Question

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

Answer 1

Given: $y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right),0<x<\frac{1}{\sqrt{2}}$

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

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

$\Rightarrow \cos y=2x^2-1$

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

Putting $x=\cos \theta$

$\Rightarrow y={\cos }^{-1}(2{\cos }^2\theta -1)$

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

$\Rightarrow y=2\theta$

Putting value of $\theta ={\cos }^{-1}x$, we get,

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

Differentiating both sides w.r.t. $x$, we get,

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

$\Rightarrow \frac{dy}{dx}=2\times \frac{d\left({\cos }^{-1}x\right)}{dx}$

$\Rightarrow \frac{dy}{dx}=2\left(\frac{-1}{\sqrt{1-x^2}}\right)[\because ({\cos }^{-1}x{)}^{\prime }=\frac{-1}{\sqrt{1-x^2}}]$

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