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

$\frac{1}{\sqrt{2}}$

We have,

$y =\sec \left(\tan ^{-1} x\right)$

$\Rightarrow y =\sec \left(\sec ^{-1} \sqrt{1+x^2}\right)$

$\Rightarrow y =\sqrt{1+x^2} \Rightarrow \frac{d y}{d x}=\frac{x}{\sqrt{1+x^2}}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{\sqrt{2}}$