Solve this
Question:

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

Solution:

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

Put $x=\tan \theta$

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

Using, $\sec ^{2} \theta=1+\tan ^{2} \theta$

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

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

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

Considering the limits

$0<x<\infty$

$0<\tan \theta<\infty$

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

Now,

$y=\theta+\theta$

$y=2 \theta$

$y=2 \tan ^{-1} x$

Differentiating w.r.t $x$ we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2 \tan ^{-1} \mathrm{x}\right)$

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