Question

Differentiate ${\tan }^{-1}\left\{\frac{\sqrt{1+a^2{x}^2}-1}{ax}\right\}$ w.r.t $x.$

Answer 1

Given $m={\tan }^{-1}\left(\frac{\sqrt{1+a^2{x}^2}-1}{ax}\right)$

Let $x=\frac{\tan \theta }{a}$ so putting this in above equation we get,

$m={\tan }^{-1}\left(\frac{\sqrt{1+a^2\frac{{\tan }^2\theta }{a^2}}-1}{a\times \frac{\tan \theta }{a}}\right)$

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

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

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

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

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

Now, putting the value of $\theta ={\tan }^{-1}ax$

$\Rightarrow m=\frac{{\tan }^{-1}ax}{2}$

So,$\frac{dm}{dx}=\frac{1}{2}\frac{1}{1+a^2{x}^2}\frac{d}{dx}\left(ax\right)=\frac{a}{2(1+a^2{x}^2)}$ .......$\left(1\right)$