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Question
If \[y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x > 0\] ,prove that \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ?
Solution
\[\text{ Here, y} = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + se c^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \]
\[\text{Put x} = \tan\theta\]
\[ \therefore y = \tan^{- 1} \left( \frac{2\tan\theta}{1 - \tan^2 \theta} \right) + \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \]
\[ \Rightarrow y = \tan^{- 1} \left( \tan 2\theta \right) + \cos^{- 1} \left( \cos 2\theta \right)\]
\[ \Rightarrow y = 2\theta + 2\theta \]
\[ \Rightarrow y = 4\theta\]
\[ \Rightarrow y = 4 \tan^{- 1} x .........\left[ \text{using, x} = tan\theta \right]\]
Differentiate it with respect to x,
\[\therefore \frac{d y}{d x} = \frac{4}{1 + x^2}\]
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