What is the derivative of $Q\_m\left(\frac{\alpha}{x^a},\frac{\beta}{x^b}\right)$ with respect to $x$, i.e,


$$\frac{\partial}{\partial x}Q\_m\left(\frac{\alpha}{x^a},\frac{\beta}{x^b}\right),
\quad
\text{ where }
\alpha,\beta,a,b \in \mathbb{R} 
\text{ and } m\in \mathbb{N} $$


where the $Q\_m(.,.)$ is a special function( <http://en.wikipedia.org/wiki/Marcum_Q-function> ) , and we have


$$Q\_m\left(\frac{\alpha}{x^a},\frac{\beta}{x^b}\right)=
\left(\frac{x^a}{\alpha}\right)^{m-1}
\int^\infty\_{\frac{\beta}{x^b}}
\: t^m
\exp\left(-\frac{t^2}{2}-\frac{\alpha^2}{2x^{2a}}\right)
I\_{m-1}\left(\frac{\alpha}{x^a}t\right)
\:\mathrm{d}t$$


where $I\_m(.)$ is the modified bessel function of the first kind with order $m $. Please note that the lower limit of the integral is $\displaystyle\frac{\beta}{x^b}$.



derivative of a special function  in integral form
This can be computed using the 'usual' rules of calculus, along with the formula
$$\frac{d}{d x} \int\_{f(x)}^{g(x)} h(t,x) dt =
 \int\_{f(x)}^{g(x)}\frac{\partial}{\partial x} h(t, x)dt + g'(x) h(g(x), x) -f'(x)h(f(x), x)$$
where $'$ denotes $\frac{d}{dx}$. If $g(x)=\infty$, you need to take a limit (over a new constant); generally this means that the second term in the formula drops out unless 
$$\lim\_{c\to\infty} h(c,x)$$
is unbounded. Then you have to work harder.


