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
$\frac{dy}{dx}=\frac{1}{2}(1+x^2{)}^{-1/2}\frac{d}{dx})(1+x^2)$
$=\frac{1}{2}\frac{1}{\sqrt{1+x^2}}\cdot 2x$
$\Rightarrow \frac{dy}{dx}=\frac{x}{\sqrt{(1+x^2)}}$
$\Rightarrow (\frac{dy}{dx}{)}_{x=1}=\frac{1}{\sqrt{(1+x^2)}}$
$=\frac{1}{\sqrt{2}}$