# "Proof" of chain rule

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!! ☺

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

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.

• 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 $\Delta u$. Nov 1, 2016 at 0:39
• @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. Nov 1, 2016 at 14:21

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

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

• 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 Nov 1, 2016 at 0:27
• @user1952009 It is not. It is the given Chain rule proof via infinitesimals written using Leibniz notation. Nov 1, 2016 at 0:29
• Yeah sure, define the hyperreal numbers for proving something trivial Nov 1, 2016 at 0:35
• @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.) Nov 1, 2016 at 0:37
• 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. Nov 1, 2016 at 0:42