Question

Find $\frac{dy}{dx}$ of $y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right),0<x<\frac{1}{\sqrt{2}}$

Answer 1

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

Put $x=\cos \theta \Rightarrow \theta ={\cos }^{-1}x$,

Then we have,

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

$\therefore \frac{dy}{dx}=2\cdot \frac{-1}{\sqrt{1-x^2}}=-\frac{2}{\sqrt{1-x^2}}$