$\frac{{f}''.g-{f}'{g}'}{{{g}^{2}}}=(\frac{{{f}'}}{g}{)}'$  $f(x)=\frac{x+1}{x+2}\Rightarrow {f}'(x)=\frac{1}{{{(x+2)}^{2}}}\,\,\,,\,\,\,g(x)=\frac{\sqrt{x}}{{{(x+2)}^{2}}}\Rightarrow \frac{{f}'(x)}{g(x)}=\frac{1}{\sqrt{x}}\xrightarrow{Moshtagh}(\frac{{f}'(x)}{g(x)}{)}'=(\frac{1}{\sqrt{x}}{)}'=\frac{-\frac{1}{2\sqrt{x}}}{x}=\frac{-1}{16}$