Find an upper bound of $\frac{d^{n+1}}{d^{n+1} x} \exp(-x) \cos(4\pi x)$? How can I find an upper bound of $\frac{d^{n+1}}{d^{n+1} x} \exp(-x) \cos(4\pi x)$? That is, for $f^{(n+1)}$? Note that x is in the interval $(0,3)$. I have been told the following hint: How many terms are in $f^{(n+1)}$ and what is an upper bound for each of them? I then spent 2 hours and finally gave up and now I'd try to reach out to some true math wizards :)
The cos and sin terms are upper bounded by 1. Also I believe that $exp(-x) \leq exp(0) = 1$

@Andrei's hint is useful. You can improve on this, but:$$\begin{align}D^{n+1}\exp(-x)\cos(4\pi x)&\stackrel{1}{=}\Re D^{n+1}\exp((-1+4\pi i)x)\\&\stackrel{2}{=}\Re((-1+4\pi i)^{n+1}\exp((-1+4\pi i)x))\\&\stackrel{3}{\le}|((-1+4\pi i)^{n+1}\exp((-1+4\pi i)x))|\\&\stackrel{4}{=}(1+16\pi^2)^{(n+1)/2}\exp(-x)\\&\stackrel{5}{<}(1+16\pi^2)^{(n+1)/2}.\end{align}$$Here 1 uses $\cos y=\exp iy$ for $y\in\Bbb R$, 2 uses $De^{zx}=ze^{zx}$ for $z\in\Bbb C$, 3 uses $\Re z\le|z|$ for $z\in\Bbb C$, 4 uses $|\exp iy|=1$ for $y\in\Bbb R$, and 5 uses $e^{-x}<1$ for $x>0$.
