# q-derivative

In mathematics, in the area of combinatorics, the q-derivative, or Jackson derivative, is a q-analog of the ordinary derivative, introduced by Frank Hilton Jackson. It is the inverse of Jackson's q-integration. For other forms of q-derivative, see (Chung et al. (1994)).

## Definition

The q-derivative of a function f(x) is defined as

${\displaystyle \left({\frac {d}{dx}}\right)_{q}f(x)={\frac {f(qx)-f(x)}{qx-x}}.}$

It is also often written as ${\displaystyle D_{q}f(x)}$. The q-derivative is also known as the Jackson derivative.

Formally, in terms of Lagrange's shift operator in logarithmic variables, it amounts to the operator

${\displaystyle D_{q}={\frac {1}{x}}~{\frac {q^{d~~~ \over d(\ln x)}-1}{q-1}}~,}$

which goes to the plain derivative ${\displaystyle \to {\frac {d}{dx}}}$ as ${\displaystyle q\to 1}$.

It is manifestly linear,

${\displaystyle \displaystyle D_{q}(f(x)+g(x))=D_{q}f(x)+D_{q}g(x)~.}$

It has product rule analogous to the ordinary derivative product rule, with two equivalent forms

${\displaystyle \displaystyle D_{q}(f(x)g(x))=g(x)D_{q}f(x)+f(qx)D_{q}g(x)=g(qx)D_{q}f(x)+f(x)D_{q}g(x).}$

Similarly, it satisfies a quotient rule,

${\displaystyle \displaystyle D_{q}(f(x)/g(x))={\frac {g(x)D_{q}f(x)-f(x)D_{q}g(x)}{g(qx)g(x)}},\quad g(x)g(qx)\neq 0.}$

There is also a rule similar to the chain rule for ordinary derivatives. Let ${\displaystyle g(x)=cx^{k}}$. Then

${\displaystyle \displaystyle D_{q}f(g(x))=D_{q^{k}}(f)(g(x))D_{q}(g)(x).}$

The eigenfunction of the q-derivative is the q-exponential eq(x).

## Relationship to ordinary derivatives

Q-differentiation resembles ordinary differentiation, with curious differences. For example, the q-derivative of the monomial is:

${\displaystyle \left({\frac {d}{dz}}\right)_{q}z^{n}={\frac {1-q^{n}}{1-q}}z^{n-1}=[n]_{q}z^{n-1}}$

where ${\displaystyle [n]_{q}}$ is the q-bracket of n. Note that ${\displaystyle \lim _{q\to 1}[n]_{q}=n}$ so the ordinary derivative is regained in this limit.

The n-th q-derivative of a function may be given as:

${\displaystyle (D_{q}^{n}f)(0)={\frac {f^{(n)}(0)}{n!}}{\frac {(q;q)_{n}}{(1-q)^{n}}}={\frac {f^{(n)}(0)}{n!}}[n]_{q}!}$

provided that the ordinary n-th derivative of f exists at x = 0. Here, ${\displaystyle (q;q)_{n}}$ is the q-Pochhammer symbol, and ${\displaystyle [n]_{q}!}$ is the q-factorial. If ${\displaystyle f(x)}$ is analytic we can apply the Taylor formula to the definition of ${\displaystyle D_{q}(f(x))}$ to get

${\displaystyle \displaystyle D_{q}(f(x))=\sum _{k=0}^{\infty }{\frac {(q-1)^{k}}{(k+1)!}}x^{k}f^{(k+1)}(x).}$

A q-analog of the Taylor expansion of a function about zero follows:

${\displaystyle f(z)=\sum _{n=0}^{\infty }f^{(n)}(0)\,{\frac {z^{n}}{n!}}=\sum _{n=0}^{\infty }(D_{q}^{n}f)(0)\,{\frac {z^{n}}{[n]_{q}!}}}$