Differentiate the function tan^-1 ≤ft[ (√(1 + x^2) + √(1 - x^2))/(√(1…

The correct answer is Option (2) → $\frac{-x}{\sqrt{1 - x^4}}$ ##

Let $y = \tan^{-1} \left[ \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right] \quad \dots(i)$

On putting $x^2 = \cos 2\theta$ in Eq. (i), we get

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

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

$[∵\cos 2\theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta]$

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

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

On dividing by $\cos \theta$ in both numerator and denominator, we get

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

$= \tan^{-1} \tan \left( \frac{\pi}{4} + \theta \right) \quad \left[ ∵ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}, \tan \frac{\pi}{4} = 1 \right]$

$= \frac{\pi}{4} + \theta = \frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2 \quad \left[ ∵2\theta = \cos^{-1} x^2 \Rightarrow \theta = \frac{1}{2} \cos^{-1} x^2 \right]$

On differentiating w.r.t. $x$, we get

$∴\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} \right) + \frac{d}{dx} \left( \frac{1}{2} \cos^{-1} x^2 \right) = 0 + \frac{1}{2} \cdot \frac{-1}{\sqrt{1 - x^4}} \cdot \frac{d}{dx} x^2 = \frac{1}{2} \cdot \frac{-2x}{\sqrt{1 - x^4}} = \frac{-x}{\sqrt{1 - x^4}}$