I am currently working through a proof of the chain rule for differentiation and it currently looks like this: Consider $f(x) = F(g(x))$
$$ \frac{df}{dx} = \lim_{h \rightarrow 0} \frac{F(g(x+h)) - F(g(x))}{h} \\ = \lim_{h \rightarrow 0} \frac{F(g(x)+hg'(x)+o(h)) - F(g(x))}{h} \\ = \lim_{h \rightarrow 0} \frac{F(g(x))+(hg'(x)+o(h))F'(g(x))+o(hg'(x)+o(h))-F(g(x)}{h} \\=\lim_{h \rightarrow 0} g'(x)F'(g(x))+\frac{o(h)}{h} \\=g'(x)F'(g(x)).$$
where $o$ is the small $o$ notation. It is the step between the third and fourth lines that I do not understand. In the third line I have the quantity $o(hg'(x)+o(h))$. As far as $o$ is concerned, I can treat $g'(x)$ as a constant because $x$ doesn't vary in the limiting process, so I have $o(ch+o(h))$ for some constant $c$, but I do not know how to proceed from here.
My questions:
- What is $o(o(h))$?
- What is $o(ch+o(h))$?
We have that by definition
$$f(x)=o(g(x)) \iff f(x)=g(x)\, \omega(x) \quad \omega(x)\to 0$$
then
$$o(o(h))=o(h)\omega_1(h)=h\,\omega_1(h)\omega_2(h)=h\,\omega_3(h)$$
$$\implies o(o(h))=o(h)$$
and
$$o(ch+o(h))=(ch+o(h))\omega_1(h)=h(c\omega_1(h)+\omega_1(h)\omega_2(h))=h\omega_3(h)$$
$$\implies o(ch+o(h))=o(h)$$