If y = ( ^ - 1 x) , then (dy)/(dx) at x = 1 is equal to

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

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

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

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

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