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Question
Differentiate \[\tan^{- 1} \left( \frac{5 x}{1 - 6 x^2} \right), - \frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}\] ?
Solution
\[\text{ Let, y } = \tan^{- 1} \left( \frac{5x}{1 - 6 x^2} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left[ \frac{3x + 2x}{1 - \left( 3x \right)\left( 2x \right)} \right]\]
\[ \Rightarrow y = \tan^{- 1} \left( 3x \right) + \tan^{- 1} \left( 2x \right) \left[ \text{Since }, \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right) \right]\]
Differentiate it with respect to x using chain rule,
\[\frac{d y}{d x} = \frac{1}{1 + \left( 3x \right)^2}\frac{d}{dx}\left( 3x \right) + \frac{1}{1 + \left( 2x \right)^2}\frac{d}{dx}\left( 2x \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{1 + 9 x^2} \times 3 + \frac{1}{1 + 4 x^2} \times 2\]
\[ \therefore \frac{d y}{d x} = \frac{3}{1 + 9 x^2} + \frac{2}{1 + 4 x^2}\]
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