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Question
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{xa}} \right)\] ?
Solution
\[\text{ Let, y } = \tan^{- 1} \left( \frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{xa}} \right)\]
\[ \Rightarrow y = \tan^{- 1} \sqrt{x} + \tan^{- 1} \sqrt{a} \left[ \text{ Since }, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \frac{x + y}{1 - xy} \right]\]
Differentiate it with respect to x using chain rule,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \tan^{- 1} \sqrt{x} \right) + \frac{d}{dx}\left( \tan^{- 1} \sqrt{a} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{1 + \left( \sqrt{x} \right)^2}\frac{d}{dx}\left( \sqrt{x} \right) + 0\]
\[ \Rightarrow \frac{d y}{d x} = \left( \frac{1}{1 + x} \right)\left( \frac{1}{2\sqrt{x}} \right)\]
\[ \therefore \frac{d y}{d x} = \frac{1}{2\sqrt{x}\left( 1 + x \right)}\]
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