Question

Find ${\displaystyle \frac{dy}{dx}}$ in the following:

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

Answer 1

The given relationship is ${\displaystyle y={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)}$

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

⇒ sec y = ${\displaystyle \frac{1}{2x^2-1}}$

⇒ cos y = ${\displaystyle 2x^2-1}$

⇒ ${\displaystyle 2x^2=1+\cos y}$

⇒ ${\displaystyle 2x^2=2{\cos }^2 \frac{y}{2}}$

⇒ ${\displaystyle x=\cos \frac{y}{2}}$

Differentiating this relationship with respect to x  , we obtain

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

⇒ ${\displaystyle 1=-\sin \frac{y}{2}.\frac{d}{dx}\left(\frac{y}{2}\right)}$

⇒ ${\displaystyle \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2}\frac{dy}{dx}}$

⇒ ${\displaystyle \frac{dy}{dx}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-{\cos }^2 \frac{y}{2}}}}$

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

Answer 2

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

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

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

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

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

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

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