# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\tan ^{-1}\left\{\frac{\sqrt{1+a^{2} x^{2}}-1}{a x}\right\}, x \neq 0$

Solution:

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

Let $a x=\tan \theta$

Now

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

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

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

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

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

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

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

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

$y=\frac{\theta}{2}$

$y=\frac{1}{2} \tan ^{-1} \mathrm{ax}$

Differentiating w.r.t $\mathrm{x}$, we get

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2} \times \frac{\mathrm{a}}{1+(\mathrm{ax})^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{2\left(1+\mathrm{a}^{2} \mathrm{x}^{2}\right)}$