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Question
Differentiate \[\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}\] ?
Solution
\[\text{Let y} = \sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}\]
\[ \Rightarrow y = \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^\frac{1}{2}\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{d}{dx} \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^\frac{1}{2} \]
\[ = \frac{1}{2} \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^{\frac{1}{2} - 1} \frac{d}{dx}\left( \tan^{- 1} \frac{x}{2} \right) \left[ \text{Using chain rule }\right]\]
\[ = \frac{1}{2} \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^\frac{- 1}{2} \times \frac{1}{1 + \left( \frac{x}{2} \right)^2} \times \frac{d}{dx}\left( \frac{x}{2} \right)\]
\[ = \frac{4}{4\left( 4 + x^2 \right)\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}}\]
\[ = \frac{1}{\left( 4 + x^2 \right)\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}}\]
\[So, \frac{d}{dx}\left\{ \sqrt{\tan^{- 1} \left( \frac{x}{2} \right)} \right\} = \frac{1}{\left( 4 + x^2 \right)\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}}\]
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