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 $x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$