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Question:

If $y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right), 0<x<1$, prove that $\frac{d y}{d x}=\frac{4}{1+x^{2}}$

Solution:

$y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sec ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$

Put $x=\tan \theta$

Using, $\sec ^{-1} x=\frac{1}{\cos ^{-1} x}$

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

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

Using $\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\sin 2 \theta$ and $\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta$

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

Considering the limits

$0<x<1$

$0<\tan \theta<1$

$0<\theta<\frac{\pi}{4}$

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

Now,

$y=2 \theta+2 \theta$

$y=4 \theta$

$y=4 \tan ^{-1} x$

Differentiating w.r.t $x$ we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(4 \tan ^{-1} \mathrm{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{1+\mathrm{x}^{2}}$