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 $\ln ^2\left(2\right)\left(\ln \left(2\right)\cdot \:2^{x+1}x+2^{x+1}\right)+\ln ^3\left(2\right)\cdot \:2^{x+1}+\ln ^2\left(2\right)\cdot \:2^{x+2}$

Let $u=x+1$ and $v=2^{x+1}$.
$u^{(1)}=1$, $u^{(k)}=0$ for $k\ge 2$.
$v^{(k)}=2^{x+1}(\ln2)^k$ for $k\ge 1$.
By the general Leibniz rule, (https://en.wikipedia.org/wiki/General_Leibniz_rule)
$$y^{(100)}=uv^{(100)}+\binom{100}{1}u^{(1)}v^{(99)}=(x+1)2^{x+1}(\ln 2)^{100}+100\cdot2^{x+1}(\ln 2)^{99}$$
$$y^{(100)}(0)=2(\ln 2)^{100}+200(\ln 2)^{99}$$
