In order to calculate the forces the derivation of the potential with respect to $x^i_n$ (the $n$th component of the position vector of atom $i$ $\equiv$ ${\bf r}_i$) has to be known.
This is gradually done in the following.
+The $n$th component of the force acting on atom $i$ is
+\begin{eqnarray}
+F_n^i & = & - \frac{\partial}{\partial x_n} \sum_{j \neq i} V_{ij} \nonumber\\
+ & = & \sum_{j \neq i} \Big( \partial_{x_n^i} f_C(r_{ij}) \big[ f_R(r_{ij}) + b_ij f_A(r_{ij}) \big] + \nonumber\\
+& & + f_C(r_{ij}) \big[ \partial_{x_n^i} f_R(r_{ij}) + b_{ij} \partial_{x_n^i} f_A(r_{ij}) + f_A(r_{ij}) \partial_{x_n^i} b_{ij} \big] \Big)
+\end{eqnarray}
The cutoff function $f_C$ derivated with repect to $x^i_n$ is
\begin{equation}
\partial_{x^i_n} f_C(r_{ij}) =
-\frac{1}{2} \sin \Big( \pi (r_{ij} - R_{ij}) / (S_{ij} - R_{ij}) \Big) \frac{\pi x^i_n}{(S_{ij} - R_{ij}) r_{ij}} \textrm { .}
+ \frac{1}{2} \sin \Big( \pi (r_{ij} - R_{ij}) / (S_{ij} - R_{ij}) \Big) \frac{\pi x^i_n}{(S_{ij} - R_{ij}) r_{ij}}
+\label{eq:d_cutoff}
\end{equation}
+for $R_{ij} < r_{ij} < S_{ij}$ and otherwise zero.
+The derivations of the repulsive and attractive part are:
+\begin{eqnarray}
+\partial_{x_n^i} f_R(r_{ij}) & = & - \lambda_{ij} A_{ij} \exp (-\lambda_{ij} r_{ij})\\
+\partial_{x_n^i} f_A(r_{ij}) & = & \mu_{ij} B_{ij} \exp (-\mu_{ij} r_{ij}) \textrm{ .}
+\end{eqnarray}
The angle $\theta_{ijk}$ can be expressed by the atom distances with the law of cosines:
\begin{eqnarray}
\theta_{ijk} & = & \arccos \Big( (r_{ij}^2 + r_{ik}^2 - r_{jk}^2)/(2 r_{ij} r_{ik}) \Big) \\
\partial_{x^i_n} \theta_{ijk} & = &
-\frac{-1}{\sqrt{1 - ((r_{ik}^2+r_{ij}^2-r_{jk}^2)/2r_{ik}r_{ij})^2}}
-\Big( \frac{4 r_{ik}r_{ij} (2 x^i_n - x^k_n - x^j_n) + 2(x^j_n - x^i_n)\frac{r_{ik}}{r_{ij}} + 2(x^k_n - x^i_n)\frac{r_{ij}}{r_{ik}} }{4 r^2_{ik} r^2_{ij}}\Big)
+\frac{-1}{\sqrt{1 - ((r_{ik}^2+r_{ij}^2-r_{jk}^2)/2r_{ik}r_{ij})^2}} \times \nonumber\\
+ & & \times \Big( \frac{4 r_{ik}r_{ij} (2 x^i_n - x^k_n - x^j_n) + 2(x^j_n - x^i_n)\frac{r_{ik}}{r_{ij}} + 2(x^k_n - x^i_n)\frac{r_{ij}}{r_{ik}} }{4 r^2_{ik} r^2_{ij}}\Big) \label{eq:d_theta}
+\end{eqnarray}
+Using the expressions \eqref{eq:d_cutoff} and \eqref{eq:d_theta} the derivation of $b_{ij}$ with respect to $x^i_n$ can be written as:
+\begin{eqnarray}
+\partial_{x^i_n} b_{ij} & = &
+- \frac{1}{2n_i} \chi_{ij} \Bigg( 1 + \beta_i^{n_i} \Bigg[ \sum_{k \ne i,j} \bigg( f_C(r_{ij}) \omega_{ik} \Big( 1 + \frac{c_i^2}{d_i^2} - \frac{c_i^2}{d_i^2 + (h_i - \cos \theta_{ijk})^2} \Big) \bigg)^{n_i} \Bigg] \Bigg)^{-\frac{1}{2n_i} - 1} \times \nonumber\\
+&& \times n_i \beta_i^{n_i} \Bigg[ \sum_{k \ne i,j} f_C(r_{ik}) \omega_{ik} \Big( 1 \frac{c_i^2}{d_i^2} - \frac{c_i^2}{d_i^2 + (h_i - \cos \theta_{ijk})^2} \Big) \Bigg]^{n_i -1} \times \nonumber\\
+&& \times \sum_{k \ne i,j} \Bigg( \omega_{ik} \Big( 1 + \frac{c_i^2}{d_i^2} - \frac{c_i^2}{d_i^2 + (h_i - \cos \theta_{ijk})^2} \Big) \partial_{x^i_n} f_C(r_{ik}) + \nonumber\\
+&& + f_C(r_{ik}) \omega_{ik} (-1) \frac{c_i^2}{(d_i^2 + (h_i - \cos \theta_{ijk})^2)^2} \times \nonumber\\
+&& \times 2 \Big( h_i - \cos \theta_{ijk} \Big) \sin \theta_{ijk} \partial_{x^i_n} \theta_{ijk} \Bigg)
\end{eqnarray}
-
\subsubsection{The Brenner potential}