Higher order derivatives of the binomial factor Let $p$,$l$ be positive integers and $\theta$ be a parameter. The question is to compute the following quantity:
\begin{equation}
\kappa^{(p)}_l := \left. \frac{\partial^p}{\partial \theta^p} \binom{\theta}{l} \right|_{\theta=0}
\end{equation}
With the help of Mathematica we found the result for consecutive values of p. We have:
\begin{equation}
\kappa^{(p)}_l = \left\{
\begin{array}{rr}
\delta_{l,0}           & \quad \mbox{if $p=0$} \\
\frac{(-1)^{l+1}}{l}   & \quad \mbox{if $p=1$} \\
2 (-1)^l \cdot \frac{H_{l-1}}{l} & \quad \mbox{if $p=2$} \\
3 (-1)^l \cdot \frac{H^{(2)}_{l-1} - H_{l-1}^2}{l} & \quad \mbox{if $p=3$} \\
4 (-1)^l \cdot \frac{2 H^{(3)}_{l-1} - 3 H^{(2)}_{l-1} H_{l-1}+H_{l-1}^3}{l} & \quad \mbox{if $p=4$} \\
5 (-1)^l \cdot \frac{6 H^{(4)}_{l-1} - 8 H^{(3)}_{l-1} H_{l-1}-3 H^{(2)}_{l-1} H^{(2)}_{l-1}+6 H^{(2)}_{l-1} H_{l-1}^2 - H_{l-1}^4}{l} & \quad \mbox{if $p=5$} \\
\end{array}
\right.
\end{equation}
if $l\ge 1$ and $\kappa^{(p)}_0=\delta_{p,0}$.
Here $H^{(p)}_l$ is the harmonic number of order $p$.
Now, the obvious question would be to find the result for generic values of $p$. 

To find the requested derivatives I would follow a different scheme, starting
by rewriting the binomial in the version that uses the falling factorial
$$
\left( \begin{gathered}
  \theta  \\ 
  l \\ 
\end{gathered}  \right) = \frac{{\theta ^{\,\underline {\,l\,} } }}
{{l!}}
$$
Now $\theta ^{\,\underline {\,l\,} } $ is a polynomial in $\theta$ of degree $l$,
which can be written in terms of powers of $\theta$ using the (unsigned) Stirling Numbers of 1st kind as
$$
\theta ^{\,\underline {\,l\,} }  = \theta \left( {\theta  - 1} \right)\; \cdots \;\left( {\theta  - l + 1} \right) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,l} \right)} {\left( { - 1} \right)^{\,l - k} \left[ \begin{gathered}
  l \\ 
  k \\ 
\end{gathered}  \right]\theta ^{\,k} } 
$$
Since the polynomial coincides with its MacLaurin series, it comes that:
$$
\left. {\frac{{\partial ^{\,p} }}
{{\partial \theta ^{\,p} }}\left( \begin{gathered}
  \theta  \\ 
  l \\ 
\end{gathered}  \right)} \right|_{\,\theta  = 0}  = \left. {\frac{{\partial ^{\,p} }}
{{\partial \theta ^{\,p} }}\frac{{\theta ^{\,\underline {\,l\,} } }}
{{l!}}} \right|_{\,\theta  = 0}  = \left( { - 1} \right)^{\,l - p} \frac{{p!}}
{{l!}}\left[ \begin{gathered}
  l \\ 
  p \\ 
\end{gathered}  \right]
$$
