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Is the composition of class $C^k$ functions also of class $C^k$?

I feel like this one should be true but I can't find any reference.

2 Answers 2

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Yes, this is a consequence of the chain rule, which allows us to calculate $(f\circ g)^{(k)}$ in terms of f,f',\ldots,f^{(k)},g,g',\ldots,g^{(k)}.

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Yes this is true.. If $f$ and $g$ are $C^1$ function, we have $d(f\circ g)_x= d f_{g(x)}\circ dg_x$

so we have $f\circ g$ is also $C^1$, Now take f' and g' which is also $C^1$, you will have for h=f'\circ g', h' is also $C^1$... proceed inductively... you will get.

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    Why does this show that $f\circ g$ is $C^1$? Can you please explain why $f\circ g$ is $C^1$ from this analysis?2018-04-05