# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\tan ^{-1}\left\{\frac{\mathrm{x}}{1+\sqrt{1-\mathrm{x}^{2}}}\right\},-1<\mathrm{x}<1$

Solution:

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

Let $x=\sin \theta$

Now

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

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

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

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

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

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

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

Considering the limits,

$-1$-1<\sin \theta<1-\frac{\pi}{2}<\theta<\frac{\pi}{2}-\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{4}$Now,$y=\tan ^{-1}\left\{\tan \frac{\theta}{2}\right\}y=\frac{\theta}{2}y=\frac{1}{2} \sin ^{-1} x$Differentiating w.r.t$\mathrm{x}$, we get$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2} \sin ^{-1} \mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \sqrt{1-\mathrm{x}^{2}}}\$