I've seen the following definition:
Theorem 5.1 $f(x),g(x)$ are the generating functions of the sequence $(a_r),(b_r)$.
$[...]$
$(iv)$ The generating function to $(ra_r)$ is equal to $xf'(x)$, where $f'(x)$ is the derivative of $f$ with respect to $x$.
There is one example of it's usage a little further in the book: To find the generating function to $a_r=r$. The author does the following:
$$f(x)=\frac{1}{1-x}=1+x+x^2+x^3+x^4+\ldots$$
Then using $(vi)$:
$$f'(x)=\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\ldots\tag{1}$$
And then:
$$xf'(x)=\frac{x}{(1-x)^2}=x+2x^2+3x^3+\ldots\tag{2}$$
I don't get why he needed to multiply by $x$ in $(2)$, doesn't $(1)$ already have the coefficients for $a_r=r$?
No, $(1)$ has the coefficients for $a_r=r+1$. Remember, $a_r$ is the coefficient of $x^r$. Here, for instance, $(1)$ has a constant term of $1$, so the coefficient of $x^0$ is $1$, not $0$, and similarly for every other term.