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