Define the following (infinite) q-Pochhammer symbols: $$(a;q)_{\infty} := (a)_{\infty} := \prod_{k=0}^{\infty}(1-aq^{k})$$ $$(a_{1}, \ldots, a_{n})_{\infty} := (a_{1})_{\infty}\cdots(a_{n})_{\infty}$$ $$[a]_{\infty} := (a, q/a)_{\infty}$$ $$[a_{1}, \ldots, a_{n}]_{\infty} := (a_{1}, q/a_{1}, \ldots, a_{n}, q/a_{n})_{\infty}$$
I am reading a paper which has an identity using these symbols, but the expression on the left of the identity is $$a\frac{(q)_{\infty}^{2}[-a, 1/a^{2}]_{\infty}}{[x, -xq/a, -ax]_{\infty}}.$$
The authors go on to differentiate the identity with respect to $a$ and thus the expression above, and then let $a \rightarrow 1$. They obtain $$4\frac{(q)_{\infty}^{4}(-q)_{\infty}^{2}}{[x, -xq, -x]_{\infty}}$$ as the derivative.
How did they obtain this? The only thing I know to do to differentiate these things is use logarithms. But the expression turns into a nasty sum of fractions of infinite series, which doesn't seem to turn back into more q-Pochhammer symbols as I see it. Does anyone have some insight?
I also have the identity $$(q;q)_{\infty} = (q)_{\infty} = \sum_{n=-\infty}^{\infty}(-1)^{n}q^{n(3n-1)/2},$$ which I believe gives us the identity $$(zq)_{\infty} = \sum_{n=-\infty}^{\infty}(-1)^{n}(zq)^{n(3n-1)/2},$$ but I did not see this as any help.
Edit: they are differentiating with respect to $a$, forgot to mention this.
What I see is that for $|a| < 1 , |q| < 1$ :
$$f_q(a) = \prod_{k \ge 0} (1- a q^k), \qquad \log f_q(a ) =\sum_{k \ge 0} \log(1- a q^k)$$
$$\frac{f_q'(a)}{f(a)} = \sum_{k \ge 0} \frac{-q^k}{1-a q^k} = -\sum_{k \ge 0} \sum_{m \ge 0} a^m q^{(m+1)k} = - \sum_{m \ge 0} a^m \sum_{k \ge 0} q^{(m+1)k} \\= \sum_{m \ge 0} \frac{a^m}{q^{m+1}-1}$$