Question

The derivative of ${\tan }^{-1}\left(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\right)$ is

• A. $\sqrt{\left(1-{\mathrm{x}}^2\right)}$
• B. $\frac{1}{\sqrt{\left(1-{\mathrm{x}}^2\right)}}$
• C. $\frac{1}{2\sqrt{\left(1-{\mathrm{x}}^2\right)}}$
• D. $\mathrm{x}$

Answer 1

The correct option is C

$\frac{1}{2\sqrt{\left(1-{\mathrm{x}}^2\right)}}$

Explanation for the correct option:

Finding the derivative of the given function:

$\mathrm{Let}y={\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

$\mathrm{Let}\mathrm{x}=\mathrm{cos\theta }\Rightarrow \mathrm{\theta }={\cos }^{-1}\mathrm{x},\mathrm{then}:$

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

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

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

$\mathrm{Or},1+\cos \theta =2{\cos }^2\left(\frac{\theta }{2}\right)$ $\mathrm{Or},1-\cos \theta =2{\sin }^2\left(\frac{\theta }{2}\right)$

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

$={\tan }^{-1}\frac{\left(\frac{\sqrt{2}\cos \frac{\theta }{2}}{\sqrt{2}\cos \frac{\theta }{2}}-\frac{\sqrt{2}\sin \frac{\theta }{2}}{\sqrt{2}\cos \frac{\theta }{2}}\right)}{\left(\frac{\sqrt{2}\cos \frac{\theta }{2}}{\sqrt{2}\cos \frac{\theta }{2}}+\frac{\sqrt{2}\sin \frac{\theta }{2}}{\sqrt{2}\cos \frac{\theta }{2}}\right)}\mathbf{}\left[\mathrm{dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\sqrt{2}\cos \left(\frac{\theta }{2}\right)\right]$

$={\tan }^{-1}\left(\frac{1-\tan \theta /2}{1+\tan \theta /2}\right)={\tan }^{-1}\left(\frac{\tan \pi /4-\tan \theta /2}{1+\tan \frac{\pi }{4}\tan \frac{\theta }{2}}\right)\mathbf{}\left[\because \tan \frac{\pi }{4}=1\right]$

$={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\frac{\theta }{2}\right)\right)=\frac{\pi }{4}-\frac{\theta }{2}=\frac{\pi }{4}-\frac{{\cos }^{-1}x}{2}$

$\mathrm{So},\frac{dy}{dx}=\frac{d}{dx}\left[\frac{\pi }{4}-\frac{{\cos }^{-1}x}{2}\right]=\frac{d}{dx}\left(\frac{\pi }{4}\right)-\frac{1}{2}\frac{d}{dx}\left({\cos }^{-1}x\right)$

$=0-\frac{1}{2}\times \frac{-1}{\sqrt{1-x^2}}$

$=\frac{1}{2\sqrt{1-{\mathrm{x}}^2}}$

Therefore, the correct answer is option (C).