Question

Differentiate \[\tan^{- 1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\}, - 1 < x < 1\] ?

Answer 1

\[\text{ Let, y } = \tan^{- 1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\}\]

\[\text{Put x } = \sin\theta\]

\[ y = \tan^{- 1} \left\{ \frac{\sin\theta}{1 + \sqrt{1 - \sin^2 \theta}} \right\}\]

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

\[ y = \tan^{- 1} \left\{ \frac{2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} \right\}\]

\[ y = \tan^{- 1} \left\{ \tan \frac{\theta}{2} \right\} . . . \left( i \right)\]

\[\text{ Here }, - 1 < x < 1\]

\[ \Rightarrow - 1 < \sin\theta < 1\]

\[ \Rightarrow - \frac{\pi}{2} < \theta < \frac{\pi}{2}\]

\[ \Rightarrow - \frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}\]

\[\text{ So, from equation } \left( i \right), \]

\[ y = \frac{\theta}{2} \left[ Since, \tan^{- 1} \left( \tan\theta \right) = \theta, \text{ if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]

\[ y = \frac{1}{2} \sin^{- 1} x \left[ \text{ Since, x } = \sin\theta \right]\]

\[\text{ Differentiating it with respect to x }, \]

\[\frac{d y}{d x} = \frac{1}{2\sqrt{1 - x^2}}\]