If $y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1-4 x^{2}},
$y=\cos ^{-1}(2 x)+2 \cos ^{-1} \sqrt{1-4 x^{2}}$
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
$-\frac{1}{2} $-1<2 x<0$ $-1<\cos \theta<0$ $\frac{\pi}{2}<\theta<\pi$ $-\frac{\pi}{2}>-\theta>-\pi$ $0>\frac{\pi}{2}-\theta>-\frac{\pi}{2}$ Now, $y=\cos ^{-1}(\cos \theta)+2 \cos ^{-1}\left(\cos \left(\frac{\pi}{2}-\theta\right)\right)$ $y=\theta+2\left\{-\left(\frac{\pi}{2}-\theta\right)\right\}$ $y=-\pi+3 \theta$ $y=-\pi+\cos ^{-1}(2 x)$ Differentiating w.r.t $x$ we get $\frac{d y}{d x}=\frac{d}{d x}\left(-\pi+3 \cos ^{-1}(2 x)\right)$ $\frac{\mathrm{dy}}{\mathrm{dx}}=0+3\left[\frac{-2}{\sqrt{1-(2 \mathrm{x})^{2}}}\right]$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-6}{\sqrt{1-4 \mathrm{x}^{2}}}$
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