100th derivative of $\frac{x^2+x}{2^x}$ at point 0 Problem:
$$\frac{\mathrm d^{100}}{\mathrm dx^{100}}\frac{x^2+x}{2^x}$$
Can anyone help me with this because I tried to use General Leibniz rule and I didn't get much better information and I also tried some proof by induction but also no success
Without Taylor
 A: Hint:
By Taylor,
$$(x^2+x)e^{-\ln(2)x}=\sum_{k=0}^\infty\frac{(-\ln(2)x)^{k+1}}{k!}+\frac{(-\ln(2)x)^{k+2}}{k!}$$
A: If we apply more than two derivatives to $x^2+x$, we get $0$. If we apply fewer than one derivative to $x^2+x$ we get $0$ because we are evaluating at $x=0$. So we only need to worry about the terms with one or two derivatives applied to $x^2+x$.
One derivative: $\binom{100}{1}\overbrace{\ \ \ \ \ \ 1\ \ \ \ \ \ }^{2x+1\text{ evaluated at }x=0}\cdot(-\log(2))^{99}$
Two derivatives: $\binom{100}{2}\overbrace{\ \ \ \ \ \ 2\ \ \ \ \ \ }^{2\text{ evaluated at }x=0}\cdot(-\log(2))^{98}$
Thus, we get
$$
\bbox[5px,border:2px solid #C0A000]{9900\log(2)^{98}-100\log(2)^{99}}
$$
A: You can use Leibniz's rule for this since all derivatives of $x^2 + x$ beyond the second derivative are zero. So your answer will be
$$(x^2 + x) {d^{100} \over dx^{100}} 2^{-x} + 100 {d \over d x} (x^2 + x) {d^{99} \over dx^{99}} 2^{-x} + {100*99 \over 2} {d^2 \over d x^2} (x^2 + x) {d^{98} \over dx^{98}} 2^{-x}$$
Each derivative landing on $2^x$ adds a factor of $-\ln 2$, so the answer is
$$ (x^2 + x)(-\ln 2)^{100} 2^{-x} + 100(2x + 1)(-\ln 2)^{99} 2^{-x} +  100*99(-\ln 2)^{98} 2^{-x}$$
Plugging in the value $x = 0$ leads to
$$100 (-\ln 2)^{99}  + 100*99(-\ln 2)^{98}$$
$$= 9900 (\ln 2)^{98} -  100 (\ln 2)^{99}$$
A: Basically, say you have $f(x) = p(x).2^{-x}$, with $p$ polynomial, then
$f'(x) = p'(x).2^{-x} - log{2}.p(x).2^{-x} = ( c.p(x)+p'(x) ).2^{-x}$
$f''(x) = ( c.p'(x)+p''(x) ).2^{-x} - log{2}.( c.p(x)+p'(x) ).2^{-x}$
where $c = -log{2}$. Then rewrite $f''(x)$
$f''(x) = ( c^2.p(x) + 2c.p'(x) + p''(x) ).2^{-x} = (\frac{d}{dx}+c)^2p(x).2^{-x}$
We can conclude that 
$\frac{d^{100}}{dx^{100}}(p(x).2^{-x}) = ((\frac{d}{dx}+c)^{100}p(x)).  2^{-x}$
