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Question
If \[y = \tan^{- 1} \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right), \text{find } \frac{dy}{dx}\] ?
Solution
\[\text{Here, y }= \tan^{- 1} \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right)\]
\[\text{Put x }= \cos2\theta\]
\[ \therefore y = \tan^{- 1} \left( \frac{\sqrt{1 + \cos2\theta} - \sqrt{1 - \cos2\theta}}{\sqrt{1 + \cos2\theta} + \sqrt{1 - \cos2\theta}} \right)\]
\[ = \tan^{- 1} \left( \frac{\sqrt{2 \cos^2 \theta} - \sqrt{2 \sin^2 \theta}}{\sqrt{2 \cos^2 \theta} + \sqrt{2 \sin^2 \theta}} \right)\]
\[ = \tan^{- 1} \left( \frac{\sqrt{2}\left( \cos\theta - sin\theta \right)}{\sqrt{2}\left( \cos\theta + sin\theta \right)} \right)\]
\[ = \tan^{- 1} \left( \frac{\frac{\cos\theta - sin\theta}{\cos\theta}}{\frac{\cos\theta + \sin\theta}{\cos\theta}} \right) \left[ \text{Dividing numerator and denominator by } \cos\theta \right]\]
\[ = \tan^{- 1} \left( \frac{\frac{\cos\theta}{\cos\theta} - \frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta} + \frac{\sin\theta}{\cos\theta}} \right)\]
\[ = \tan^{- 1} \left( \frac{1 - \tan\theta}{1 + \tan\theta} \right)\]
\[ = \tan^{- 1} \left( \frac{\tan\frac{\pi}{4} - \tan\theta}{1 + \tan\frac{\pi}{4} \times \tan\theta} \right) \]
\[ = \tan^{- 1} \left[ \tan\left( \frac{\pi}{4} - \theta \right) \right] \]
\[ = \frac{\pi}{4} - \theta\]
\[ = \frac{\pi}{4} - \frac{1}{2} \cos^{- 1} x \left( \text{ Using x }= \cos2\theta \right)\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = 0 - \frac{1}{2}\left( \frac{- 1}{\sqrt{1 - x^2}} \right)\]
\[ \therefore \frac{d y}{d x} = \frac{1}{2\sqrt{1 - x^2}}\]
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