Official solution
Let \[y = \sec \left( {{{\tan }^{ - 1}}x} \right)\]
and \[{\tan ^{ - 1}}\,\,x = \theta .\]
\[ \Rightarrow x = \tan \theta \]
Thus, we have \[y = \sec \,\theta \]
\[ \Rightarrow y = \sqrt {1 + {x^2}} \]
\[\left( {\,\,} \right.\] As \[\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta \] \[\left. {\,\,} \right)\]
\[ \Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x\]
At \[x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.\]