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

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).
