Solve this

Question:

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

Solution:

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

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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