Use the principle that \[\frac{d}{d2^2} 2\cdot5 = \frac{d}{d2}\frac{d2}{d2^2} 2\cdot5 = \frac{5}{4}\]
\[\begin{array}{|c|c c c c } \hline & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline 2 & 1 & 0 & 4 & 0 & 3 & 0 & 12 & 0 & 5\\ 3 & 0 & 1 & 0 & 0 & 2 & 0 & 0 & 6 & 0\\ 4 & \frac{1}{4} & 0 & 1 & 0 & 0 & 0 & 8 & 0 & \frac{5}{4}\\ 5 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 2\\ 6 & \frac{1}{3} & ? & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 7 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 8 & \frac{1}{12} & 0 & ? & 0 & 0 & 0 & 1 & 0 & 0\\ 9 & 0 & ? & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 10& ? & 0 & 0 & ? & 0 & 0 & 0 & 0 & 1\\ \hline \end{array}\]