Question

Find derivative of $y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right),where0<x<\frac{1}{\sqrt{2}}$.

Answer 1

$y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$

Let $x=\cos \theta$

then

$y={\sec }^{-1}\left(\frac{1}{2co{s}^2\theta -1}\right)$

$y={\sec }^{-1}\left(\frac{1}{\cos 2\theta }\right)$

$y={\sec }^{-1}\left(\sec 2\theta \right)$

$y=2\theta$

$y=2{\cos }^{-1}x$

Now,

$\frac{dy}{dx}=2\left(\frac{-1}{\sqrt{1-x^2}}\right)$

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