# Solve this

Question:

If $y=\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}$, find $\frac{d y}{d x}$

Solution:

$y=\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}$

Put $x=\cos 2 \theta$

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

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

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

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

Dividing by $\cos \theta$ both numerator and denominator,

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

$y=\tan ^{-1}\left\{\frac{1-\tan \theta}{1+\tan \theta}\right\}$

$y=\tan ^{-1}\left\{\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\right\}$

$y=\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\theta\right)\right\}$

$y=\frac{\pi}{4}-\theta$

$y=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x$

Differentiating w.r.t $x$, we get

$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right)$

$\frac{d y}{d x}=0-\frac{1}{2}\left(-\frac{1}{\sqrt{1-x^{2}}}\right)$

$\frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^{2}}}$