<s>[INST] Formula for the large derivative Is there any formula for the large number derivative?
I need to find $y^{(100)}$ at $x=0$, if $y=(x+1)2^{x+1}$
I tried to find a pattern, but 2nd and 3rd derivatives are already too hairy. I see no pattern, how it evolves.
1st derivative $2^{x+1}+\ln \left(2\right)\cdot \:2^{x+1}\left(x+1\right)$
or $2^{x+1}(1+ln(2)(x+1))$
2nd $\ln ^2\left(2\right)\cdot \:2^{x+1}x+\ln ^2\left(2\right)\cdot \:2^{x+1}+\ln \left(2\right)\cdot \:2^{x+2}$
3rd $\ln ^2\left(2\right)\left(\ln \left(2\right)\cdot \:2^{x+1}x+2^{x+1}\right)+\ln ^3\left(2\right)\cdot \:2^{x+1}+\ln ^2\left(2\right)\cdot \:2^{x+2}$
 [/INST] Use Taylor series:
$$y=2(x+1)\left(1+x\ln 2+\frac{x^2}{2!}\ln^2 2+\cdots \right)=\\
[2x+\cdots +\frac{2x^{100}}{99!}\ln^{99} 2+O(x^{101})]+\\
[2+\cdots+\frac{2x^{100}}{100!}\ln^{100} 2+O(x^{101})].$$
$$y^{(100)}(0)=200\ln^{99} 2+2\ln^{100} 2.$$
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