Solve this

$y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1-4 x^{2}}$

Put $2 x=\cos \theta$

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

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

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

Considering the limits

$0

$0<2 x<1$

$0<\cos \theta<1$

$0<\theta<\frac{\pi}{2}$

$0>-\theta>-\frac{\pi}{2}$

$\frac{\pi}{2}>\frac{\pi}{2}-\theta>0$

Now,

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

$y=\theta+2\left(\frac{\pi}{2}-\theta\right)$

$y=\pi-\theta$

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

Differentiating w.r.t $\mathrm{x}$ we get

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

$\frac{d y}{d x}=0-\left[\frac{-2}{\sqrt{1-(2 x)^{2}}}\right]$

$\frac{d y}{d x}=\frac{2}{\sqrt{1-4 x^{2}}}$