Formula for the large derivative

Question

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

You should state the 2nd and 3rd derivatives here if you've found them. Saves us some work, and shows your effort (even though you've said you've done them, the whole effort thing seems to be sacred on this website) — , Mar 21 '18 at 15:36
maybe substitute variables $t=x+1$ and find derivative at $t=-1$? — Vasili, Mar 21 '18 at 15:43

score 1 · Accepted Answer · answered Mar 21 '18 at 15:43

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)

$$y^{(100)}=uv^{(100)}+\binom{100}{1}u^{(1)}v^{(99)}=(x+1)2^{x+1}(\ln 2)^{100}+100\cdot2^{x+1}(\ln 2)^{99}$$

$$y^{(100)}(0)=2(\ln 2)^{100}+200(\ln 2)^{99}$$

score 0 · Answer 2 · answered Mar 21 '18 at 15:45

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 $$200\log(2)^{99}+2\log(2)^{100}$$

score 0 · Answer 3 · answered Mar 21 '18 at 20:18

0

Use Taylor series: $$y=2(x+1)\left(1+x\ln 2+\frac{x^2}{2!}\ln^2 2+\cdots \right)=\\ [2x+\cdots +\frac{2x^{100}}{99!}\ln^{99} 2+O(x^{101})]+\\ [2+\cdots+\frac{2x^{100}}{100!}\ln^{100} 2+O(x^{101})].$$ $$y^{(100)}(0)=200\ln^{99} 2+2\ln^{100} 2.$$

answered Mar 21 '18 at 20:18

farruhota

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Formula for the large derivative

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