I created this proof of the chainrule. Being a (relative) beginner at math I have a few questions.

  1. Is the proof below correct? I was especially in doubt about the use of $h$ on both sides.
  2. Is the (Langrange?) notation correct this way?
  3. How to write the same proof using Leibniz's notation? I wrestled writing this proof in Leibniz notation, because what would in that case be the meaning of $dg$? Is it $g(x+h)-g(x)$ or $k$ or $h$?

To be proved:

If $f(u)$ is differentiable at $u=g(x)$, and $g(x)$ is differentiable at $x$ then:

$$f(g(x))'\stackrel{?}{=}f'(g(x))g'(x)$$ Or similarly $$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}\stackrel{?}{=}\lim \limits_{k \to 0}\frac{f(g(x)+k)-f(g(x))}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$

Case 1: if $h$ has a value such that $g(x+h)=g(x)$ then: $$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=0$$ And $$\lim \limits_{k \to 0}\frac{f(g(x)+k)-f(g(x))}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}=0$$

Both sides of the equation to prove equal zero, therefore the equation holds in this case.

Case 2: if $h$ has a value such that $g(x+h)\ne g(x)$ then:

We multiply the lefthandside by $\frac{g(x+h)-g(x)}{g(x+h)-g(x)}$ $$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$ Taking $$u=g(x)$$ $$k=g(x+h)-g(x)$$ We get $$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=\lim \limits_{h \to 0}\frac{f(u+k)-f(u)}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$ And as $h\to 0, k\to 0$, therefore $$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=\lim \limits_{k \to 0}\frac{f(u+k)-f(u)}{k}\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}$$ Thus $$f(g(x))'=f'(u)g'(x)=f'(g(x))g'(x) \tag*{$\blacksquare$}$$

  • 3
    $\begingroup$ Be careful when you say $g(x+h)\neq g(x)$ what do you mean ? If you mean that $g$ is not a constant function near $x$ then the proof incomplete. You do not know what happens to the function $h\mapsto g(x+h)-g(x)$ for all "small" $h$. This function can oscillate, i.e., hit zero at some point while $h$ approaches zer, and this is the difficulty of the proof. Please see how to go around this in any standard book, line Bartle's introduction to real analysis. $\endgroup$ – Medo Oct 11 '18 at 20:02
  • $\begingroup$ @Medo I am going around the problem (I think) by first considering the case where $g(x+h)-g(x)=0$ $\endgroup$ – GambitSquared Oct 11 '18 at 20:04
  • 2
    $\begingroup$ @Medo My high school book jumped over the obstacle by setting $k=g(x+h)-g(x)$ and saying that if $k=0$ the case would be trivial. Yes, really! I always show that to my students of “Mathematics Teaching” as a memento. $\endgroup$ – egreg Oct 11 '18 at 20:05
  • 2
    $\begingroup$ @GambitSquared. Yes, but $g(x+h)-g(x)=0$ is an identity. It means $g(x+h)-g(x)=0$ for every $h$ which happens exclusively if $g$ is constant near zero. $\endgroup$ – Medo Oct 11 '18 at 20:06
  • 1
    $\begingroup$ @egreg: this is unfortunately the case with many textbooks. A simple observation is that if every neighborhood of $h=0$ contains some $h$ with $g(x+h) =g(x) $ then $g'(x) =0$ and then we need to show that $(f\circ g) '(x) =0$. $\endgroup$ – Paramanand Singh Oct 12 '18 at 6:46

One thing that's worth learning is the notation for the composition of $f$ and $g.$ We use $f\circ g$ to denote this function. I.e., $(f\circ g)(x) = f(g(x)).$

In your To be proved, we can use this notation. You wrote $f(g(x))' = f'(g(x))g'(x).$ The problem with this is that on the left you have $'$ followed by blank space, whereas on the right the $'$ symbols are followed by further notation. It's an inconsistency that can be corrected by writing

$$(f\circ g)'(x)=f'(g(x))g'(x).$$

On your paragraph that starts "Or similarly": You have

$$\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}$$ $$=\lim \limits_{k \to 0}\frac{f(g(x)+k)-f(g(x))}{k}\cdot\lim \limits_{h \to 0}\frac{g(x+h)-g(x)}{h}.$$

That is fine. You asked about having $h$ on both sides, but that is no problem whatsoever. In fact you could replace $k$ by $h$ on the right. The variable $h$ is called a "dummy variable", meaning any symbol could be used there (except for the symbols that already have meaning, like $x.$)

That's the small stuff. Others have pointed out the big mistakes, where you divide the proof into two cases: i) There is an $h\ne 0$ such that $g(x+h)-g(x) =0,$ and ii) There is an $h\ne 0$ such that $g(x+h)-g(x) \ne 0.$ Huge problem here: a few values of $h$ cannot tell you anything about about a limiting process, where we are letting $h\to 0$ through infinitely many values.

A better division into cases is this: case i) $g'(x)=0;$ case ii) $g'(x)\ne 0.$ How would the proofs go in these cases?

Proof for case i): Observe the following: There is a constant $C$ such that

$$|f(y)-f(g(x))|\le C|y-g(x)|$$

for all $y$ sufficiently close to $g(x).$ This follows from the existence of $f'(g(x)).$ Thus for small $h\ne 0,$

$$\left |\frac{f(g(x+h))-f(g(x))}{h} \right | \le C\frac{|g(x+h))-g(x)|}{|h|} \to C\cdot |g'(x)|=0.$$

Thus $(f\circ g)'(x) = 0,$ which is exactly what we want in this case.

Proof for case ii): This is the easy case. We need only observe that $g'(x)\ne 0$ implies $g(x+h)-g(x)\ne 0$ for all small nonzero $h.$ For such $h$ we can do what all beginners crave to do:

$$\frac{f(g(x+h))-f(g(x))}{h} = \frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\frac{g(x+h)-g(x)}{h} \to f'(g(x))g'(x).$$

I've been brief in these proofs. Please ask if you have questions.

  • $\begingroup$ Also @user21820 pointed out that even if $g(x+h)-g(x) \ne 0$ doesnt necessarily mean that $\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}$ exists. Or am I missing something here? $\endgroup$ – GambitSquared Oct 18 '18 at 8:22
  • $\begingroup$ @GambitSquared: I did not say we cannot assume existence of $f'(g(x))$. I also wrote "for some $h$", whereas zhw. wrote "for all small nonzero $h$". Understanding completely how such arguments work would require a proper grasp of logic, because concepts in real analysis involve a number of alternating quantifiers. This post left out significant details of that sort (such as what "small" means), but anyway that is what a textbook proof is supposed to be for. $\endgroup$ – user21820 Oct 18 '18 at 8:49
  • $\begingroup$ @user21820 pointed out that we cannot assume the existence of $(f \circ g)'(x)$ $\endgroup$ – GambitSquared Oct 18 '18 at 9:21
  • $\begingroup$ @GambitSquared: zhw. didn't assume that. Symbolically, zhw. showed that, if $g(x+h)-g(x) \ne 0$ as $h \to 0$ (A), then as $h \to 0$ we also have the following: $\frac{f(g(x+h))-f(g(x))}{h} = \frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\frac{g(x+h)-g(x)}{h}$; $g(x+h) ≈ g(x)$ by differentiability of $g$ at $x$, so $g(x+h) \to g(x)$ by (A), and hence $f(g(x+h)-g(x))/(g(x+h)-g(x)) ≈ f'(g(x))$ by differentiability of $f$ at $g(x)$; $(g(x+h)-g(x))/h ≈ g'(x)$. Therefore, we also have $\frac{f(g(x+h))-f(g(x))}{h} ≈ f'(g(x))·g'(x)$ (as $h \to 0$). $\endgroup$ – user21820 Oct 18 '18 at 9:32
  • $\begingroup$ @zhw What actually is a "limiting process"? Why do you need all values for a succesful limiting process? $\endgroup$ – GambitSquared Feb 12 at 21:51

Your case 1 is irreparably flawed.$ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $ In that case you claimed that if $g(x+h)=g(x)$ for some $h$ then $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{h} = 0$. That is false. For example let $f$ be the identity function, and $g = \sin$ and $x = 0$ and $h = π$. Then $g(x+h) = g(x)$ but $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{h}$ $= \lim_{h\to0} \lfrac{\sin(h)-\sin(0)}{h} = 1$, contradicting your claim.

Your case 2 is also completely broken for the same logical reason, because even if $g(x+h) \ne g(x)$ for some $h$ it does not mean that $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{g(x+h)-g(x)}$ exists, so your first line in that case is already wrong. For example let $f$ be the identity function again, and $g(t) = |t-1|+|t+1|$ for every real $t$, and $x = 0$. Then $g(2) \ne g(0)$ but $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{g(x+h)-g(x)}$ does not exist because $g(x+h)-g(x) = 0$ for every $h \in [-1,1]$. Note that $g'(0) = 0$ and $f'(2) = 1$, and the chain-rule still holds, but the limit you wrote down does not exist.

  • $\begingroup$ Isn't the limit of $\lim_{h\to0} \lfrac{f(g(x+h))−f(g(x))}{g(x+h)-g(x)}=1$ Since the nominator and denominator are equal if $f$ is the identity function? $\endgroup$ – GambitSquared Oct 14 '18 at 9:17
  • $\begingroup$ And about the second case: we know that $\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}$ exists, because it's a condition of the chainrule. Then why can't I multiply that limit with $\frac{g(x+h)-g(x)}{g(x+h)-g(x)}$? (and switch the denominators) $\endgroup$ – GambitSquared Oct 14 '18 at 14:03
  • $\begingroup$ @GambitSquared: Think slowly and carefully about the example I gave in my post. Given the particular $g$ in my example, does the expression you wrote in your last comment make sense? Also, in mathematics we do not anyhow switch anything; everything must be justified 100% logically. $\endgroup$ – user21820 Oct 14 '18 at 15:13
  • $\begingroup$ @GambitSquared: And you're wrong about it being a condition of the chain rule. The conditions are that $f$ is differentiable at $g(x)$ and $g$ is differentiable at $x$, but the existence of the limit you wrote is totally different; it is if and only if $f \circ g$ is differentiable at $x$. $\endgroup$ – user21820 Oct 14 '18 at 15:17
  • $\begingroup$ I still don't get why I can't multiply with $\frac{g(x+h)-g(x)}{g(x+h)-g(x)}$ Isn't this equal to $1$ as long as $g(x+h)-g(x) \ne 0$, which was the condition for that case? $\endgroup$ – GambitSquared Oct 18 '18 at 8:07

This answer will attempt to elaborate further on the problem that Medo pointed out in the comments:

Fix some value of $x$ and consider the sets $$ H=\{h\mid\ g(x+h)= g(x)\} \\ H^c=\{h\mid\ g(x+h)\neq g(x)\} $$ Here $H$ corresponds to your case 1, and $H^c$ corresponds to your case 2.

  • Either set may be infinite (countable or uncountable) or empty and $H$ may also be finite.
  • So first of all, there is no guarantee that $h=0$ is a limit point of both sets meaning $h\to 0$ may not even make sense within both sets. In that case one can simply argue using the set in which $h=0$ is in fact a limit point, and everything should be fine.
  • Second of all, if $h=0$ is a limit point of both $H$ and $H^c$ so that $h\to 0$ makes senses for both, we need an extra argument to make sure that the two limits are equal. This is where some work remains to be done.

The way I have developed for dealing with the problem uses a more Leibnizian approach, namely define: $$ \frac{\Delta f(g)}{\Delta x} = \begin{cases} 0 & \text{for }\Delta g=0 \\ \quad\\ \frac{\Delta f}{\Delta g}\cdot\frac{\Delta g}{\Delta x} & \text{for }\Delta g\neq 0 \end{cases} $$ where $\Delta g$ and $\Delta f$ denote the corrsponding changes of $g(x)$ and $f(g(x))$ when $x$ is changed by $\Delta x$. This can be shown to be continuous and always equal to $$ \frac{f(g+\Delta g)-f(g)}{\Delta x} $$ and so it provides a continuous alternative to the problematic fatorization by "filling in" the missing values when $\Delta g=0$.

  • $\begingroup$ What do you mean by $h=0$? Since it is a limit where $h\to 0$, then $h$ will never be equal to $0$ isn't it? However, the value of $g(x+h)$ may be equal to $g(x)$. What am I missing here? $\endgroup$ – GambitSquared Oct 12 '18 at 6:58
  • $\begingroup$ @GambitSquared: You are correct that plugging in $h=0$ would not make sense. But $h=0$ can still be a limit point for either set. $\endgroup$ – String Oct 12 '18 at 7:07

The main reason that this is not written well is that you are manipulating limits. This is very hard to read because there is a non-trivial claim built into each use of "$\lim$".

Unless you have separately verified it, the fact that the function $f \circ g$ is differentiable at $x$ is really part of the conclusion, so if you start with this limit on the left-hand side and then manipulate that expression, in my book you have committed an unforgivable analysis sin immediately.

If you are just doing algebra, then just write the algebra without $\lim$ in front of everything.

If you are taking a limit, justify it exists first, and then take the limit.

  • $\begingroup$ $f(g(x))'=\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}$ is the definition of the derivative is it not? $\endgroup$ – GambitSquared Oct 11 '18 at 21:33
  • $\begingroup$ But derivatives don't always exist. My point is that you cannot write down $\lim_{h \to 0}$ of something until you know that the limit exists. Its like what if $f(x) = |x|$ and I write $f'(x) = \lim_{h \to 0} \tfrac{f(h)}{h}$ and start manipulating the right-hand side as if it is a number? $\endgroup$ – T_M Oct 11 '18 at 22:31
  • $\begingroup$ Ok, good point. I fixed it by adding the conditions for which the chainrule is valid. $\endgroup$ – GambitSquared Oct 12 '18 at 6:34

(Note: the proof has been edited so the comment below no longer applies.)

In case 1, you are assuming that $g(x+h) = g(x)$ for all real numbers $h$. In case 2, you are assuming that $g(x+h) \neq g(x)$ for all nonzero real numbers $h$. However, there is a third case that you have not covered, which is the case where $g(x+h)=g(x)$ for some but not all nonzero real numbers $h$.

By the way, the assumptions you made in each case could have been stated more clearly by inserting phrases such as "for all real numbers $h$". You could also use phrasing like, "if $h$ is a nonzero real number then $g(x+h) \neq g(x)$."

Follow-up comment: The proof has been revised to say:

if $h$ has a value such that $g(x+h)=g(x)$ then: $$\tag{1}\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}=0$$

But, how does equation (1) follow from the fact that there is a value of $h$ such that $g(x + h) = g(x)$? That is a non-sequitur.

Equation (1) would be obviously true if $g(x + h) = g(x)$ for all real numbers $h$. But case 1 (as written currently) only assumes that there exists a value of $h$ such that $g(x + h) = g(x)$.

  • $\begingroup$ What I mean in case one is only the cases for $h$ where $g(x+h)=g(x)$, so not all real numbers. Similarly in case 2. Therefore I don't understand your comment. $\endgroup$ – GambitSquared Oct 14 '18 at 7:01
  • $\begingroup$ @GambitSquared I see that you edited the proof to clarify that point, thanks. I edited my answer to state a different objection that I have now to the revised proof. $\endgroup$ – littleO Oct 14 '18 at 8:35

An easy way to avoid the problem with the case $g'(p)=0$ is to perturb $g$ by a linear function: we can evaluate the derivative $$ \frac{d}{dx}f(g(x)+\epsilon x)\Big|_{x=p} $$ using the "Leibniz way" since $g'(p)+\epsilon\neq 0$. Hence, $$ \frac{d}{dx}f(g(x)+\epsilon x)\Big|_{x=p}=f'(g(p)+\epsilon p)(0+\epsilon). $$ The left hand side depends continuously on $\epsilon$, so letting $\epsilon\to 0$ gives $$ \frac{d}{dx}f(g(x))\Big|_{x=p}=0=f'(g(p))g'(p). $$


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