$f(x)=x^3e^x, \; g(x)=\exp(x^{999}) ,\; h(x)=\frac{1}{2+\exp(i4x)}$. Computing $f^{(1000)}(x)$ ,$g^{(1000)}(0)$,$h'(x)$ without technology

Question

$f: \mathbb{R} \rightarrow \mathbb{R}$, $g: \mathbb{R} \rightarrow \mathbb{R}$, $h: \mathbb{R} \rightarrow \mathbb{C}$

$f(x)=x^3e^x, \; g(x)=\exp(x^{999}) ,\; h(x)=\frac{1}{2+\exp(i4x)}$

Are there some efficient way to compute (without using technology):

$f^{(1000)}(x)=$

$g^{(1000)}(0)=$

$h'(x)=$

Answer 1

If without using technology means just computation by hand, this is the solution:

1. Using $(uv)^{(n)}(x) = \sum_{k=0}^nC^n_k u^{(n-k)}(x)v^{(k)}(x)$, only a few terms survive,$f^{(1000)}(x)=997002000 e^x+2997000x e^x +3000 x^2 e^x +x^3e^x$

2. Since $g(x)=\exp(x^{999}) = \sum_{n=0} \frac{x^{999n}}{n!}$, there is no $x^{1000}$ term, so $g^{(1000)}(0)=0$

3. This one is too easy, you can figure it out.