If y = sec tan - 1 x , then d y d x at x = 1 is equal to

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

Differentiating both sides w.r.t. $x$, we get, 

$\frac{dy}{dx}=\sec \left({\tan }^{-1}x\right)\tan \left({\tan }^{-1}x\right)·\frac{1}{1+x^2}$

At  $x=1$, we have, ${\tan }^{-1}x=\frac{\pi }{4}$

$\frac{dy}{dx}=\frac{\sec \left({\displaystyle \frac{\pi }{4}}\right)\times \tan \left({\displaystyle \frac{\pi }{4}}\right)}{1+1^2}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$