6
$\begingroup$

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 $$f(g(x))'=f'(u)g'(x)=f'(g(x))g'(x) \tag*{$\blacksquare$}$$

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
+200
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $\Delta g=0$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $f(g(x))'=\lim \limits_{h \to 0}\frac{f(g(x+h))-f(g(x))}{h}$ 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
1
$\begingroup$

(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 $g(x + h) = g(x)$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$ \frac{d}{dx}f(g(x))\Big|_{x=p}=0=f'(g(p))g'(p). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.