Exponential Generating Function for $ng_{n-1}$ given some recurrence
Suppose I want to find the exponential generating function for $ng_{n-1}$ given some recurrence.
Let $g(x) = \sum_{n=0}^{\infty} g_n \frac{x^n}{n!}$ be the generating function for $g_n$.
I know that the generating function for $g_{n-1}$ is $xg(x)$, but I'm not quite sure how to find the generating function of $n g_{n-1}$
It looks like $n g_{n-1} = \frac{d}{dx}xg(x)$ but I'm not entirely certain, and I'm also not sure how to find a generating function from this since we'll have a $\frac{d}{dx} g(x)$ hanging around.
I think you may be mixing up ordinary and exponential generating functions.
Ordinary g.f.
$f(x) = \sum_n a_n x^n$, so $x f(x) = \sum_n a_n x^{n+1} = \sum_n a_{n-1} x^{n}$ is the o.g.f. for the sequence shifted down one index; and $x\frac{\textrm{d}}{\textrm{d}x} f(x) = x\sum_n n a_n x^{n-1} = \sum_n n a_n x^n$ is the o.g.f. for the sequence weighted by index.
Exponential g.f.
$g(x) = \sum_n g_n \frac{x^n}{n!}$ so $\frac{\textrm{d}}{\textrm{d}x} g(x) = \sum_n n g_n \frac{x^{n-1}}{n!} = \sum_n g_n \frac{x^{n-1}}{(n-1)!} = \sum_n g_{n+1} \frac{x^n}{n!}$ is the e.g.f. for the sequence shifted up one index and $x \frac{\textrm{d}}{\textrm{d}x} g(x) = \sum_n g_{n+1} \frac{x^{n+1}}{n!} = \sum_n (n+1)g_{n+1} \frac{x^{n+1}}{(n+1)!} = \sum_n n g_n \frac{x^n}{n!}$ is the e.g.f. for the sequence weighted by index.