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 $\ln ^2\left(2\right)\left(\ln \left(2\right)\cdot \:2^{x+1}x+2^{x+1}\right)+\ln ^3\left(2\right)\cdot \:2^{x+1}+\ln ^2\left(2\right)\cdot \:2^{x+2}$
 A: 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)
$$y^{(100)}=uv^{(100)}+\binom{100}{1}u^{(1)}v^{(99)}=(x+1)2^{x+1}(\ln 2)^{100}+100\cdot2^{x+1}(\ln 2)^{99}$$
$$y^{(100)}(0)=2(\ln 2)^{100}+200(\ln 2)^{99}$$
A: I see all kinds of fancy techniques in the comments, no need for that. There's a pretty simple pattern here.
First define $g(x)=2^{x+1}$ and $f(x)=x2^{x+1}=xg(x)$. Now $y=f+g$, so to find $y^{(100)}$, we have to find $f^{(100)}$ and $g^{(100)}$.
For $g$, this isn't too hard; $g^{(n)}(x)=\log(2)^n2^{x+1}$, and you can prove this by induction easily.
For $f(x)$, we see $f'(x)=g(x)+xg'(x)$, and $f^{(2)}(x)=g'(x)+g'(x)+xg^{(2)}(x)$. From this, it's easy enough to see a pattern; $f^{(n)}(x)=ng^{(n-1)}(x)+xg^{(n)}(x)$. This too is easy enough to prove with induction.
In the end,
\begin{align}
y^{(100)}(x)&=f^{(100)}(x)+g^{(100)}(x)\\
&=100g^{(99)}(x)+xg^{(100)}(x)+g^{(100)}(x)\\
&=100\log(2)^{99}2^{x+1}+x\log(2)^{100}2^{x+1}+\log(2)^{100}2^{x+1}\\
\end{align}
Now at $x=0$, we get
$$200\log(2)^{99}+2\log(2)^{100}$$
A: Use Taylor series:
$$y=2(x+1)\left(1+x\ln 2+\frac{x^2}{2!}\ln^2 2+\cdots \right)=\\
[2x+\cdots +\frac{2x^{100}}{99!}\ln^{99} 2+O(x^{101})]+\\
[2+\cdots+\frac{2x^{100}}{100!}\ln^{100} 2+O(x^{101})].$$
$$y^{(100)}(0)=200\ln^{99} 2+2\ln^{100} 2.$$
