I am looking into understanding the proofs for the Chain Rule. I came across this video by Dr. Peyam on youtube.
When proving the chain rule he states that:
$$\frac{g(y+\widetilde{h})-g(y)}{\widetilde{h}} = g'(y) + \sigma(\widetilde{h})$$ Where he refers to $\sigma$ as a junk term. And later he says that $\sigma \rightarrow 0$ as $\widetilde{h} \rightarrow 0$. Why can one use this alternative description of the derivative and why does the junk term become $0$?
Thank you very much.
This is a common and useful way to handle proofs involving derivatives. We know that $$ \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = f'(x) $$ If we define a funciton $\sigma(h)$ (I have seen $\varepsilon(h)$ used as function name for the same thing) as $$ \sigma(h) = \frac{f(x+h)-f(x)}{h} - f'(x) $$ then $$ \lim_{h\to 0}\sigma(h) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} - f'(x) = 0 $$ which proves why the "junk term" goes to zero.
Put in words, the "junk term" is the difference between the derivative and the approximation given by the difference quotiont. As $h$ gets smaller, the approximation gets better.
Bonus: It works the other way around too! If $$ \frac{f(x+h)-f(x)}{h} = g(x) + \sigma(h) $$ and $\lim_{h\to0}\sigma(h)=0$, then $g(x)=f'(x)$, which is equally useful for making proofs.
EDIT. By the way, $\sigma$ is really a function of both $x$ and $h$, but we tend to work with a fixed $x$ (or $y$ in your case), so we don't write $\sigma(x,h)$.