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If $f(x):\mathbb{R}\rightarrow\mathbb{C}$ is $C^\infty$-smooth. Is $1/f(x)$ also $C^\infty$-smooth? $f(x)\neq0$

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If $f$ is differentiable and non-zero at some point $a$, then $1/f$ is differentiable at $a$, and $(1/f)'=-f'/f^2$. This is a "base case" of an induction argument for the following statement:

$1/f$ is $n$ times differentiable, and $(1/f)^{(n)}$ equals a polynomial in the functions $f,f',f'',\ldots,f^{(n-1)}$ divided by a power of $f$.

The induction step is just applying the quotient rule.

Have fun!

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