Finding the $n$-th derivatives of $x^n \ln x$ and $\frac{\ln x}{x}$. How can I prove the following identities:
$$ \left( x^n \ln x \right)^{(n)}=  n! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{n} \right), \quad x>0, \quad n\ge 1, \tag{a}$$
$$ \left( \frac{\ln x}{x} \right)^{(n)}= (-1)^n \ n! \ x^{-n-1}  \left( \ln x - 1 - \frac{1}{2} - \cdots -\frac{1}{n} \right), \quad x>0,\quad n\ge 1, \tag{b}$$
Here $(\cdot )^{(n)}$ indicates the $n$-th derivative with respect to $x$.
I don't know where to start!
 A: I prefer using induction to prove the statement. For (a), when $n=1$, by product rule
$$\left( x \ln x \right)'=\ln x+x\cdot\frac{1}{x}=1!(\ln x+1),$$
which shows that (a) is true for $n=1$. Now suppose that (a) is true for $n=k$, i.e. 
$$\left( x^k \ln x \right)^{(k)}=  k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right).$$
Now, for $n=k+1$, we have
$$\left( x^{k+1} \ln x \right)^{(k+1)}=\left[ \big(x^{k+1} \ln x\big)' \right]^{(k)}=
\left[ (k+1)x^k \ln x+x^{k+1}\cdot\frac{1}{x}\right]^{(k)}$$
$$=(k+1)\left( x^k \ln x \right)^{(k)}+(x^k)^{(k)}.$$
Note that $(x^k)^{(k)}=k!$, by induction assumption, we have
$$\left( x^{k+1} \ln x \right)^{(k+1)}=(k+1)k! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k} \right)+k!$$
$$=(k+1)! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{k}+\frac{1}{k+1} \right),$$
which shows that (a) is true for $n=k+1$.
I will let you try (b) using induction. 
A: To finish off Yuval's argument, we try an inductive proof on
$$\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}=-\sum_{k=1}^n \frac1{k}=-H_n$$
The equality is easily verified for $n=1$. We now try to simplify
$$\sum_{k=1}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}$$
which goes a bit like this:
$$\begin{align*}
&\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \binom{n+1}{k}\frac{(-1)^k}{k}-\sum_{k=1}^n \binom{n}{k}\frac{(-1)^k}{k}\\
&\frac{(-1)^{n+1}}{n+1}+\sum_{k=1}^n \left(\binom{n+1}{k}-\binom{n}{k}\right)\frac{(-1)^k}{k}\\
&\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\sum_{k=1}^n (-1)^k\binom{n+1}{k}\\
&\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+\sum_{k=0}^{n+1} (1)^{n-k+1}(-1)^k\binom{n+1}{k}\right)\\
&\frac{(-1)^{n+1}}{n+1}+\frac1{n+1}\left(-1-(-1)^{n+1}+(1-1)^{n+1}\right)=-\frac1{n+1}\\
\end{align*}$$
and since we also have $-H_{n+1}-(-H_n)=-\dfrac1{n+1}$, the equality is established.

Here's an explicit application of Leibniz's product rule for the second problem, which uses the identity $\dfrac{\mathrm d^n}{\mathrm dx^n}\dfrac1{x}=\dfrac{(-1)^n n!}{x^{n+1}}$:
$$\begin{align*}
\frac{\mathrm d^n}{\mathrm dx^n}\frac{\ln\,x}{x}&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\binom{n}{k}\left(\frac{\mathrm d^k}{\mathrm dx^k}\ln\,x\right)\left(\frac{\mathrm d^{n-k}}{\mathrm dx^{n-k}}\frac1{x}\right)\\
&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\sum_{k=1}^n\frac{n!}{\color{blue}{k!}\color{magenta}{(n-k)!}}\left(\frac{\color{red}{(-1)^{k-1}}\color{blue}{(k-1)!}}{\color{orange}{x^k}}\right)\left(\frac{\color{red}{(-1)^{n-k}}\color{magenta}{(n-k)!}}{\color{orange}{x^{n-k+1}}}\right)\\
&=\frac{(-1)^n n!}{x^{n+1}}\ln\,x+\frac{\color{red}{(-1)^{n-1}} n!}{\color{orange}{x^{n+1}}}\sum_{k=1}^n\frac1{\color{blue} k}=\frac{(-1)^n n!}{x^{n+1}}\left(\ln\,x-\sum_{k=1}^n\frac1{k}\right)
\end{align*}$$
A: The familiar formula $(fg)' = f'g + fg'$ generalizes to the $n$th derivative in the following way: given a product of $m$ functions $f_1,\ldots,f_m$, the $n$th derivative is a sum of $m^n$ terms, each corresponding to a sequence of length $n$ listing which function is differentiated at every "step".
Write $x^n\ln x$ as a product of $n+1$ factors, $n$ of them equal to $x$ and the other one equal to $\ln x$. The $n$th derivative is a sum of terms, each corresponding to a sequence of length $n$. If any of the first $n$ functions appears twice, then the corresponding term is zero, since $x''=0$. So the general term lists $k$ different functions out of the first $n$, and the other one ($\ln x$) appears in the remaining $n-k$ positions. Call that a term of type $T_k$.
The term $T_n$ appears $n!$ times, and contributes $n!\ln x$ to the total sum. For $k < n$, the term $T_k$ appears $n(n-1)\cdots(n-k+1) \binom{n}{k}$ times. It is easy to prove by induction that for $m > 0$, $(\ln x)^{(m)} = (-1)^{m+1} (m-1)! x^{-m}$. Substituting $m = n-k$, we see that each appearance of $T_k$ contributes $(-1)^{n-k+1} (n-k-1)!$ to the sum. So all terms $T_k$ contribute $(-1)^{n-k+1} n!/(n-k) \binom{n}{k}$. Therefore the $n$th derivative is
$$
\begin{align*}
&n!\ln x + n!\sum_{k=0}^{n-1} (-1)^{n-k+1} \frac{1}{n-k} \binom{n}{k} \\
=&n!\ln x + n!\sum_{k=1}^n (-1)^{k+1} \frac{1}{k} \binom{n}{k}.
\end{align*}
$$
Now all that remains is to prove the identity
$$
\sum_{k=1}^n (-1)^{k+1} \frac{1}{k} \binom{n}{k} = \sum_{k=1}^n \frac{1}{k}.
$$
That should prove you with a good start on the first one. I trust that the second one is very similar (perhaps after substituting $y = 1/x$).
