Question

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

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

Answer 1

Let $co{s}^{-1}x=\theta i.e.,x=cos\theta$

$\therefore y=se{c}^{-1}\frac{1}{2x^2-1}\Rightarrow =se{c}^{-1}\left(\frac{1}{2co{s}^2\theta -1}\right)\Rightarrow y=se{c}^{-1}\left(\frac{1}{cos2\theta }\right)\because cos2\theta =2co{s}^2\theta -1$

$\Rightarrow y=se{c}^{-1}\left(sec2\theta \right)\Rightarrow =2\theta \Rightarrow y=2co{s}^{-1}x)$

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

$\Rightarrow \frac{dy}{dx}=2\frac{d}{dx}\left(co{s}^{-1}x\right)=\frac{-2}{\sqrt{1-x^2}}(\because \frac{d}{dx}(co{s}^{-1}x)=\frac{-1}{\sqrt{1-x^2}})$