# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

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

Solution:

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

Let $x=a \sin \theta$

Now

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

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

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

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

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

$-a$-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} \frac{x}{a}$Differentiating w.r.t$\mathrm{x}$, we get$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1}{2} \sin ^{-1} \frac{x}{a}\right)\frac{d y}{d x}=\frac{a}{2 \sqrt{a^{2}-x^{2}}} \times \frac{1}{a}\frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}}\$