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If $f(x)$ is differentiable $n$ times, how to show $f(x^k)$ is also differentiable $n$ times. It is intuitively true, but chain rule does not work out very well, any advice?

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  • $\begingroup$ Faa di Bruno's formula is useful. It basically says that the $n$th derivative of $f\circ g$ is a combination of the lower derivatives. So if all of the lower derivatives of $f$ and $g$ exist, $f\circ g$ is $n$ times differentiable. $\endgroup$ Mar 20, 2015 at 2:07
  • $\begingroup$ But we are not allowed to use theorems we didn't prove in class. So I don't think I can use this one... Is there any other ways to prove existence? $\endgroup$
    – Kun
    Mar 20, 2015 at 2:08
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    $\begingroup$ I'm not saying use the formula but gain some insight from it. Try proving that the $m$th derivative of $f\circ g$ only contains derivatives of each of at most order $m$. (Induction might be useful here.) Since $f$ is $n$ times differentiable and $x^k$ is infinitely differentiable, $f(x^k)$ is $n$ times differentiable. $\endgroup$ Mar 20, 2015 at 2:09
  • $\begingroup$ I will try. Thanks $\endgroup$
    – Kun
    Mar 20, 2015 at 2:10
  • $\begingroup$ By "$f(x)$ is differentiable $n$ times," do you mean that $f$ is $n$-times differentiable at $x$, or that $f$ is $n$-times differentiable everywhere? $\endgroup$
    – Math1000
    Mar 20, 2015 at 2:11

1 Answer 1

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Hint: try to prove it by induction on $n$ ...

NB: you can replace $f(x^k)$ by $f \circ g$ with $g$ differentiable $n$ times

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