<s>[INST] How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$ 

How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$


I have tried this 
$$f(x)=x^2\ln(x+1)$$
$$ f'(x) = 2x\ln(x+1) + \frac{x^2}{x+1}$$
$$ f''(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
However I do not see any pattern :( 
 [/INST] Let $u=x^2$ and $v=\ln (1+x)$ then,
$$u'=2x,u''=2,u'''=0$$
And
$v^{(n)}=\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}$ for $n \geq 1$.
Let's compute $(uv)^{(3)}=\sum_{k=0}^{3} {3 \choose k} u^{(k)}v^{(n-k)}={3 \choose 0} x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+{3 \choose 1}2x\frac{(-1)^{n-2}(n-2)!}{(1+x)^{n-1}}+{3 \choose 2}2\frac{(-1)^{n-3}(n-3)!}{(1+x)^{n-2}}+{3 \choose 3}0$
With $n=3$ by the general product rule
Now it's easy to see that for $n \geq 3$ we have 
$$(uv)^{(n)}={n \choose 0} x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+{n \choose 1}2x\frac{(-1)^{n-2}(n-2)!}{(1+x)^{n-1}}+{n \choose 2}2\frac{(-1)^{n-3}(n-3)!}{(1+x)^{n-2}}$$
As if the third derivative of $u$ is $0$ ,and the fourth, the fifth, the sixth, etc. 
This simplifies to,
$$x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+2x\frac{(-1)^{n-2}n!}{(n-1)(1+x)^{n-1}}+\frac{(-1)^{n-3}n!}{(n-2)(1+x)^{n-2}}$$
</s>