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!

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. 
