If $$y = \sec(\tan^{-1} x),$$​ then $$\frac{dy}{dx}$$​ at x = 1 is equal to:

Correct option is D

Given:
​$$y = \sec(\tan^{-1} x)$$​​
Formula used:
If $$\theta = \tan^{-1} x$$​, then
​$$\sec \theta = \sqrt{1 + \tan^{2}\theta} = \sqrt{1 + x^{2}}$$​​
Also,
​$$\frac{dy}{dx} = \frac{d}{dx}\left(\sqrt{1 + x^{2}}\right)$$​​
Solution:
​$$y = \sec(\tan^{-1} x)= \sqrt{1 + x^{2}}$$​​
Differentiate:
​$$\frac{dy}{dx}= \frac{1}{2}(1 + x^{2})^{-1/2} \cdot 2x= \frac{x}{\sqrt{1 + x^{2}}}$$​​
At x = 1:
​$$\frac{dy}{dx} = \frac{1}{\sqrt{2}}$$​​
The correct answer is (d) $$\frac{1}{\sqrt{2}}.$$​​