It is not hard to find easy formulas for the derivative of the function $\binom{x}{k}$, for instance it is not too hard to see (for $k$ an integer) that

$\frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \frac{1}{i} \binom{x-i}{k-i}.$

Now I was wondering if there is a nice q-analog of this formula. So if I assume that $\frac{d}{dx}$ is still the right thing to so and we write $\frac{d}{dx} \binom{x}{k}_q$ as a linear combination of $\binom{x-i}{k-i}_q$'s, then we have

$\frac{d}{dx} \binom{x}{1}_q = \frac{q^x \log(q)}{q-1} \binom{x-1}{0}_q.$

$\frac{d}{dx} \binom{x}{2}_q = \frac{q^x \log(q)}{q(q^2-1)} \left[ 2q \binom{x-1}{1}_q + \binom{x-2}{0}_q \right].$

$\frac{d}{dx} \binom{x}{3}_q = \frac{q^x \log(q)}{(q^2-1)(q^3-1)} \cdot \left[ 3(q^2-1) \binom{x-1}{2}_q + (q-1) \binom{x-2}{1}_q+ q^{-3} (2q^x+q^3-2q^2-q) \binom{x-3}{0}_q \right].$

This seems to turn out ugly. For $q \rightarrow 1$ it does not even seem to have the right limit for $k=3$ which is a property that a "nice q-analog" should have. So at this point I concluded that this is not the right generalization (if there is any). The derivative of $\binom{x}{k}$ is used very often, so I feel like that someone should have thought about this before me.