Question

If $y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right),$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{\sqrt{1-{{x}^{2}}}}$
• B. $\frac{2}{\sqrt{1-{{x}^{2}}}}$
• C. $\frac{1}{\sqrt{1+{{x}^{2}}}}$
• D. $\frac{2}{\sqrt{1+{{x}^{2}}}}$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{1-{{x}^{2}}}}$

Answer 2

Given, $y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{x}^{2}}}} \right)$
Put, $x=\sin \theta$
$\therefore$ $y={{\sec }^{-1}}\left( \frac{1}{\sqrt{1-{{\sin }^{2}}\theta }} \right)$
$\Rightarrow$ $y={{\sec }^{-1}}(\sec \theta )=\theta$
$\Rightarrow$ $y={{\sin }^{-1}}x$
$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{\sqrt{1-{{x}^{2}}}}$