Determine $f^{(22)}(0)$, when $f(x)=x^{19}\ln(1+2x)$ "Determine $f^{(22)}(0)$, when $f(x)=x^{19}\ln(1+2x)$."
Here is my attempt. First, I wrote $f=gh$, where $g=x^{19}$ and $h=\ln(1+2x)$. The derivatives of both of these functions behave predictably at $x=0$. For example, $g^{19} = 19!$, taking further derivatives simply yields zero, and previous derivatives at $x=0$ are also zero, since they contain $x$ as a multiplier. The derivatives of $h$ behave according to this sequence $$\frac{{x_n}}{(1+2x)^n}$$
where ${x_n}=\{2,-4,16,-96,768,...\}.$ The denominator is $(1+2x)^n=(1+2 \cdot 0)^n=1^n =1$, so the value of $h'$ is simply the value of the sequence at $n\in \Bbb{N}.$ Next I tried to find a pattern in the application of the product rule with $f=gh$. Multiple applications yields $$f^1=g^1h+gh^1$$ $$f^2=g^2h+g^1h^1+g^1h^1+gh^2$$ $$f^3 = g^3h+g^2h^1+g^2h^1+g^1h^2+g^2h^1+g^1h^2+g^1h^2+gh^3$$ $$...$$
The idea was to find a way to connect $g^{19}$ with the correct values of $h'$, then compute the value of $f^{22}$ with the help of these simple derivatives.
I haven't found one yet, but I think it might be possible to find a pattern in repeated applications of the product rule, but before that I wanted to ask whether there is some flaw in my work. Also, I'm not entirely sure about this but might this https://en.wikipedia.org/wiki/General_Leibniz_rule help here?
 A: $f^{(22)}$ probably means the $22$th derivative of $f$, not $f$ raised to the $22$th power. With this in mind, write $\mathrm{ln}(1+2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \cdots$. Thus
$$
f(x) = x^{19}\mathrm{ln}(1+2x) = 2x^{20} - \frac{4x^{21}}{2} + \frac{8x^{22}}{3} - \cdots,
$$
from which we conclude that $f^{(22)}(0) = 22!\frac{8}{3}$, since the $n$th coefficient of the Taylor series at $x = 0$ for a function $f(x)$ is $\frac{f^{(n)}(0)}{n!}$.
A: Using your notation for $g(x)= x^{19}$ and $h(x)= ln(1+2x)$
From Leibmitz theorem on repeated derivatives:
$$f^n(a)=(gh)^n(a)= g^n(a)h(a)+ ^nC_1 f^{n-1}(a)g^1(a)+^nC_2f^{n-2}(a)g^2(a)+•••+^nC_rf^{n-r}(a)g^r(a)+•••+f(a)g^n(a)$$
Now ,we see that, $f^{19}(0)= (19)! $,$ f^n(0)=0 $ for $n= 1,2,..,18,20,21,21$
And 
The equation becomes,$ f^{22}(0)=(gh)^{22}(0)= ^{22}C_3(19)!g^3(0)= ^{22}C_3 (19)! (2^4)$ 
Since$ \ g^3(0)=2^4$
Thus $$f^{22}(0)= \frac{8}{3}(22!)$$
A: After applying the Leibniz Rule to this problem, I was able to determine that $$f^{22}=2.997335274\cdot 10^{21}$$
which I believe is the correct answer. Comments are still appreciated.
A: The link you provided should certainly help and you are correct that there is a flaw in your work. $f^{(22)}(0)$ is the 22nd derivative of $f(x)$ evaluated at $x=0$.
