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If $\mathrm{y}=\cos ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to
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The correct answer is:
$-\frac{2}{1+x^{2}}$ for all $|x| < 1$
$\mathrm{y}=\cos ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)$
Put $\mathrm{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \mathrm{x}$
$\mathrm{y}=\cos ^{-1}\left(\frac{2 \tan \theta}{\left(1+\tan ^{2} \theta\right)}\right)=\cos ^{-1}(\sin 2 \theta)$
$=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)=\frac{\pi}{2}-2 \theta .$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \theta\right)$
$=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \tan ^{-1} \mathrm{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{1+\mathrm{x}^{2}}$, when $|\mathrm{x}| < 1$
Put $\mathrm{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \mathrm{x}$
$\mathrm{y}=\cos ^{-1}\left(\frac{2 \tan \theta}{\left(1+\tan ^{2} \theta\right)}\right)=\cos ^{-1}(\sin 2 \theta)$
$=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)=\frac{\pi}{2}-2 \theta .$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \theta\right)$
$=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \tan ^{-1} \mathrm{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{1+\mathrm{x}^{2}}$, when $|\mathrm{x}| < 1$
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