Question

If $y={\cos }^{-1}\left(2x\right)+2{\cos }^{-1}\sqrt{1-4x^2},-\frac{1}{2}<x<0,\mathrm{find}\frac{dy}{dx}.$

Answer 1

$$\begin{gathered} \mathrm{Here},y={\cos }^{-1}\left(2x\right)+2{\cos }^{-1}\sqrt{1-4x^2} \\ \mathrm{Put}2x=\cos \theta \\ \therefore y={\cos }^{-1}\left(\cos \theta \right)+2{\cos }^{-1}\sqrt{1-{\cos }^2\theta } \\ \Rightarrow y={\cos }^{-1}\left(\cos \theta \right)+2{\cos }^{-1}\left(\sin \theta \right) \\ \Rightarrow y={\cos }^{-1}\left(\cos \theta \right)+2{\cos }^{-1}\left[\cos \left(\frac{\mathrm{\pi }}{2}-\theta \right)\right]...\left(\mathrm{i}\right) \\ \mathrm{Now},-\frac{1}{2}<x<0 \\ \Rightarrow -1<2x<0 \\ \Rightarrow -1<\cos \theta <0 \\ \Rightarrow \frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi } \\ \mathrm{And} \\ \Rightarrow -\frac{\mathrm{\pi }}{2}>-\mathrm{\theta }>-\mathrm{\pi } \\ \Rightarrow \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}\right)>\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)>\left(\frac{\mathrm{\pi }}{2}-\mathrm{\pi }\right) \\ \Rightarrow 0>\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)>-\frac{\mathrm{\pi }}{2} \\ \Rightarrow -\frac{\mathrm{\pi }}{2}<\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)<0 \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ \mathrm{y}=\mathrm{\theta }+2\left[-\left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)\right]\left[\begin{array}{c}\mathrm{Since},{\cos }^{-1}\cos \left(\mathrm{\theta }\right)=\mathrm{\theta },\mathrm{if}\mathrm{\theta }\in \left[0,\mathrm{\pi }\right]\\ {\cos }^{-1}\cos \left(\mathrm{\theta }\right)=-\mathrm{\theta },\mathrm{if}\mathrm{\theta }\in \left[-\mathrm{\pi },0\right]\end{array}\right] \\ \mathrm{y}=\mathrm{\theta }-2\times \frac{\mathrm{\pi }}{2}+2\mathrm{\theta } \\ \mathrm{y}=-\mathrm{\pi }+3\mathrm{\theta } \\ \mathrm{y}=-\mathrm{\pi }+3{\cos }^{-1}\left(2\mathrm{x}\right)\left[\mathrm{Since},2\mathrm{x}=\mathrm{cos\theta }\right] \end{gathered}$$

Differentiate it with respect to x using chain rule,

$$\begin{gathered} \frac{dy}{dx}=0+3\left(\frac{-1}{\sqrt{1-{\left(2x\right)}^2}}\right)\frac{d}{dx}\left(2x\right) \\ \Rightarrow \frac{dy}{dx}=\frac{-3}{\sqrt{1-4x^2}}\times 2 \\ \therefore \frac{dy}{dx}=-\frac{6}{\sqrt{1-4x^2}} \end{gathered}$$