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)

$\frac{1}{\sqrt{2}}$

Answer 2

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