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The chain rule. If $g$ is a function that is differentiable at $x$ and $f$ is a function that is differentiable at $g(x)$, then $f \circ g$ is differentiable at $x$, and $(f \circ g)'(x) = f'(g(x))g'(x).$

Question: I know the definition of the little-o notation. I also know that "$f$ is differentiable at $x$ $\iff$ there exists a real number $f'(x)$ such that $f(x+h) = f(x) + f'(x)h + o(h)$ as $h \rightarrow 0$". But could someone show how to give an alternative proof of the chain rule using the little-o notation? (I think I am stuck at understanding the algebraic properties of the little-o notation.) Also, is it possible to give an alternative proof of the inverse function rule using the little-o notation? Thanks in advance!

TCHuang
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    Straightforward. $f(g(x+h))=f(g(x)+hg'(x)+o(h))=f(g(x))+f'(g(x))(hg'(x)+o(h))+o(hg'(x)+o(h))= f(g(x))+f'(g(x))g'(x)h+o(h)$. – A.S. Dec 05 '15 at 09:14
  • @A.S. Could you teach me why $(hg'(x)+o(h))$ becomes $g'(x)h$ and why $o(hg'(x)+o(h))$ becomes $o(h)$? I am stuck at understanding the algebraic properties of the little-o notation... – TCHuang Dec 05 '15 at 09:26
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    $o(ch)=o(h)$ for any constant $c$ and $g'(x)$ and such are just constants and $o(h)+o(h)=o(h)$. – A.S. Dec 05 '15 at 09:41
  • See here. – psie Jul 12 '23 at 18:09

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