Showing that if $xf(x)=\log x$ for $x>0$, then $f^{(n)}(1)=(-1)^{n+1}n!\left(1+\frac12+\cdots+\frac{1}{n}\right)$

Question

Let $f(x)$ be a function satisfying $$xf(x)=\log x$$ for $x>0$. Show that $$f^{(n)}(1)=(-1)^{n+1}n!\left(1+\frac12+\frac13+\cdots+\frac{1}{n}\right),$$ where $f^{(n)}(x)$ denotes the $n$th derivative of the function $f$ evaluated at $x$.

I attempted plain differentiation of the function. But, that seemed to go the complex way. Please help. Thank you.

Answer 1

Hint. We know that $$ (f\: g)^{(n)}=\sum_{k=0}^n{n\choose k}(f)^{(k)}(g)^{(n-k)} $$ giving $$ \left(\frac{1}{x} \times \ln x\right)^{(n)}=\sum_{k=0}^n{n\choose k}\left(\frac{1}{x} \right)^{(k)}(\ln x)^{(n-k)} $$ then use $$ \begin{align} \left(\frac{1}{x} \right)^{(k)}&=(-1)^k\frac{k!}{x^{k+1}}\\\\ (\ln x)^{(n-k)}&=\left(\frac{1}{x} \right)^{(n-k-1)}. \end{align} $$

Answer 2

Hint: First, write out the first few derivatives of each side of $$x f(x) = \log x$$ (rather than the derivatives of $f(x) = \frac{\log x}{x}$), identify the patterns of the derivatives, and prove that they are correct (using, say, induction).