let $f(x) = e^{-x^2} \cos(e^{x^2})$ Prove/Disprove $\forall k \in N, \lim_{|x| \to \infty} |x^k f'(x)|=0$

Question

let $f(x) = e^{-x^2} \cos(e^{x^2})$

Prove/Disprove $\forall k \in N, \lim_{|x| \to \infty} |x^k f'(x)|=0$

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getting that $$x^pf'(x) = -2x^{p+1} e^{x^2}\sin(e^{x^2})-2x^{p+1} e^{-x^2} \cos(e^{x^2}) $$

But can't recall a way to show it diverges or converges. Leaning towards divegence. This is part of a problem showing that its derivative is not rapidly decaying

Answer 1

Actually,

$$f'(x) = \left(e^{-x^2}\right)' cos(e^{x^2}) + e^{-x^2} \left(cos(e^{x^2})\right)' = \\ = -2xe^{-x^2}cos(e^{x^2}) - e^{-x^2}sin(e^{x^2})e^{x^2}2x = \\ = -2xe^{-x^2}cos(e^{x^2}) - sin(e^{x^2})2x $$

Now, the part with $e^{-x^2}$ goes to $0$ even after multiplying with $x^k$, since $e^{-x^2}$ decays to zero faster than any power of $x$ grows to infinity. However, the $sin(e^{x^2})2x$ part is unbounded, and the limit cannot be zero.