# Math Help - Question 13

1. ## Question 13

(I might be wrong on this historical note). The famous mathematician, G.H. Hardy wrote in the book A course in Pure Mathematics a proof of the chain rule.

If $f(x),g(x)$ are differenciable at $c$. Then, $f(g(c))'=g'(c)f'(g(c))$. He made the most simple proof.
He wrote,
$\frac{\Delta y}{\Delta x}=\frac{\Delta y}{\Delta u}\cdot \frac{\Delta u}{\Delta x}$.
$\lim_{x\to c}\frac{f(g(x))-f(g(c))}{x-c}=\lim_{x\to c}\frac{f(g(x))-f(g(c))}{g(x)-g(c)}\cdot \frac{g(x)-g(c)}{x-c}$.
The first fraction is the derivative of $f(x)$ only evaluated at $g(c)$. The second fraction is derivative of $g(x)$ evaluated at $c$.
Thus,
$f'(g(c))g'(c)$.

But in the next edition of the book he omitted the proof.
Why? What ailed him?

2. As far as I can see, you introduce a new denominator but you don't à priori know if that expression isn't zero. In this case, if g(x) = g(c), you have a problem. If g(x)-g(c) is not zero, you can indeed take the limit and you're done. However, to allow the case where g(x) = g(c), it's a bit more subtle. This problem can be resolved by defining another function, which doesn't change the case where g(x) doesn't equal g(c), but which takes the desired derivative by definition in the case of g(x) = g(c).

3. Originally Posted by TD!
As far as I can see, you introduce a new denominator but you don't à priori know if that expression isn't zero. In this case, if g(x) = g(c), you have a problem. If g(x)-g(c) is not zero, you can indeed take the limit and you're done. However, to allow the case where g(x) = g(c), it's a bit more subtle. This problem can be resolved by defining another function, which doesn't change the case where g(x) doesn't equal g(c), but which takes the desired derivative by definition in the case of g(x) = g(c).
I think I understand what you are saying. It could have been said a little better. Let me re-edit what you said.

There is a theorem that says that if two functions are agree on some open interval except possibly at the limit point then the two limits are equal.
Hence if,
$A(x)=\frac{f(g(x))-f(g(c))}{x-c}$
And,
$B(x)=\frac{f(g(x))-f(g(c))}{g(x)-g(c)}\cdot \frac{g(x)-g(c)}{x-c}$.
Thus, we are assuming what,
$A(x)=B(x), \forall x\in I-\{c\}$
What TD! is saying (I think) we do not know that.
Because if $g(x)=g(c)$ on some open interval except possible at $c$. And what we said fails.

Good job, TD!
-------
But there is another thing that makes it faulty.

4. Originally Posted by TD!
As far as I can see, you introduce a new denominator but you don't à priori know if that expression isn't zero. In this case, if g(x) = g(c), you have a problem. If g(x)-g(c) is not zero, you can indeed take the limit and you're done. However, to allow the case where g(x) = g(c), it's a bit more subtle. This problem can be resolved by defining another function, which doesn't change the case where g(x) doesn't equal g(c), but which takes the desired derivative by definition in the case of g(x) = g(c).
Which is what Hardy said the problem with the original proof was.
(I have posted the text from the 10th edition somewhere where
you cannot see it and this is the gist of what Hardy says was wrong)

IPH says:

I however have a different explanation why there is a flaw.
RonL

5. Originally Posted by CaptainBlack
where
you cannot see it

6. Originally Posted by TD!
Well this is supposedly the proble of the week and I did not want to give anything away. But that seems to have happeded anyways,m so here it is:

This is item (6) from section 114 of Hardy's "A Course of Pure Mathematics" 10th Ed, pages 217-218:

(6) If $\phi(x)=F\{f(x)\}$, then $\phi(x)$ has a derivative

........ $\phi'(x)=F'\{f(x)\}f'(x)$

The proof of this theorem requires a little care $^*$.
We write $f(x)=y,\ f(x+h)=y+k$, so that $k \to 0$ when $h \to 0$and

................. $k/h \to f'(x)\ .....................(1).$

We must now distinguish two cases.

(a) Suppose that $f'(x) \ne 0$, and that $h$ is small, but not zero. Then $k \ne 0$ because $(1)$, and

........ $\frac{\phi(x+h)-\phi(x)}{h}=\frac{F(y+k)-F(y)}{k}\,\frac{k}{h}$ $\to F'(y)f'(x)$.

(b) Suppose that $f'(x)=0$, and that $h$ is small, but not zero. There are now two possibilities. If $k=0$, $^{\dagger}$ then

........ $\frac{\phi(x+h)-\phi(x)}{h}=0$.

If $h \ne 0$, then

........ $\frac{\phi(x+h)-\phi(x)}{h}=\frac{F(y+k)-F(y)}{k}\,\frac{k}{h}$.

The first factor in nearly $F'(y)$, and the second is small, because $k/h \to 0$. Hence $\{\phi(x+h)-\phi(x)\}/h$ is small in any case, and

........ $\frac{\phi(x+h)-\phi(x)}{h}\to 0 = F'(y)f'(x)$

$^*$ The proofs in many text-books (and in the first three editions of this book) are inaccurate. See a note by Prof H. S. Carslaw in vol XXIX of the Bulletin of the American Mathematical Society.
$^{\dagger}$The fallecy in the inaccurate proofs lies in overlooking this possibility

RonL

7. What TD! said was the "official" answer. That we need to show that the two functions (after the manipulation) agree on some open interval which is not necessarily true.

I have my own answer. Let us assume that we can write them like that, we will still arrive at a problem.
Look at the limits again,
$\lim_{x\to c} \frac{f(g(x))-f(g(c))}{g(x)-g(c)}\cdot \frac{g(x)-g(c)}{x-c}$
Now, the theorem said that the limit of the product is the product of the limits whenever both limit exists.
The second function,
$\frac{g(x)-g(c)}{x-c}$
Certainly exists when $x\to c$. Because it is differenciable.
But what about the first function.
I said it is the derivative only evaluated at $g(c)$.
Mathematically I used the limit composite rule.
$\lim_{x\to c}F\circ G$
$G=g(x)$
$F=\frac{f(x)-f(g(c))}{x-f(g(c))}$
Thus, we have,
$\lim_{x\to c}G=g(c)$
Since the function is differenciable it is continous.
And hence we need to show that, (limit composite rule)
$\lim_{x\to g(c)}\frac{f(x)-f(g(c))}{x-f(g(c))}$.
But we DO NOT know that.
Because we are claiming that $f(x)$ is differenciable at $x=g(c)$. It is differenciable at $x=c$ but we do not know whether or not $x=g(c)$.