# If $y=tan^{-1}\left \{\large\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right \}$,find $\large\frac{dy}{dx}.$

Toolbox:
• $1+\cos 2\theta=2\cos^2\theta$
• $1-\cos 2\theta=2\sin^2\theta$
• $\tan\large\frac{\pi}{4}$$=1 Step 1: y=\tan^{-1}\left \{\large\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right \} Put x^2=\cos 2\theta \therefore y=tan^{-1}\bigg(\large\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\bigg) 1+\cos 2\theta=2\cos^2\theta 1-\cos 2\theta=2\sin^2\theta \therefore y=\tan^{-1}\bigg(\large\frac{\sqrt 2(\cos\theta-\sin\theta)}{\sqrt 2(\cos\theta+\sin\theta)}\bigg) Step 2: Dividing by \cos\theta we get, y=\tan^{-1}\bigg(\large\frac{1-\tan \theta}{1+\tan\theta}\bigg) \tan\large\frac{\pi}{4}$$=1$
$\;\;=\tan^{-1}\bigg(\large\frac{\tan \pi/4-\tan\theta}{\tan\pi/4+\tan\theta}\bigg)$
This is of the form $\tan(A-B)$
$\;\;\;=\tan^{-1}\big(\tan(\large\frac{\pi}{4}$$-\theta)\big) Step 3: y=\large\frac{\pi}{4}$$-\theta$
But $\theta=\cos^{-1}(x^2)$
$\Rightarrow y=\large\frac{\pi}{4}$$-\cos^{-1}(x^2) \therefore \large\frac{dy}{dx}$$=0-\big(-\large\frac{1}{2}.\frac{2x}{\sqrt{1-x^4}}\big)$
$\Rightarrow \large\frac{x}{\sqrt{1-x^4}}$