Problem
Find the q-derivative of $(a-x)_q^n$ for $n \ge 1$.
Answer
This is actually equation (3.11) of Kac and Cheung's Quantum Calculus $$D_q(a-x)_q^n=-[n](a-qx)_q^{n-1}.\tag{3.11}$$
My question
I've got a different answer $-[n](a-x)_q^{n-1}$ by applying the fact that
$$D_q(x-a)_q^n=-[n](x-a)_q^{n-1}.\tag{3.5}$$
with $a$ replaced by $-a$ (since $a$ can be any number) and the chain rule for $u(x)=\alpha x^\beta$
$$D_q f(u(x)) = (D_{q^\beta} f) (u(x)) \; D_q u(x) \tag{1.15}$$
with $u(x)=-x$ and $f(x)=(x+a)_q^n$.
Where's my mistake?
I understand the book's derivation (3.11) line by line, so please don't repeat the book's proof. Instead, please point out my mistakes.
To clear this question from the unanswered queue, I am going to answer my own question.
It turns out that I had overlooked the line $(a-x)_q^n \ne (-1)^n(x-a)_q^n$, where
$$(x-a)_q^n = \begin{cases} 1 & \text{ if } n = 0, \tag{3.4} \\ (x-a)(x-qa) \cdots (x-q^{n-1} a) & \text{ if } n \ge 1. \end{cases}$$
My attempt is actually addressing $D_q (-x + a)_q^n$ instead of $D_q (a-x)^n$.
N.B. My attempted answer "$D_q (-x + a)_q^n = -[n](a-x)_q^{n-1}$" is not even correct due to the same reason. It should be "$D_q (-x + a)_q^n = -[n](-x+a)_q^{n-1}$" instead.