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