Evaluation of q-Pochhammer series Let $f_q(x)$ be the generating function of the sequence $(q;q)_n$:
$$f_q(x):=\sum_{n=0}^{\infty} (q;q)_n x^n,$$
where $(q;q)_n: = (1-q)(1-q^2) \cdots (1-q^n)$ with convention $(q;q)_0$:=1. 
Let $g_q(x):=1/f_q(x)$.
Question: Are there closed form expressions for
$$\lim_{x \rightarrow 1} g'_q(x) \quad \mbox{and} \quad \lim_{x \rightarrow 1} g''_q(x) \quad ?$$
By a rather tricky argument, I get $\lim_{x \rightarrow 1} g'_q(x) = -(q;q)_{\infty}$. But I don't get anything simple for $\lim_{x \rightarrow 1} g''_q(x)$.
 A: We have
$$f_q(x)=(q;q)_{\infty}\sum_{n\ge 0}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right)x^n+(q;q)_{\infty}\frac{1}{1-x}.$$
Note that
\begin{equation}
\frac{1}{(q^{n+1};q)_{\infty}}-1=\exp\left(\sum_{\ell\ge 1}\frac{q^{(n+1)\ell}}{\ell(1-q^{\ell})}\right)-1\ll \frac{|q|^n}{(|q|;|q|)_{\infty}}
\end{equation}
for sufficiently large $n$. Thus
$$h_q(x):=\sum_{n\ge 0}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right)x^n$$
is analytic for $|x|<|q|^{-1}$.
Hence for $|q|<1$,
$$(q;q)_{\infty}g_q(1-x)=\frac{x}{xh_q(1-x)+1}$$
is analytic for $|1-x|<1/|q|$. Further for $|1-x|<1/|q|$,
\begin{align}
h_q(1-x)&=\sum_{n\ge 0}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right)\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}x^r\\
&=\sum_{r\ge 0}(-x)^r\sum_{n\ge r}(-1)^n\binom{n}{r}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right):=\sum_{r\ge 0}\alpha_r(q)x^r.
\end{align}
In particular, we have
\begin{align}
(q;q)_{\infty}g_q(1-x)&=\frac{x}{1+\sum_{r\ge 1}\alpha_{r-1}x^r}\\
&=x-\alpha_1(q)x^2+(\alpha_1(q)^2-\alpha_2(q))x^3+\dots
\end{align}
The following is easy!
A: From $q$-binomial theorem we have
\begin{align}
f_q(x)&=(q;q)_{\infty}\sum_{n\ge 0}\frac{x^n}{(q^{n+1};q)_{\infty}}\\
&=(q;q)_{\infty}\sum_{n\ge 0}x^n\sum_{k\ge 0}\frac{q^{(n+1)k}}{(q;q)_{k}}=(q;q)_{\infty}\sum_{k\ge 0}\frac{q^{k}}{(q;q)_{k}}\frac{1}{1-xq^k}
\end{align}
for all $|q|<1$ and $|x|<1$. Therefore for $|1+x|<1$ and $|x|<|q^{-k}-1|, k\ge 1$, 
\begin{align}
\frac{f_q(1+x)}{(q;q)_{\infty}}&=-\frac{1}{x}+\sum_{k\ge 1}\frac{q^{k}}{(q;q)_{k}}\frac{1}{1-q^k-xq^k}\\
&=-\frac{1}{x}+\sum_{k\ge 1}\frac{q^{k}}{(q;q)_{k}}\frac{1}{1-q^k}\sum_{r\ge 0}\left(\frac{q^k}{1-q^k}\right)^rx^r\\
&=-\frac{1}{x}+\sum_{r\ge 0}x^r\sum_{k\ge 1}\frac{1}{(q;q)_{k}} \left(\frac{q^k}{1-q^k}\right)^{r+1}.
\end{align}
Now, we see that
$$\frac{f_q(1+x)}{(q;q)_{\infty}}+\frac{1}{x}$$
is an analytic function for 
$$x\in\{z\in\mathbb{C}:|z|<|q^{-k}-1|, k=1,2,\dots\}:=\Omega_q.$$
It is clear that $\Omega_q\neq\emptyset$ is an open set for $|q|<1$. We further have
\begin{align}
\frac{xf_q(1+x)}{(q;q)_{\infty}}&=-1+\sum_{r\ge 1}x^{r}\sum_{k\ge 1}\frac{1}{(q;q)_{k}} \left(\frac{q^k}{1-q^k}\right)^{r}:=-1+\sum_{r\ge 1}A_r(q)x^r.
\end{align}
Hence
$$(q;q)_{\infty}g_q(1+x)=\frac{-x}{1-\sum_{r\ge 1}A_r(q)x^r}=-x-A_1(q)x^2-(A_2(q)+A_1(q)^2)x^3-\dots$$
is an analytic function at $x=0$. Thus,
$$g_q'(1)=-\frac{1}{(q;q)_{\infty}}\quad\mbox{and}\quad g_q''(1)=-\frac{2}{(q;q)_{\infty}}\sum_{k\ge 1}\frac{1}{(q;q)_{k}}\frac{q^k}{1-q^k}.$$
