mathbb R^n \rightarrow \mathbb R$, the existence of the derivative at the origin would require that for each vector $v$, there is $\phi_v$ such that $f(tv) = \phi_v(t)f(v)$. There would then have to be linearity, so for any scalars $a,b$ and vectors $v,w$, $\phi_{av+bw}(t) = a\phi_v(t)+b\phi_w(t)$. The gradient is then $[\phi_{e1}(0)...\phi_{en}(0)]$ where $e_i$ are the elementary vectors. For the derivative at other points, you'd have to translate these equations. $\endgroup$
    – Acccumulation
    Commented Nov 22, 2022 at 4:37
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If $f,g$ are continuously differentiable, then it is possible to prove directly.
Here is an argument:
Using mean value theorem: $g(a+h) = g(a) + g'(x) h$ with $x \in [a,a+h]$, $$ \lim_{h \rightarrow 0} \frac{f(g(a+h))-f(g(a))}{h} = \lim_{h \rightarrow 0} \frac{f(g(a)+g'(x)h)-f(g(a))}{h} $$ Again using mean value theorem, now in $f$: $f(g(a)+h) = f(g(a)) + f'(y) h$ with $y \in [g(a),g(a)+h]$, $$ = \lim_{h \rightarrow 0} \frac{f(g(a)) + f'(y) g'(x) h - f(g(a))}{h} = \lim_{h \rightarrow 0} \frac{f'(y)g'(x)h}{h} = f'(g(a)) g'(a)$$

The last step: $\lim_{h \rightarrow 0} f'(y)g'(x) = f'(g(a)) g'(a)$ uses the fact that derivatives are continuous.

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