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I wanted a midly rigorous proof to make the chain rule more intuitive to me, and this is what I found: http://web.mit.edu/wwmath/calculus/differentiation/chain-proof.html

However, I am not sure how how going from step two to three on this part of the proof is justified: http://web.mit.edu/wwmath/calculus/differentiation/chaineq/chaineq31.gif

High school sophomore in calc ab

Thanks for any help!! ☺

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    $\begingroup$ the use of infinitesimal is blurring and ambiguous $\endgroup$
    – janmarqz
    Oct 8, 2015 at 2:10
  • $\begingroup$ I know that, but if infinitesimals weren't being used, how would going from step 2 to 3 be justified? $\endgroup$
    – Silas Olsen
    Oct 8, 2015 at 2:15
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    $\begingroup$ take a glimpse at en.wikipedia.org/wiki/Chain_rule#Proofs $\endgroup$
    – janmarqz
    Oct 8, 2015 at 2:18
  • $\begingroup$ @janmarqz: This proof does not use infinitesimals. $\endgroup$
    – Paramanand Singh ♦
    Oct 8, 2015 at 4:19
  • $\begingroup$ what is this you mention? @ParamanandSingh $\endgroup$
    – janmarqz
    Oct 8, 2015 at 14:35

2 Answers 2

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The proof given in your link is wrong, but in a very subtle way. The main problem with the proof is that there may be cases where $\Delta u = 0$ identically when $\Delta x \to 0$. And then you can't do division and multiplication by $\Delta u$. This case however is possible only when $du/dx = 0$ and further it is easy to show that $dy/dx = 0$ in this case so that the chain rule $dy/dx = dy/du \times du/dx$ remains valid here. Apart from this small gap, the proof in your link is OK.

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  • $\begingroup$ There are also cases where $\Delta u = 0$ along a sequence of values of $\Delta x$ going to zero. So you still cannot multiply and divide by $\Delta u$. $\endgroup$
    – GEdgar
    Nov 1, 2016 at 0:39
  • $\begingroup$ @GEdgar: I meant to include this case also, but perhaps the language was not explicit/clear enough. The derivative $du/dx$ is $0$ in this case. $\endgroup$
    – Paramanand Singh ♦
    Nov 1, 2016 at 14:21
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Here is an intuitive quasi-proof:

Let $\frac{dF(x)}{dx}=f(x)$ and $g=g(x)$

$$\frac{dF(g)}{dx}=\frac{dF(g)}{dx}\frac{dg}{dg}=\frac{dF(g)}{dg}\frac{dg}{dx}=f(g)g'(x)$$

Since $\frac{dF(g)}{dg}=f(g)$ and $\frac{dg}{dx}=g'(x)$

If one does not feel justified, we can use the following:

$$\frac{F(g(x))-F(g(h))}{x-h}=\frac{F(g(x))-F(g(h))}{x-h}\frac{g(x)-g(h)}{g(x)-g(h)}=\frac{F(g(x))-F(g(h))}{g(x)-g(h)}\frac{g(x)-g(h)}{x-h}$$

So we have

$$\frac{dF(g)}{dx}=\lim_{h\to x}\frac{F(g(x))-F(g(h))}{x-h}=\lim_{h\to x}\frac{F(g(x))-F(g(h))}{g(x)-g(h)}\frac{g(x)-g(h)}{x-h}\\=f(g)g'(x)$$

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  • $\begingroup$ this is clearly the worst thing to say. If you want an intuitive proof, use instead $F(x) = ax+b, g(x) = cx+d$, and make it rigorous by adding the $o(x)$ terms $\endgroup$
    – reuns
    Nov 1, 2016 at 0:27
  • $\begingroup$ @user1952009 It is not. It is the given Chain rule proof via infinitesimals written using Leibniz notation. $\endgroup$
    – Simply Beautiful Art
    Nov 1, 2016 at 0:29
  • $\begingroup$ Yeah sure, define the hyperreal numbers for proving something trivial $\endgroup$
    – reuns
    Nov 1, 2016 at 0:35
  • $\begingroup$ @user1952009 Or we can use limit definitions of the derivative, but the message is the same. (and I agree seemingly trivial things are too complicated.) $\endgroup$
    – Simply Beautiful Art
    Nov 1, 2016 at 0:37
  • $\begingroup$ Or, you can use that $f'(a) = \alpha \Leftrightarrow f(a+h) = f(a)+ \alpha h+o(h)$ as $h\to 0$, which is useful, intuitive, and makes all the derivatives formulas trivial. $\endgroup$
    – reuns
    Nov 1, 2016 at 0:42

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