Question

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

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

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

$y=\sec \left({\tan }^{-1}x\right)$
$\frac{dy}{dx}=\sec \left({\tan }^{-1}x\right)\cdot \tan \left({\tan }^{-1}x\right)\cdot \frac{1}{1+x^2}$
$\Rightarrow \frac{dy}{dx}=\frac{x\sec \left({\tan }^{-1}x\right)}{1+x^2}$
$\Rightarrow \frac{dy}{dx}\text{at}(x=1)=\frac{1\cdot \sec \left({\tan }^{-1}1\right)}{1+1^2}$
$=\frac{\sec \left(\pi /4\right)}{2}$
$=\frac{1}{\sqrt{2}}$