$\frac{\mathrm{d}}{\mathrm{dx}}\left\{{\cos }^2\left[{\mathrm{Tan}}^{-1}\left(\sin {\mathrm{Cot}}^{-1}\mathrm{x}\right)\right]\right\}=$

$$\begin{gathered} \mathrm{y}={\cos }^2\left[{\tan }^{-1}\left(\sin \left({\cot }^{-1}\mathrm{x}\right)\right)\right] \\ \mathrm{Put} {\cot }^{-1}\mathrm{x}=\mathrm{\theta } \\ \mathrm{x}=\cot \mathrm{\theta } \\ \sin \mathrm{\theta }=\frac{1}{\sqrt{1+{\mathrm{x}}^2}} \\ \mathrm{y}={\cos }^2\left[{\tan }^{-1}\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)\right] \\ ={\cos }^2 \mathrm{\beta }, \mathrm{where} \mathrm{\beta }={\tan }^{-1}\frac{1}{\sqrt{1+{\mathrm{x}}^2}} \\ \begin{array}{c}\mathrm{y}=\frac{{\mathrm{x}}^2+1}{{\mathrm{x}}^2+2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{({\mathrm{x}}^2+2)2\mathrm{x}-({\mathrm{x}}^2+1)2\mathrm{x}}{{({\mathrm{x}}^2+2)}^2}\end{array} \overline{) \begin{array}{c}\tan \mathrm{\beta }=\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\\ \cos \mathrm{\beta }=\sqrt{\frac{1+{\mathrm{x}}^2}{2+{\mathrm{x}}^2}}\end{array}} \\ =\frac{2\mathrm{x}}{{({\mathrm{x}}^2+2)}^2} \end{gathered}$$