The value of $f'(n+1)-f'(1)$

Question

Let $g(x) = e^{f(x)}$ where $g(x)$ is a differentiable function on $(0,\infty)$ such that $g(x+1) = (x+1)g(x)$. Then for $n = 1,2,3,\dots$,

What is the value of $f'(n+1)-f'(1)$?

My try :

Answer 1

$e^{f(x+1)}=(x+1)e^{f(x)}$. So $ f(x+1)=\log (x+1) +f(x)$ and $ f(x+1)-f(x)=\log (x+1)$. Hence $f'(x+1)-f'(x)=\frac 1 {x+1}$. Put $x=n$ to get $f'(n+1)-f'(n)=\frac 1 {n+1}$.