If x is so small. That ${\mathrm{x}}^5$ and higher power of x may be neglected, then the coefficient of ${\mathrm{x}}^4$ in the expansion of $\sqrt{{\mathrm{x}}^2+4}-\sqrt{{\mathrm{x}}^2+9}$ is

$\begin{array}{rcl} & & {\left({\mathrm{x}}^2+4\right)}^{\frac{1}{2}}-{\left({\mathrm{x}}^2+9\right)}^{\frac{1}{2}}\\ & & 2{\left(1+\frac{{\mathrm{x}}^2}{4}\right)}^{\frac{1}{2}}+3{\left(1+\frac{{\mathrm{x}}^2}{9}\right)}^{\frac{1}{2}}\\ & & 2\left[1+\frac{1}{2} \frac{{\mathrm{x}}^2}{4}+\frac{\left({\displaystyle \frac{1}{2}}\right) \left({\displaystyle \frac{1}{2}}\right)-1}{2!} {\left(\frac{{\mathrm{x}}^2}{4}\right)}^2+.....\right]\\ & & -3\left[1+\frac{1}{2} \left(\frac{{\mathrm{x}}^2}{\mathrm{a}}\right)+\frac{\left({\displaystyle \frac{1}{2}} \left({\displaystyle \frac{1}{2}}-1\right)\right)}{2!} {\left(\frac{{\mathrm{x}}^2}{9}\right)}^2+....\right]\\ \mathrm{Coeff} {\mathrm{x}}^4& =& \left(2\right) \left(\frac{-1}{8}\right) \left(\frac{1}{16}\right)-3\left(\frac{-1}{8}\right) \left(\frac{1}{81}\right)\\ & =& \frac{-1}{64}+\frac{1}{216}\\ & =& \frac{-216+64}{\left(64\right) \left(216\right)}=\frac{-152}{64\times 216}\\ & =& \frac{-19}{1728}\end{array}$