"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!! ☺
 A: 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.
A: 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
$$\frac{dF(g)}{dx}=\lim_{h\to x}\frac{F(g(x))-F(g(h))}{x-h}=\lim_{h\to x}\frac{F(g(x))-F(g(h))}{g(x)-g(h)}\frac{g(x)-g(h)}{x-h}\\=f(g)g'(x)$$
