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Q.
The derivative of $\tan^{-1} \left(\frac{\sqrt{1+x^{2}} - 1}{x}\right)$ w.r.t. $ \tan^{-1} \left(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}}\right) $ at x = 0 is
Limits and Derivatives
Solution:
Let $y =\tan^{-1} \left(\frac{\sqrt{1+x^{2}} - 1}{x}\right)$ Put $ x =\tan\theta $
$\therefore y =\tan^{-1} \left(\frac{\sqrt{1+\tan^{2} \theta} -1}{\tan\theta}\right)$
$ = \tan^{-1} \left(\frac{\sec\theta -1}{\frac{\sin\theta}{\cos\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) $
$= \tan^{-1} \left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2} \tan^{-1} x$
$ \therefore \frac{dy}{dx} = \frac{1}{2\left(1+x\right)^{2}}$
Let $ z =\tan^{-1} \left(\frac{2x \sqrt{1-x^{2}}}{1-2x^{2}}\right) $ Put $x =\sin\phi $
$ \therefore z =\tan^{-1} \left(\frac{2 \sin\phi \cos\phi}{1- 2 \sin^{2} \phi}\right) $
$=\tan^{-1} \left(\frac{\sin 2 \phi}{\cos2 \phi}\right) = 2\phi$
$ \therefore z =2 \sin^{-1} x \therefore \frac{dz}{dx} = \frac{2}{\sqrt{1-x^{2}}} $
$\therefore \frac{dy}{dz} = \frac{ \frac{1}{2\left(1+x^{2}\right)}}{\frac{2}{\sqrt{1-x^{2}}}} = \frac{\sqrt{1-x^{2}}}{4\left(1+x^{2}\right)}$
At $ x =0 , \frac{dy}{dz} = \frac{\sqrt{1-0}}{4\left(1+0\right)} =\frac{1}{4}$