Question

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

• A. $ \frac{1}{2} $
• B. $ \frac{1}{\sqrt{2}} $
• C. $ \sqrt{2} $
• D. $ 1 $

Answer 1

Answer: (B)

We have, $y = \sec (\tan ^{-1} x)$
$ = \sec[\sec^{-1} ( \sqrt{1 + x^2})]$
$[\because \tan^{-1} \theta = \sec^{-1} \sqrt{( 1 + \theta^2})]$
$\Rightarrow y = \sqrt{ 1 + x^2}$
On differentiating both sides w.r.t ‘$x$’, we get
$\frac{dy}{dx} = \frac{1}{2} (1 + x^2) ^{-1/2} \frac{d}{dx} )( 1 + x^2)$
$ = \frac{1}{2} \frac{1}{\sqrt{1 + x^2}} \cdot 2x$
$\Rightarrow \frac{dy}{dx} = \frac{x}{\sqrt{(1 + x^2)}} $
$\Rightarrow (\frac{dy}{dx}) _{x = 1} = \frac{1}{\sqrt{(1+x^2)}} $
$= \frac{1}{\sqrt{2}}$