Maple seems to suggest the following formula for $n>0$, $p \le q$:


\begin{align}
\frac{d^n}{d x^n} & {}\_p F\_q (a\_1,\ldots,a\_p;b\_1,\ldots,b\_q;1/x) \\[8pt]
= {} & (-1)^n \hspace{1pt} n!\hspace{2pt} \frac{\prod\_j a\_j}{\prod\_k b\_k}\hspace{2pt} x^{-n-1} {}\_{p+1} F\_{q+1} (n+1, a\_1+1, \ldots, a\_p+1; 2, b\_1+1, \ldots, b\_q+1; 1/x)
\end{align}


It also holds for $n=0$ if $1$ is added to the right hand side. As an example, for $n\ge 1$:


$$ \frac{d^n}{d x^n}\hspace{1pt} \cosh\big(2/\hspace{-1pt}\sqrt{x}\big) = (-1)^n \hspace{1pt} n!\hspace{2pt} 2\hspace{1pt} x^{-n-1} {}\_1 F\_2 (n+1\hspace{1pt}; 2,3/2\hspace{1pt}; 1/x)
$$


Does anyone know how this is derived? I haven't seen that formula in classic refs, wikis, or Wolfram. I think it must be based on Cauchy's integral formula, then a change of variable to turn it into a sort of Euler's integral formula for $\_{p+1} F\_{q+1}$ as a Beta scale mixture of $\_p F\_q$. I'm not sure how the details of the complex integration would work.


For $n=1$ it's the ordinary derivative formula with the chain rule. So I guess an induction argument would work.


However there is a very interesting more general formula the relates:  

$$ \frac{d^n}{d x^n}\hspace{1pt}\normalsize \_p F\_q (\ldots;\ldots;x^{-m})$$
to a single term involving:  

$$ \_{p+m} F\_{q+m} (\ldots;\ldots;x^{-m})$$
The additional $2 m$ parameters follow a simple pattern. Maybe this would follwo from an iteration for multiple poles in the Cauchy integral due to $m>1$? It doesn't seem to be easily amenable to induction.


There doesn't seem to be a similar formula for integral $m>1$, just the usual increase in terms with $n$. Though it's not clear--Maple returns a single term for the general $n$th derivative with positive integral $m$, but it evaluates to zero (and is incorrect).


For $x^{m/2}$ argument, with $m$ a positive integer, there seems to be an even more interesting single term formula as well, relating:  

$$ \frac{d^n}{d x^n}\hspace{1pt}\normalsize \_p F\_q (\ldots; \ldots; x^{m/2})$$
to a single term involving:  

$$ \_{2p+m-1}\hspace{-1pt} F\hspace{0.5pt}\_{2q+m} (\ldots;\ldots;4^{p-q-1} x^m)$$
where the *F* parameters are halved and incremented by 1 and 1/2, doubling their number, in addition to $2m-1$ patterned fraction parameters. It would be very interesting to know how that is derived.


Edit: the $x^{m/2}$ derivative actually has two terms, not 1. This seems to arise from doing termwise differentiation and then dividing the series into even and odd parts so that the $(mk/2+...)$ coefficients become $(mk+...)$ and $(mk+1/2+...)$ in the new series, which can then be expanded using the Gauss multiplication theorem. Similarly any rational power $m/l$.



$n$th Derivative of $_p F_q(a_1,...,a_p; b_1,...,b_q;x^{-m})$, $p \le q$
The first displayed identity can be verified in a straightforward manner, by differentiating the power series for ${}\_{p} F\_{q} (a\_1,...,a\_p;b\_1,...,b\_q;1/x)$ (in the powers of $1/x$) term-wise $n$ times in $x$, and then comparing the coefficients of the resulting power series with the coefficients of the power series (in the powers of $1/x$) on the right-hand side of the identity. (Let me know if details on this are needed.)


Expression for the derivatives of any order of ${}\_{p} F\_{q} (a\_1,...,a\_p;b\_1,...,b\_q;x^c)$ can be obtained similarly, by term-wise differentiation.


