If ${\mathrm{y}}_{\mathrm{k}}$ is the $k^{th}$ derivative of y with respect to x, $\mathrm{y}=\cos (\sin \mathrm{x})\text{ then }{\mathrm{y}}_1\sin \mathrm{x}+{\mathrm{y}}_2\cos \mathrm{x}=$

$\begin{array}{l}\mathrm{y}=\cos (\sin \mathrm{x})\Rightarrow {\mathrm{y}}_1=-\sin (\sin \mathrm{x})\cos \mathrm{x}\Rightarrow {\mathrm{y}}_2=\sin (\sin \mathrm{x})\sin \mathrm{x}-\cos (\sin \mathrm{x}){\cos }^2\mathrm{x}\\ {\mathrm{y}}_1\sin \mathrm{x}+{\mathrm{y}}_2\cos \mathrm{x}=-\sin (\sin \mathrm{x})\cos \mathrm{xsin}\mathrm{x}+\sin (\sin \mathrm{x})\sin \mathrm{xcos}\mathrm{x}-\cos (\sin \mathrm{x}){\cos }^3\mathrm{x}=-{\mathrm{ycos}}^3\end{array}$