If \(y=\sec \left(\tan ^{-1} x\right)\), then \(\frac{d y}{d x}\) at \(x=1\) is

\(
\begin{array}{l}
y=\sec \left(\tan ^{-1} x\right) \\
\therefore \frac{d y}{d x}=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \cdot \frac{1}{1+x^{2}} \\
\text { At } x=1, \frac{d y}{d x}=\sec \frac{\pi}{4} \tan \frac{\pi}{4}\left(\frac{1}{2}\right) \\
=\frac{1}{2} \times \sqrt{2} \times 1=\frac{1}{\sqrt{2}}
\end{array}
\)
This problem can also be solved as follows :
\(
\begin{array}{l}
y=\sec \left(\tan ^{-1} x\right)=\sec \left(\sec ^{-1} \sqrt{1+x^{2}}\right)=\sqrt{1+x^{2}} \\
\therefore \frac{d y}{d x}=\frac{1}{2 \sqrt{1+x^{2}}} \times 2 x=\frac{x}{\sqrt{1+x^{2}}} \\
\therefore\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{\sqrt{2}}
\end{array}
\)