How do I find the $n$th derivative of $x^n\ln\left(x\right)$? Using Leibniz's theorem $y^{(n)}=\sum _{k=0}^n\displaystyle {\tbinom {n}{k}}\cdot (x^n)^{(n-k)}\cdot (\ln\left(x\right))^{(k)}$. But I am unable to simplify this any further. I also took the first couple of derivatives by hand but couldn't see any pattern. Thanks for any help!
 A: Leibnitz's theorem basically employs the binomial theorem with powers representing orders of derivatives.
$\dfrac{d^n}{dx^n}\left(x^n\ln(x)\right)=\sum_{k=0}^{n}\binom{n}{k}(x^n)^{(n-k)}\ln^{(k)}(x).$ Now, note that
$\begin{equation}
(x^{n})^{(n-k)}=\dfrac{d^{n-k}}{dx^{n-k}}x^{n}=\dfrac{x^kn!}{k!},\,\,(x^{n})^{n}=n!
\end{equation}$ and
$\begin{equation}
\ln^{(k)}(x)=\dfrac{d^k}{dx^k}\ln(x)=\dfrac{(-1)^{k+1}(k-1)!}{x^{k}},\,\,\ln^{(0)}(x)=\ln(x)
\end{equation}$
Therefore
$\begin{equation}
\begin{aligned}
\dfrac{d^n}{dx^n}\left(x^n\ln(x)\right)&=n!\ln(x)+\sum_{k=1}^{n}\binom{n}{k}\dfrac{x^kn!}{k!}\dfrac{(-1)^{k+1}(k-1)!}{x^{k}}\\
&=n!\ln(x)+\sum_{k=1}^{n}\binom{n}{k}\dfrac{n!}{k}(-1)^{k+1}.
\end{aligned}
\end{equation}$
This is the best you can do I think.
A: We can consider $x^{n+\epsilon}= x^n e^{\epsilon \log(x)}$, because Taylor-expanding for small $\epsilon$ yields $e^{\epsilon \log(x)} = 1 +\epsilon \log(x) + \mathcal O(\epsilon^2)$ and thus
$$
x^{n+\epsilon} = x^n  + \epsilon \, x^n \log(x) + \mathcal O(\epsilon^2)\,.
$$
Now, clearly
$$
\frac{d^k}{dx^k} x^{n+\epsilon} = (n+\epsilon)(n+\epsilon-1)\cdots (n+\epsilon-k+1)x^{n+\epsilon-k}
$$
and to linear order in $\epsilon$,
$$
\frac{d^k}{dx^k} x^{n+\epsilon}
=
\frac{n!}{(n-k)!}\, x^{n-k}
\\
+
\epsilon\, \frac{n!}{(n-k)!} \left(\frac{1}{n} + \frac{1}{n-1}+\cdots +\frac{1}{n-k+1} + \log(x)\right)x^{n-k} + \mathcal O(\epsilon^2)\,.
$$
In conclusion,
$$
\frac{d^k}{dx^k} x^{n}\log(x) = \frac{n!}{(n-k)!} \left(\frac{1}{n} + \frac{1}{n-1}+\cdots +\frac{1}{n-k+1} + \log(x)\right)x^{n-k} \,,
$$
and in particular
$$
\frac{d^n}{dx^n} x^{n}\log(x) = n!\,\left(\frac{1}{n} + \frac{1}{n-1}+\cdots +\frac12 + 1 + \log(x)\right) 
$$
A: Hint:
$$(x^n\log x)'=nx^{n-1}\log x+x^{n-1}$$
$$(x^n\log x)''=n(n-1)x^{n-2}\log x+(2n-1)x^{n-2}$$
$$(x^n\log x)'''=n(n-1)(n-2)x^{n-2}\log x+(3n^2-6n+2)x^{n-2}$$
The pattern is visibly
$$(x^n\log x)^{(m)}=\frac{n!}{(n-m)!}x^{n-m}\log x+p_m(n)x^{n-m}$$
where $p_m$ is a polynomial. We have the recurrence
$$p_{m+1}(n)=\frac{n!}{(n-m)!}+(n-m)p_m(n).$$
