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Question
Differentiate \[\tan^{- 1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\}, - 1 < x < 1\] ?
Solution
\[\text{ Let, y } = \tan^{- 1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\}\]
\[\text{Put x } = \sin\theta\]
\[ y = \tan^{- 1} \left\{ \frac{\sin\theta}{1 + \sqrt{1 - \sin^2 \theta}} \right\}\]
\[ y = \tan^{- 1} \left( \frac{\sin\theta}{1 + \cos\theta} \right) \]
\[ y = \tan^{- 1} \left\{ \frac{2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} \right\}\]
\[ y = \tan^{- 1} \left\{ \tan \frac{\theta}{2} \right\} . . . \left( i \right)\]
\[\text{ Here }, - 1 < x < 1\]
\[ \Rightarrow - 1 < \sin\theta < 1\]
\[ \Rightarrow - \frac{\pi}{2} < \theta < \frac{\pi}{2}\]
\[ \Rightarrow - \frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}\]
\[\text{ So, from equation } \left( i \right), \]
\[ y = \frac{\theta}{2} \left[ Since, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]
\[ y = \frac{1}{2} \sin^{- 1} x \left[ \text{ Since, x } = \sin\theta \right]\]
\[\text{ Differentiating it with respect to x }, \]
\[\frac{d y}{d x} = \frac{1}{2\sqrt{1 - x^2}}\]
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